id string | solution string | answer string | metadata dict | problem string | candidates list | relevance_scores list | relevance_scores_full null |
|---|---|---|---|---|---|---|---|
aops_3125762 | We use the principle of inclusion and exclusion and stars and bars!\\
first set $|S|:$ total number of ways of distributing $12$ identical apples to $4$ children\\
$|\mathcal{A}_{i}|:$ total number of ways of distributing 12 identical apples such that student $i$ gets $\geqslant 4$ apples for each $i$ from $1$ to $4$... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "In how many ways can 12 identical apples be distributed among four children if each gets at least 1 apple and not more than 4 apples?\n\n[hide]31[/hide]",
"content_html": "In how many ways can 12 ide... | In how many ways can 12 identical apples be distributed among four children if each gets at least 1 apple and at most 4 apples? | [
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numina_10148687 | {1, 2, 3, 5, 8} has five elements and all pairs with a different sum. If there is a subset with 6 elements, then it has 15 pairs, each with sum at least 1 + 2 = 3 and at most 8 + 9 = 17. There are only 15 numbers at least 3 and at most 17, so each of them must be realised. But the only pair with sum 3 is 1,2 and the on... | 5 | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} | 34th CanMO 2002 Problem 1 What is the largest possible number of elements in a subset of {1, 2, 3, ... , 9} such that the sum of every pair (of distinct elements) in the subset is different? | [
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aops_511420 | That's if you care about the order of the prime factors. If you don't, this is your standard multichoose/stars-and-bars/balls-into-boxes kind of problem and the answer is a binomial coefficient. If either of these is what was meant, this should have been placed somewhere like HS Intermediate Topics. (Actually the me... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Let $\\mathbb{P}={p_1,p_2,...,p_n}$, for $n \\in \\mathbb{N}$, be set of primes. Find the number of combinations of elements of $\\mathbb{P}$ out of which we can build a $k$-divisor number (by multiplying ... | Let \(\mathbb{P}=\{p_1,p_2,\dots,p_n\}\) be a set of primes, with \(n\in\mathbb{N}\). Find the number of combinations (multisets) of elements of \(\mathbb{P}\) whose product is a \(k\)-divisor number, i.e. the number of multisets of size \(k\) chosen from \(\mathbb{P}\) (elements may repeat), for \(k\in\mathbb{N}\).
E... | [
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numina_10227498 | ### Part (a)
1. **Determine the row sum:**
- The sum of the numbers from 1 to 8 is:
\[
1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = 36
\]
- Let the sum of each row be \( S \). Since there are 3 rows, the total sum of the numbers in the rows is \( 3S \).
- Including the number \( M \), the total sum of the n... | M = 6, 9 | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "aops_forum"
} | Given a board of $3 \times 3$ you want to write the numbers $1, 2, 3, 4, 5, 6, 7, 8$ and a number in their boxes positive integer $M$, not necessarily different from the above. The goal is that the sum of the three numbers in each row be the same
$a)$ Find all the values of $M$ for which this is possible.
$b)$ For whic... | [
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aops_301931 | I was a bit confused by their explanation, wouldn't it make more sense to say that
8+5=13
13-12=1 slice of pizza | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "A 12-slice pizza was made with only pepperoni and mushroom toppings, and every slice has at least one topping. Only five slices have pepperoni, and exactly eight slices have mushrooms. How many slices have... | A 12-slice pizza was made with only pepperoni and mushroom toppings, and every slice has at least one topping. Only five slices have pepperoni, and exactly eight slices have mushrooms. How many slices have both pepperoni and mushrooms? | [
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numina_10150413 | We prove the general case. Let the first sequence be a 1 , a 2 , ... , a m and the second sequence be b 1 , b 2 , ... , b n , were 0 t n (if they are equal, then we are done). Let f(i) be the smallest k such that s k >= t i . If it is equal, we are done, so assume s k > t i . Now consider the n numbers s f(i) - t i . ... | proof | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} | 22nd ASU 1988 Problem 4 Given a sequence of 19 positive integers not exceeding 88 and another sequence of 88 positive integers not exceeding 19. Show that we can find two subsequences of consecutive terms, one from each sequence, with the same sum. | [
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aops_17991 | [hide]Well, there are 10 numbers that have a ones digit of 5, and 10 numbers with atens digit of 5. SInce one of the numbers with a 5 is in both categories (55), you need to add 10 and 10, and subtract 1, getting 19.
19/100 * 100 = 19%[/hide] | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "What percent of the natural numbers from 1 to 100, inclusive, have at least one digit which is a 5?",
"content_html": "What percent of the natural numbers from 1 to 100, inclusive, have at least one ... | What percent of the natural numbers from \(1\) to \(100\), inclusive, have at least one digit which is a \(5\)? | [
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ours_26557 | Part (a): Each inhabitant can receive either an autograph from each player or no autograph. Thus, there are a total of \(2^{11} = 2048\) different possibilities for receiving autographs.
Part (b): We need to distribute these 2048 possibilities among the 1111 inhabitants, such that each inhabitant receives a different ... | null | {
"competition": "swiss_mo",
"dataset": "Ours",
"posts": null,
"source": "firstRoundSolution2018.md"
} | The SMO country has 1111 inhabitants. The eleven players of the Liechtenstein national team distribute autographs to all inhabitants, with no inhabitant receiving an autograph twice (i.e., each inhabitant receives either no autograph or one autograph from each player).
(a) How many possibilities are there for which au... | [
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numina_10148979 | : 2002C9 = 1,398,276,498,745,133,921,413,500. Let mCn represent the binomial coefficient m!/(n! (m-n)! ). We prove first a simple lemma: kCk + (k+1)Ck + (k+2)Ck + ... + nCk = (n+1)C(k+1). This is an almost trivial induction. It is obviously true for n = k. Suppose it is true for n. Then the sum up to (n+1)Ck = (n+1)C(k... | 1,398,276,498,745,133,921,413,500 | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} | 10th Balkan 1993 Problem 2 How many non-negative integers with not more than 1993 decimal digits have non-decreasing digits? [For example, 55677 is acceptable, 54 is not.] Solution | [
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ours_19730 | This is a direct result of repeatedly applying the Hockey Stick Identity.
\(\boxed{}\) | null | {
"competition": "nt_misc",
"dataset": "Ours",
"posts": null,
"source": "IntroCombinatorics.md"
} | Prove that:
$$
\begin{gathered}
1=\binom{n}{0} \\
\sum_{i=1}^{n} 1=\binom{n}{1} \\
\sum_{i=1}^{n-1} \sum_{j=i+1}^{n} 1=\binom{n}{2} \\
\sum_{i=1}^{n-2} \sum_{j=i+1}^{n-1} \sum_{k=j+1}^{n} 1=\binom{n}{3}
\end{gathered}
$$
and so on. | [
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aops_135586 | The first one is incorrect, but the second one is correct
[hide="1"] You were right until the part where it said that the last 4 numbers must be a palindrome. So there are 25 choices for the first letter, 24 choices for the second letter, 9 choices for the first number, 9 choices for the second number, but the third ... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Can any one please check my following answers are right or not?\r\n\r\nHow many 6 character license plates start with two letters and end with 4 numbers if the letters must be different and neither is an O... | Can any one please check my following answers are right or not?
1. How many 6-character license plates start with two letters and end with four numbers if the letters must be different and neither is an O and the numbers form a palindrome and the digit 0 is not used?
2. A committee of four students needs to be chosen... | [
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numina_10136431 | 3. $A$
For the general case, consider arranging the numbers $1,2, \cdots, n$ in a circle in a clockwise direction. Let the last remaining number according to the problem's operation be $a_{n}$. When $n$ is an even number $2 m$, $a_{2 m}=2 a_{m}-1$. When $n$ is an odd number $2 m+1$, then $a_{2 m+1}=2 a_{m}+1$. Therefo... | 1955 | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} | 3. Arrange the numbers $1,2,3, \cdots, 2001$ in a clockwise direction on a circular diagram. First, cross out 2, then continue in a clockwise direction, crossing out every second number until only one number remains. The last remaining number is ( ).
A. 1955
B. 2001
C. 81
I. 163 | [
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aops_559264 | That's a very unhelpful answer Dr. Graubner.
[hide="As a hint for how to solve this"] Look at the middle coefficient of $(1-x)^{2n}(1+x)^{2n}$ to find a closed form expression for A, and then look at the middle coefficient of $x(1-x)^{2n} \cdot \frac{d}{dx}(1+x)^{2n}$ to find a comparable closed form expression for the... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "If $\\sum_{k=0}^{2n}(-1)^k\\cdot \\binom{2n}{k}^2 = A$, Then value of $\\sum_{k=0}^{2n}(-1)^k\\cdot (k-2n)\\cdot \\binom{2n}{k}^2$ in terms of $n$ and $A$, is",
"content_html": "If <span style=\"whit... | If \(\displaystyle \sum_{k=0}^{2n}(-1)^k\binom{2n}{k}^2 = A\), find the value of
\[
\sum_{k=0}^{2n}(-1)^k (k-2n)\binom{2n}{k}^2
\]
in terms of \(n\) and \(A\). | [
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ours_31149 | There are \(\binom{3}{2} = 3\) pairs of people, so there are 3 handshakes in total.
\(\boxed{3}\) | 3 | {
"competition": "bmt",
"dataset": "Ours",
"posts": null,
"source": "IFa2019S.md"
} | In a group of 3 people, every pair of people shakes hands once. How many handshakes occur? | [
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aops_1360325 | [Hide=Solution]
Pretty sure you can just do this using some quick and fast casework.
We need to keep this organized in order to know what we are doing so lets define this 4 digit number as $ABCD$ where the letters stand for digits.
Case 1: $A=1 B=1 C=1$
4 Arrangements
Case 2:$A=1 B=1 C=2$
3 Arrangements
And so and so ... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": " How many 4-digit numbers are there which obey all three of the following rules: (i) no digit other than 1, 2, 3, or 4 is to be used; (ii) a\n digit may occur more than once; (iii) as you read the number f... | How many 4-digit numbers are there which obey all three of the following rules?
(i) No digit other than 1, 2, 3, or 4 is used.
(ii) A digit may occur more than once.
(iii) As you read the number from left to right, digits never decrease (for example, 1134 is allowed, but 3314 is not). | [
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aops_546838 | Danica has a mansion, with floors 1 to 9. Danica wants to paint the wall of each floor with either pink, yellow, or sky blue. In how many ways can this be done if :
(a) At least one floor is painted pink, at least one floor is painted yellow, and at least one floor is painted sky blue.
(a) Exactly four floors ar... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Danica has a mansion, with floors 1 to 9. Danica wants to paint the wall of each floor with either pink, yellow, or sky blue. In how many ways can this be done if : \n\n(a) At least one floor is painte... | Danica has a mansion with floors \(1\) to \(9\). Danica paints the wall of each floor either pink, yellow, or sky blue. In how many ways can this be done if:
(a) At least one floor is painted pink, at least one floor is painted yellow, and at least one floor is painted sky blue?
(b) Exactly four floors are painted ye... | [
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aops_1859371 | [hide="Answer"]$N=75582$. So sum of digits of $N$ is $\boxed{27}$ [/hide] | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "There are $N$ numbers of $11$ digit positive integers such that the digits from left to right are non-decreasing. (For example, $12345678999, 55555555555,23345557889$). Find the sum of the digits of $N$.",... | There are \(N\) numbers of 11-digit positive integers such that the digits from left to right are non-decreasing (for example, \(12345678999\), \(55555555555\), \(23345557889\)). Find the sum of the digits of \(N\). | [
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aops_2143452 | If $a+b=k,$ then $0 \le c+d+e \le 4-k,$ so there are $k+1$ choices for $a,b$ and $\binom{6-k}{2}$ choices for $c,d,e.$ The cardinality is $\binom{6}{2} + 2 \cdot \binom{5}{2} + 3 \cdot \binom{4}{2} = 15+20+18=53.$ | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Let $N = \\{0, 1, 2, 3, . . .\\}$. Find the cardinality of the set \n$$\\{(a, b, c, d, e) \\in N^5 \\colon 0 \\leq a+b \\leq 2, 0 \\leq a+b+c+d+e \\leq 4 \\}.$$",
"content_html": "Let <span style=\"w... | Let \(N=\{0,1,2,3,\dots\}\). Find the cardinality of the set
\[
\{(a,b,c,d,e)\in N^5 \mid 0\le a+b\le 2,\ 0\le a+b+c+d+e\le 4\}.
\] | [
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numina_10115729 | Solution: Let $I$ be the set of all paths from $O$ to $P$;
$A_{1}$ be the set of all paths from $O$ to $P$ passing through $A B$;
$A_{2}$ be the set of all paths from $O$ to $P$ passing through $C D$;
$A_{3}$ be the set of all paths from $O$ to $P$ passing through $E F$;
$A_{4}$ be the set of all paths from $O$ to $P$ ... | \mathrm{C}_{15}^{5}-\mathrm{C}_{4}^{2}\mathrm{C}_{10}^{3}-\mathrm{C}_{6}^{2}\mathrm{C}_{8} | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} | Example 2.3.6 Find the number of shortest paths from point $O(0,0)$ to point $P(10,5)$ in a coordinate grid that do not pass through any of the segments $AB, CD, EF, GH$, where the coordinates of $A, B, C, D, E, F, G, H$ are $A(2,2), B(3,2)$,
$$
\begin{array}{l}
C(4,2), D(5,2), E(6,2), F(6,3), \\
G(7,2), H(7,3) .
\end{... | [
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0.401... | null |
numina_10092095 | In the numbers between 1000 and 9999, all are ordered arrangements of 4 digits. Therefore, to find the number of certain arrangements, we need to make 4 choices: the units, tens, hundreds, and thousands digits. Since the number we want is odd, the units digit can be any one of 1, 3, 5, 7, 9. The tens and hundreds digit... | 2240 | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} | Example 2: How many odd numbers with all different digits are there between 1000 and 9999? | [
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0.3903... | null |
aops_620173 | [hide]
Let $S_k=\sum_{r=1}^k\frac{C^k_r}r\ \ (k\ge1,S_0=0)$
$S_k-S_{k-1}=\sum_{r=1}^{k-1}\frac{C^k_r-C^{k-1}_r}r+\frac1k
=\sum_{r=1}^{k-1}\frac{C^{k-1}_{r-1}}r+\frac1k
=\sum_{r=1}^{k-1}\frac{C^k_r}k+\frac1k
=\frac{2^k-1}k$
$\sum_{k=1}^n(S_k-S_{k-1})=S_n=\sum_{k=1}^n\frac{2^k-1}k.$
[/hide] | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Prove that $\\sum_{r=1}^{n} \\frac{1}{r} \\binom{n}{r} = \\sum_{r=1}^{n} \\frac{1}{r} (2^{r}-1)$.",
"content_html": "Prove that <span style=\"white-space:nowrap;\"><img src=\"//latex.artofproblemsolv... | Prove that
\[
\sum_{r=1}^{n} \frac{1}{r}\binom{n}{r}
=
\sum_{r=1}^{n} \frac{1}{r}\bigl(2^{r}-1\bigr).
\] | [
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... | null | null |
numina_10084982 | The two members of the equality count in two different ways the number of ways to choose $A$ included in $B$ included in $C$ a set $E$ of cardinality $n$ such that $A$ is of cardinality $s$, $B$ of cardinality $r$, and $C$ of cardinality $k$. Indeed, it is equivalent to choose $A$ in $E$, $B \backslash A$ in $E \backsl... | proof | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} | Show that for all integers $n, k, r, s$ we have:
$$
\binom{n}{k}\binom{k}{r}\binom{r}{s}=\binom{n}{s}\binom{n-s}{r-s}\binom{n-r}{k-r}
$$ | [
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2... | null | null |
numina_10124750 | 【Analysis】A total of $12+26+23-5-2-4+1=51$ people visited at least one hall, so there is 1 person who did not visit any of the three halls. | 1 | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} | 4. A total of 52 students participated in the garden tour activities, among which 12 visited the Botanical Pavilion, 26 visited the Animal Pavilion, 23 visited the Science and Technology Pavilion, 5 visited both the Botanical Pavilion and the Animal Pavilion, 2 visited both the Botanical Pavilion and the Science and Te... | [
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aops_1297858 | We will first count the entire number of arrangements. By PIE, we know that this is $9!-8!-8!+7!=7!*57$.
Now, we set the third term to be $3$ and count the amount of cases: $8!-7!-7!+6!=6!*43$. Putting the second result over the first gives us $a$. | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "The numbers $1,2,\\ldots,9$ are arranged so that the $1$st term is not $1$ and the $9$th term is not $9$. What is the probability that the third term is $3$?\n\n$\\text{(A) }\\frac{17}{75}\\qquad\\text{(B)... | The numbers \(1,2,\ldots,9\) are arranged so that the first term is not \(1\) and the ninth term is not \(9\). What is the probability that the third term is \(3\)?
(A) \(\tfrac{17}{75}\) (B) \(\tfrac{43}{399}\) (C) \(\tfrac{127}{401}\) (D) \(\tfrac{16}{19}\) (E) \(\tfrac{6}{7}\) | [
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aops_90215 | From 1-9, there are 9 one-digit numbers. From 10-99, there are 99-10+1=90 two-digit numbers. From 100-150, there are 150-100+1=51 three-digit numbers. So there are $9+2(90)+3(51)=\boxed{342}$ pieces of type. | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "A printer has to numberthe pages of a book from 1 to 150. Suppose the printer uses a seperate piece of type for each digit in each number. How many pieces of type will the printer have to use?",
"co... | A printer has to number the pages of a book from \(1\) to \(150\). Suppose the printer uses a separate piece of type for each digit in each number. How many pieces of type will the printer have to use? | [
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numina_10160164 | 19. Ans: 49950
Number of three-digit integers formed $=5^{3}=125$. Observe that each of the five digits $1,2,3,5,7$ appears 25 times in the first, second and third digits of the integers formed. Thus,
$$
\begin{aligned}
\text { sum }= & 25 \times(1+2+3+5+7) \times 100 \\
& +25 \times(1+2+3+5+7) \times 10 \\
& +25 \time... | 49950 | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} | 19. Different positive 3-digit integers are formed from the five digits $1,2,3,5,7$, and repetitions of the digits are allowed. As an example, such positive 3-digit integers include 352, 577, 111, etc. Find the sum of all the distinct positive 3-digit integers formed in this way. | [
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... | null | null |
aops_517362 | Let's look at the change in the number of flat cups. Let's choose x flat cups at any step. Then we chose 4-x reverse cups. Then change in the number of flat cups = -x + (4-x) = 4-2x.
So the value of the number of flat cups in mod 2 does not change. At first, number of flat cups is odd. Then the number of flat cups can... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "There are $24$ cups on a table. In the beginning, only three of them placed upside-down. At each step, we are turning four cups. Can we turn all the cups right-side up in at most $100$ steps?",
"cont... | There are \(24\) cups on a table. Initially, exactly three of them are placed upside-down. At each step, four cups are flipped. Can all the cups be turned right-side up in at most \(100\) steps? | [
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... | null | null |
numina_10014710 | Answer: 18 blue beads.
Solution. Let's find the number of beads that have both a blue and a green bead next to them: $26+20-30=$ 16. The number of beads that have only blue beads next to them is $26-16=10$. The number of blue beads is $\frac{10 \cdot 2+16}{2}=18$. Here is an example of such an arrangement. Let's denot... | 18 | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} | 2. Thirty beads (blue and green) were laid out in a circle. For 26 beads, the neighboring one was blue, and for 20 beads, the neighboring one was green. How many blue beads were there? | [
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29... | null | null |
aops_1612848 | For part 1) the answer is $\frac{1}{24}$
We can easily study the favourable outcomes as:
$(1,1,6), (1,6,1),(6,1,1),(1,2,3),(1,3,2),(2,3,1),(2,1,3)(3,2,1),(3,1,2)$ | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "When three dice are rolled,\n\n(1) What is the probability of getting a product of 6 ? \n(2) Determine all values of $k$ such that the probability of the product being $k$ is $\\frac{1}{36}.$",
"cont... | When three fair six-sided dice are rolled:
(1) What is the probability that the product of the three outcomes is 6?
(2) Determine all values of \(k\) such that the probability that the product of the three outcomes is \(k\) equals \(\tfrac{1}{36}\). | [
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... | null | null |
aops_422711 | well here let the ones passed in mathematics be of set A and of science be of set B.
68 passed in at least one means that n(A U B) = 68
but n(A U B) = n(A) + n(B) - n(A intersection B)
so we get n(A intersection B) or students passed in both to be 93 - 68 = 25.
number of students passed in science will be n(B) - n( A i... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "out of 80 students who appeared in a combined test in science and mathematics, 68 passed in atleast one subject. If 43 passed in science, 50 passed in mathematics then find \n\n(i) Number of students passe... | Out of 80 students who appeared in a combined test in Science and Mathematics, 68 passed in at least one subject. If 43 passed in Science and 50 passed in Mathematics, find:
(i) Number of students who passed in both subjects.
(ii) Number of students who passed in Science only.
(iii) Number of students who failed in ... | [
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aops_258191 | [hide]
When we add the number of students with dogs and the number of students with cats, we get $ 46$. When we subtract it form the original number of students, we will find out how many have both, or $ \boxed{\textbf{(A)}\ 7}$[/hide] | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Each of the $ 39$ students in the eighth grade at Lincoln Middles School has one dog or one cat or both a dog and a cat. Twenty students have a dog and $ 26$ students have a cat. How many students have b... | Each of the $39$ students in the eighth grade at Lincoln Middle School has one dog or one cat or both a dog and a cat. Twenty students have a dog and $26$ students have a cat. How many students have both a dog and a cat?
\[
\textbf{(A)}\ 7\qquad \textbf{(B)}\ 13\qquad \textbf{(C)}\ 19\qquad \textbf{(D)}\ 39\qquad \tex... | [
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aops_1809588 | Put the $n$ balls out in a row. Now, put $k-1$ dividers, which split the $n$ balls into $k$ groups, which each go in one the boxes. Now, there are $n+(k-1)$ spots, and you have to chose $n$ of them to be balls, So it is indeed $(n+k-1)C(n)$. | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Prove that number of ways of distributing n identical balls to k different boxes where each boxes can be empty is (n+k-1)C(n)",
"content_html": "Prove that number of ways of distributing n identical... | Prove that the number of ways of distributing \(n\) identical balls into \(k\) distinct boxes (boxes may be empty) is \(\binom{n+k-1}{n}\). | [
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numina_10085976 | A5. D) 32 The number of blue socks in the drawer we call $b$. To be sure that he has two blue socks, Frank has to take 12 socks. In the worst case, he first picks the ten red socks. To be sure that he has two red socks, he has to take $b+2$ socks. In the worst case, he first picks the $b$ blue socks. It is given that t... | 32 | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} | A5. Frank has a drawer filled with loose socks. There are 10 red socks in it, and the rest of the socks are blue. He is going to blindly pick a number of socks from the drawer and wants to end up with two socks of a certain color. To be sure of having at least two red socks, he needs to pick twice as many socks as to b... | [
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aops_355543 | She has $4$ choices for a shirt, $3$ choices for a pair of pants of a skirt, and then $2$ choices for a pair of shoes, which makes:
$4\times3\times2=\boxed{24}\text{ outfits}$
EDIT: gah, beaten by professordad... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Mattie looks in her closet and finds 4 shirts, 2 pairs of pants, 1 skirt, and 2 pairs of shoes. If she must wear (a) one shirt, (b) one pair of pants or a skirt, and (c) one pair of (matching!) shoes, how ... | Mattie has 4 shirts, 2 pairs of pants, 1 skirt, and 2 pairs of shoes. She must wear (a) one shirt, (b) either one pair of pants or the skirt, and (c) one pair of matching shoes. How many possible complete outfits can she choose? | [
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aops_1571042 | [Hide=easier solution]
This equals the number of non-negative integer solutions to $a' + b' + c' + d' \le 8$.
A classic technique for this problem is to add a dummy variable $w = 8-a'-b'-c'-d'$, so it follows that the answer equals the number of non-negative solutions to $a'+b'+c'+d'+w = 8$, or $\binom{12}{4}$.[/hide] | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "The number of positive integer solution of $a+b+c+d\\leq 12$",
"content_html": "The number of positive integer solution of <img src=\"//latex.artofproblemsolving.com/8/8/d/88d6de149896ee86c93084da048... | Find the number of positive integer solutions to
\[
a+b+c+d \le 12,
\]
where \(a,b,c,d\) are positive integers. | [
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ours_13659 | The number of ways Kate can divide the four red socks into two pairs is \(\binom{4}{2} / 2 = 3\). Similarly, the number of ways she can divide the four blue socks into two pairs is also \(3\). Therefore, the number of ways she can form two pairs of red socks and two pairs of blue socks is \(3 \cdot 3 = 9\).
The total ... | 38 | {
"competition": "hmmt",
"dataset": "Ours",
"posts": null,
"source": "guts_feb_2004_1.md"
} | Kate has four red socks and four blue socks. If she randomly divides these eight socks into four pairs, what is the probability that none of the pairs will be mismatched? That is, what is the probability that each pair will consist either of two red socks or of two blue socks? If the answer is of the form of an irreduc... | [
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aops_24979 | I think i can help.
Let $S$ be our sample space (this is just the set of all of the different possible outcomes), then |$S$|=$6^3$.
Now, let $A$ be the set of all outcomes such that the sum of the 3 dice is greater than or equal to 10. Let $B$ be the set of all outcomes such that the sum of the 3 dice is less tha... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Can anybody help me on this problem:\r\nIf 3 dices are thrown, then what is the probability of getting the total numbers as equal to or more than 10? How can I solve this problem?",
"content_html": "... | If three fair dice are thrown, what is the probability that the sum of the faces is at least 10? | [
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aops_3309096 | [quote=BackToSchool][quote=miyukina]Answer
= 20 + 24 + 27 – 2 × 30
= 11[/quote]
Can you give a full solution? or explain your thought?[/quote]
Yes, let’s conveniently forget that this is only 30 people at first
Summing the roles / characteristics, there are 20 + 24 + 27 = 71 different ones, but now it is time... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "There are 30 students in Andrew's class. Out of all students, 20 are good at swimming, 24 are good at cycling, and 27 are good at running. At least how many students are good at all three sports?",
"... | There are 30 students in Andrew's class. Out of all students, 20 are good at swimming, 24 are good at cycling, and 27 are good at running. At least how many students are good at all three sports? | [
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numina_10130943 | 7. 3985 We consider the $k$-th 1 and the $2^{k-1}$ 2s that follow it as the $k$-th group, consisting of $2^{k-1}+1$ terms.
Let the 1998th term be in the $k$-th group, then $k$ is the smallest positive integer satisfying
$k+\left(1+2+2^{2}+\cdots+2^{k-1}\right)=2^{k}+k-1 \geqslant 1998$.
Since $2^{10}+10-1=1033<1998$ a... | 3985 | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} | 7. Given a sequence whose terms are 1 or 2, the first term is 1, and there are $2^{k-1}$ 2's between the $k$-th 1 and the $(k+1)$-th 1, i.e., $1,2,1,2,2,1,2,2,2,2,1,2,2,2,2,2,2,2,2,1, \cdots$, then the sum of the first 1998 terms of this sequence is $\qquad$ . | [
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aops_2627485 |
$x_{1} = \prod_{r=1}^{5} \cos\frac{r\pi}{11}$.
$x_{1} = \cos\frac{\pi}{11}\cos\frac{2\pi}{11}\cos\frac{3\pi}{11}\cos\frac{4\pi}{11}\cos\frac{5\pi}{11}$.
$2\sin\frac{\pi}{11} \cdot x_{1} = 2\sin\frac{\pi}{11}\cos\frac{\pi}{11}\cos\frac{2\pi}{11}\cos\frac{3\pi}{11}\cos\frac{4\pi}{11}\cos\frac{5\pi}{11}$.
$2\sin\frac{\pi... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Let $x_1 = \\prod_{r=1}^5 \\cos\\frac{r\\pi}{11}$ and $x_2 = \\sum_{r=1}^5 \\cos\\frac{r\\pi}{11}$ then show that $x_1\\cdot x_2$ $=$ $\\frac{1}{64} \\left(\\csc\\frac{\\pi}{22} - 1\\right)$",
"cont... | Let
\[
x_1=\prod_{r=1}^5\cos\frac{r\pi}{11},\qquad
x_2=\sum_{r=1}^5\cos\frac{r\pi}{11}.
\]
Show that
\[
x_1\cdot x_2=\frac{1}{64}\left(\csc\frac{\pi}{22}-1\right).
\] | [
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aops_1747922 | [quote=martinba314]That's one way of looking at it, although the machines would be distinguishable, while the people are indistinguishable.
$\binom{7+5-1}{5-1}$ is the correct expression by stars and bars.[/quote]
So that is for where both people and machines are distinguishable?
[quote=NikoIsLife]I think its actual... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "So the number of nonnegative integer solutions to the equation a+b+c+d+e=7 is basically the number of ways for 7 people to line up at 5 machines, right?\nIs it (5+7-1 5)*7! or is it (7+5-1 5-1)?",
"c... | Find the number of nonnegative integer solutions to the equation
\[
a+b+c+d+e = 7.
\]
(Equivalently, the number of ways to assign 7 people to 5 machines, where multiple people may be assigned to the same machine.) | [
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aops_2934494 | [hide=Solution]
We use complementary counting. The probability that Petunia picks a different desert than Prajwal is $\tfrac23$, and the probability that Pete picks a different desert than Prajwal and Petunia is $\tfrac13$. As well, the probability that Petunia picks a different subset of pizza toppings that Prajwal is... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "At Patty the panda's pizza parlor, the possible pizza toppings are pesto, pepperoni, peppers, and pineapple, and the possible desserts are pumpkin pie, powdered-sugar pretzels, and papaya popsicles. If Pra... | At Patty the panda's pizza parlor, the possible pizza toppings are pesto, pepperoni, peppers, and pineapple, and the possible desserts are pumpkin pie, powdered-sugar pretzels, and papaya popsicles. If Prajwal the porcupine, Petunia the peacock, and Pete the parakeet each randomly choose a subset of the pizza toppings ... | [
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numina_10266231 | To find the number of positive integers less than or equal to $2017$ that have at least one pair of adjacent digits that are both even, we will use casework based on the number of digits and the positions of the even digits.
### Case 1: One-digit numbers
There are no one-digit numbers that satisfy the condition since ... | 738 | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "aops_forum"
} | Find the number of positive integers less than or equal to $2017$ that have at least one pair of adjacent digits that are both even. For example, count the numbers $24$, $1862$, and $2012$, but not $4$, $58$, or $1276$. | [
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aops_264452 | [quote="isabella2296"]Wouldn't it be 5? You can even count the segments on the diagram. However, the answer on the review is 6...glitch or am I missing something?[/quote]
Assume they are all equidistent with length 1. There are 3 of length 1, 2 of length 2, and 1 of length 3.
OR
Call the points A,B,C,D. The d... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "How many segments are determined by 4 points on a line?\n\n[asy]size(10cm,10cm);\n\ndraw((0,0)--(20,0),linewidth(2bp), Arrows(20bp));\n\ndot((3,0),linewidth(6bp));\n\ndot((8,0),linewidth(6bp));\n\ndot((13,... | How many segments are determined by 4 points on a line?
\[
\begin{asy}
size(10cm,10cm);
draw((0,0)--(20,0),linewidth(2bp), Arrows(20bp));
dot((3,0),linewidth(6bp));
dot((8,0),linewidth(6bp));
dot((13,0),linewidth(6bp));
dot((17,0),linewidth(6bp));
\end{asy}
\] | [
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... | null | null |
numina_10012593 | Solution: Suppose it was possible to arrange the numbers 1 and 3 in the table such that their sum in each of the 5 rows and in each of the 8 columns is divisible by 7. The sum in any row is not less than $8 \cdot 1=8$ and not more than $8 \cdot 3=24$. The numbers in the range from 8 to 24 that are divisible by 7 are 14... | impossible | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} | 7.4 Can the cells of a $5 \times 8$ table be filled with the numbers 1 and 3 so that the sum of the numbers in each row and each column is divisible by 7? | [
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aops_390813 | Case 1:
2 is the last digit.
Then we can have 5 has the hundred's place.
We have 4 choices for the middle digit.
We have a total 4 possibilities.
Case 2:
8 is the last digit.
We have 2 choices for the first digit.
We have 4 choices for the middle digit.
We have a total of 8 possibilities.
4+8=12 A. | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "How many even integers are there between 200 and 700 whose digits are all different and come from the set {1, 2, 5, 7, 8, 9}?\n\n$ \\textbf{(A)}\\ 12 \\qquad\n\\textbf{(B)}\\ 20 \\qquad\n\\textbf{(C)}\\ ... | How many even integers are there between 200 and 700 whose digits are all different and come from the set \(\{1,2,5,7,8,9\}\)?
\(\textbf{(A)}\ 12 \qquad
\textbf{(B)}\ 20 \qquad
\(\textbf{(C)}\ 72 \qquad
\(\textbf{(D)}\ 120 \qquad
\(\textbf{(E)}\ 200\) | [
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... | null | null |
aops_1101423 | [quote=Banana_]How would you do this? I know the answer is [hide=answer]28090/6949[/hide] but what is an easy way to sum up the digits? And how would you know how many digits are there in total?[/quote]
[hide]Do casework for 1-digit numbers, 2-digit, 3, and 4.[/hide] | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Elli writes down all the digits of the positive integers less than 2015. Then she randomly selects a digit. What is the expected value of this digit? Express your answer as a common fraction.\n",
"co... | Elli writes down all the digits of the positive integers less than 2015. Then she randomly selects a digit. What is the expected value of this digit? Express your answer as a common fraction. | [
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... | null | null |
aops_442246 | [quote="osmosis92"]not sure if i made a mistake but:
The total is $\binom{12}{3}$.
We count the number of sets with no white is $\binom{8}{3}$, the number of sets with no red is $\binom{7}{3}$, and the number of sets with neither red nor white is $\binom{3}{3}$.
Using complementary, the answer is $1-\frac{\binom{7}{... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "A combination of 3 balls is picked at random from a box containing 5 red, 4 white, and 3 blue balls. Find the probability that the set chosen contains at least 1 white and 1 red ball.\n\n(My answer is 13/2... | A combination of 3 balls is picked at random from a box containing 5 red, 4 white, and 3 blue balls. Find the probability that the set chosen contains at least one white and one red ball. | [
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aops_3225648 | I got an answer of 165 (the sum of the first 9 triangular numbers) but i might be wrong. The number of 3-digit ordered numbers which begin with the digit $k$ is the $(10-k)$th triangular number where $k$ is a positive integer from 1 to 9. | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "A number is said to be ordered if each digit is greater than or equal to the digit on its\nleft.\nHow many three-digit numbers are ordered?\nIs the answer to this 291?",
"content_html": "A number is ... | A number is said to be ordered if each digit is greater than or equal to the digit on its left. How many three-digit numbers are ordered? Is the answer to this 291? | [
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aops_2645319 | i got f(n)=2f(n-2)+2f(n-1)
if the last number is a 0, then the second to last number must be a 1 or 2 and the other numbers are unaffected. if the last number is 1 or 2 the other numbers are unaffected and we have two cases | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Let $f(n) $ be the number of $ n$ length words without neighbouring zeroes from the alphabet ${0,1,2}$.Find a recursion for $f(n)$.",
"content_html": "Let <img src=\"//latex.artofproblemsolving.com/... | Let \(f(n)\) be the number of length-\(n\) words over the alphabet \(\{0,1,2\}\) that contain no two consecutive zeros. Find a recurrence relation for \(f(n)\). | [
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22701... | null | null |
aops_623035 | Hello,
we immediately see that the answer is $ \binom{19}{9}=92378$.
[hide="If you're not Sonnhard"] Notice that choosing all the possibilities for the variables is essentially the same as the number of ways to divide 20 objects into 10 groups, with each group having at least one object in it. If we treat the object... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "How many solutions are there to \n$a+b+c+d\\cdots+n=20$ (10 variables (don't let letters mislead you)) such that all variables are integers >0?",
"content_html": "How many solutions are there to<br>\... | How many solutions are there to
\[
a+b+c+d+\cdots+n=20
\]
(10 variables) such that all variables are integers \(>0\)? | [
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aops_1086937 | Make a hypothetical $f(0)=1, f(4)=5$. Then, we choose 4 non-negative numbers that sum to 4 to represent $f(1)-f(0), f(2)-f(1), f(3)-f(2), f(4)-f(3)$. The number of ways to do this is $\binom73=35$, and these define any $f(1),f(2),f(3)$ with that condition, so there are 35 maps. | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "The number of maps f from the set {1,2,3} into the set {1,2,3,4,5} such that f(i)is less than or equal to f(j) whenever i<j ?",
"content_html": "The number of maps f from the set {1,2,3} into the set... | Find the number of maps \(f\) from the set \(\{1,2,3\}\) into the set \(\{1,2,3,4,5\}\) such that \(f(i)\le f(j)\) whenever \(i<j\). | [
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4... | null | null |
aops_144167 | \substack is a useful command for this:
$\sum_{\substack{0 \leq n \leq r \\ 6 \leq r \leq 12 \\ 6-n \leq 12-r}}\binom{r}{6}\binom{12-r}{6-n}$
Usually, this means you sum over all choices of the variables which meet the conditions. In this case, it might be better to write it like this:
$f(r) = \sum_{\substack... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Yeah, there was an error in the Advanced Topics test at PUMaC, and the summation was somewhat harder, because Vandermonde trivialized the intended one. But I'm pretty sure of my answer on this one.\r\n\r\n... | Evaluate the sum
\[
\sum_{\substack{0\le n\le r \\ 6\le r\le 12 \\ 12-r\ge 6-n}} \binom{r}{6}\binom{12-r}{6-n}.
\] | [
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numina_10092779 | Prove that given $n a_{n}=a_{1}+a_{2}+\cdots+a_{n-2} \quad(n \geqslant 3)$
$$
(n-1) a_{n-1}=a_{1}+a_{2}+\cdots+a_{n-3} \quad(n \geqslant 4),
$$
(1) - (2) yields: $n a_{n}-(n-1) a_{n-1}=a_{n-2}$
$$
(n \geqslant 4) \text {, }
$$
Rearranging gives $n\left(a_{n}-a_{n-1}\right)=-\left(a_{n-1}-a_{n-2}\right)$
If there exist... | proof | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} | Example 3 The sequence $\left\{a_{n}\right\}$ satisfies the relation $a_{n}=\frac{1}{n}\left(a_{1}+a_{2}+\cdots+a_{n-2}\right), n>2$ and $a_{1}=1, a_{2}=\frac{1}{2}$, prove that: $a_{n}=\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\frac{1}{5!}+\cdots+\frac{(-1)^{n}}{n!} \quad(n \geqslant 2)$ | [
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numina_10104635 | 3. $\frac{15}{2}$.
Notice,
$$
\begin{array}{l}
1+x_{k}^{2}=1+\tan ^{2} \frac{k \pi}{17}=\frac{1}{\cos ^{2} \frac{k \pi}{17}} . \\
\text { Then } \sum_{k=1}^{16} \frac{1}{1+x_{k}^{2}}=\sum_{k=1}^{16} \cos ^{2} \frac{k \pi}{17} \\
=\sum_{k=1}^{16} \frac{1+\cos \frac{2 k \pi}{17}}{2} \\
=8+\frac{1}{2} \sum_{k=1}^{16} \co... | \frac{15}{2} | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} | 3. Let $x_{k}=\tan \frac{k \pi}{17}(k=1,2, \cdots, 16)$. Then $\sum_{k=1}^{16} \frac{1}{1+x_{k}^{2}}=$ $\qquad$ | [
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aops_11592 | hmm... i got something different (dunno if it's right though)
[hide]
if first digit is 4 or 6...
then there are 4 choices for the last digit
and 8*7 choices for the middle ones
2*4*8*7 = 448 beginning with 4 or 6
if first digit is 5...
then there are 5 choices for the last digit
and 8*7 choices for the midd... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "How many even integers between 4000 and 7000 have all digits different?",
"content_html": "How many even integers between 4000 and 7000 have all digits different?",
"post_id": 77336,
"pos... | How many even integers between \(4000\) and \(7000\) have all digits different? | [
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aops_281875 | R : advance to the right
D : advance to the down
The route supports that I arrange five R and fore D to one line.
Therefore,
$ \dfrac{(5 \plus{} 4)!}{5! \ 4!} \equal{} \binom{9}{5} \equal{} \boxed{126}$ | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "How many 9-step paths are there from $ E$ to $ G$?[asy]size(4cm,4cm);int w=6;int h=5;int i;for (i=0; i<h; ++i){draw((0,i) -- (w-1,i));}for (i=0; i<w; ++i){draw((i, 0)--(i,h-1));}label(\"$G$\", (w-1,0... | How many 9-step paths are there from \(E\) to \(G\)?
size(4cm,4cm);
int w=6;
int h=5;
int i;
for (i=0; i<h; ++i){ draw((0,i) -- (w-1,i)); }
for (i=0; i<w; ++i){ draw((i,0) -- (i,h-1)); }
label("$G$", (w-1,0), SE);
label("$E$", (0,h-1), NW); | [
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numina_10086606 | First method:
In the 1st possible case, the competitors live in 18 or more different settlements.
Then we select 1 student from each of the 18 different settlements, thus completing the required selection.
In the 2nd possible case, the competitors live in 17 different settlements.
In this case, it is not possible t... | proof | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} | 5. At a certain State Competition, 290 students participated. Prove that it is possible to select 18 students among them who live in the same settlement or live in 18 different settlements.
STATE COMPETITION IN MATHEMATICSPrimošten, April 4-6, 2016. | [
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aops_3014856 | First of all if $T=\sum_{i=0}^nif(i)$ where $f(i)=f(n-i)$ holds then we can write $T=\sum_{i=0}^n(n-i)f(i)$ and get that $2T=\sum_{i=0}^nnf(i)\implies T=\frac{n}{2}\sum_{i=0}^nf(i)$.
We also know that $\sum_{i=0}^n\binom{n}{i}^2=\binom{2n}{n}$ and $\sum_{i=0}^n\binom{n}{i}=2^n$.
\begin{align*}
S&=\sum_{i=0}^n\sum_{j=... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Find \n\\[\\sum_{0 \\le i < j \\le n} \\sum (i+j) \\left(\\binom{n}{i}^2 + \\binom{n}{j}^2 + \\binom ni \\binom nj\\right)\\]",
"content_html": "Find<br>\n<img src=\"//latex.artofproblemsolving.com/a... | Find
\[
\sum_{0\le i<j\le n}\sum (i+j)\left(\binom{n}{i}^2+\binom{n}{j}^2+\binom{n}{i}\binom{n}{j}\right).
\] | [
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aops_531066 | For most elementary algebraic proofs it really should suffice to have $n - x$ be a nonnegative integer. (Actually the stated result should be true more generally if we replace the finite sum with an infinite sum over $i \geq 1$; of course in that case we should ask that $n$ [i]not[/i] be a positive integer since we do... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Inspired from a related WOOT PoTD by [b]MellowMelon[/b]. \n\nProve that\n\n$\\sum_{i=1}^{n-x+1}{\\frac{x\\binom{n-x}{i-1}}{\\binom{n}{i}}}=\\frac{n+1}{x+1}$\n\nFor all $x,n$ such that $x \\leq n$ are both ... | Prove that for positive integers n and x with x ≤ n,
\[
\sum_{i=1}^{\,n-x+1}\frac{x\binom{n-x}{\,i-1}}{\binom{n}{i}}=\frac{n+1}{x+1}.
\] | [
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aops_1900271 | $2^{2n}=\binom{2n}{0}+\binom{2n}{1}+\ldots+\binom{2n}{n}+\ldots+\binom{2n}{2n-1}+\binom{2n}{2n}$
So, $2^{2n}>\binom{2n}{n}$ | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Let $n$ is positive integer. Prove that $(2n)!<2^{2n}(n!)^2.$",
"content_html": "Let <img src=\"//latex.artofproblemsolving.com/1/7/4/174fadd07fd54c9afe288e96558c92e0c1da733a.png\" class=\"latex\" al... | Let \(n\) be a positive integer. Prove that
\[
(2n)! < 2^{2n}\,(n!)^2.
\] | [
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aops_1536130 | Sorry, I misread my calculator (it's $462$). To solve this, imagine 6 vertical lines that form 7 different spaces. Each of those spaces represents one of the numbers from 1 to 7. In those 7 spaces, you have to fit 5 objects (let's call them *). So, if you draw this, you have something like *|||*|**|*|. In all, you... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "How many increasing functions (counting even those that are weakly growing)\nthere are from $A=\\{1,2,3,4,5\\} $ to $B=\\{1,2,3,4,5,6,7\\} $?",
"content_html": "How many increasing functions (counti... | How many increasing functions (including weakly increasing functions) are there from \(A=\{1,2,3,4,5\}\) to \(B=\{1,2,3,4,5,6,7\}\)? | [
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ours_33318 | From left to right, let \(A\) be the event that the first three students are seated in increasing age order, let \(B\) be the event that the middle three students are seated in increasing age order, and let \(C\) be the event that the last three students are seated in increasing age order.
Using the principle of inclu... | 19 | {
"competition": "smt",
"dataset": "Ours",
"posts": null,
"source": "combo-solutions.md"
} | Five students, all with distinct ages, are randomly seated in a row at the movies. The probability that, from left to right, no three consecutive students are seated in increasing age order is \(\frac{m}{n}\), where \(m\) and \(n\) are relatively prime positive integers. Find \(m+n\). | [
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aops_338387 | [quote="pi man"]How many arrangements of the letters in the word ARRANGE contain no double-letters?[/quote]
Please specify the source of your problems, especially during the MATHCOUNTS competition season.
[quote="joshisunny16"]no of ways = factorial of total letters/factorial of repeated letters
$ \equal{} \frac {7!}{... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "How many arrangements of the letters in the word ARRANGE contain no double-letters?",
"content_html": "How many arrangements of the letters in the word ARRANGE contain no double-letters?",
"pos... | How many arrangements of the letters in the word \(\text{ARRANGE}\) contain no double letters? | [
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aops_145574 | My reasoning is that the variables are the urns, and 10=1+1+1+1+1+1+1+1+1+1, each of the "1" is a ball.. so 10 balls and n urns.
And the balls-and-urn argument, if i remember correctly, is C ( balls + urns - 1, urns - 1), which is C (10 + n - 1, n - 1) = C (9 + n, n - 1).
Sorry for not knowing Latex.. | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "how many solutions $(x_{1},x_{2},...,x_{n})$ in nonnegative integers are there to the equation\r\n\r\n$x_{1}+x_{2}+....+x_{n}=10$?",
"content_html": "how many solutions <img src=\"//latex.artofproble... | How many solutions \((x_1,x_2,\dots,x_n)\) in nonnegative integers are there to the equation
\[
x_1+x_2+\cdots+x_n=10 ?
\] | [
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56931,
40079,
2203,
6919,
55439,
16072,
47927,
61791,
28017,
45788,
25223,
51646,
... | null | null |
numina_10057242 | The first digit can be chosen in 9 ways, and each of the six remaining digits can be chosen in 10 ways. In total, $9 \cdot 10^{6}$ ways.
## Answer
$9 \cdot 10^{6}$ numbers. | 9\cdot10^{6} | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} | 2 [ Decimal numeral system ]
How many different seven-digit telephone numbers exist (assuming that the number cannot start with zero)?
# | [
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24389,
63450,
28905,
1619,
46909,
2... | null | null |
aops_3287025 | a nice counting argument
[hide]Let $S = {1, 2, 3, \cdots n}$. We pick a subset of 2 elements, then we pick a pair of those 2-element subsets. Now, either they share one element, or they share 2 elements. We proceed via casework:
(1): There is one common element within the pair of subsets
If we have one common element, ... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Prove that for all positive integers $n$, we have that $3\\binom{n}{3} + 3\\binom{n}{4} = \\binom{\\binom{n}{2}}{2}$.\n\nEdit: just realized the problem statement can be re-written as $3\\binom{n+1}{4} = \... | Prove that for all positive integers n,
\[3\binom{n}{3} + 3\binom{n}{4} = \binom{\binom{n}{2}}{2}.\] | [
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49882,
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64693,
35013,
51642... | null | null |
numina_10174035 | It is known that
$$
\begin{gathered}
\binom{n}{k}=\binom{n+1}{k+1}-\binom{n}{k+1} \\
\binom{n+1}{k}=\binom{n+2}{k+1}-\binom{n+1}{k+1} \\
\binom{n+2}{k}=\binom{n+3}{k+1}-\binom{n+2}{k+1} \\
\ldots \\
\ldots \\
\ldots \\
\binom{n+m-1}{k}=\binom{n+m}{k+1}-\binom{n+m-1}{k+1} \\
\binom{n+m}{k}=\binom{n+m+1}{k+1}-\binom{n+m... | \frac{(+1)(+2)(+3)\ldots(n++1)}{n+1} | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} | Show that
$$
\binom{n}{k}+\binom{n+1}{k}+\binom{n+2}{k}+\ldots+\binom{n+m}{k}=\binom{n+m+1}{k+1}-\binom{n}{k+1}
$$
and based on this, sum the following series:
$$
1 \cdot 2 \cdot 3 \ldots n+2 \cdot 3 \cdot 4 \ldots(n+1)+\ldots+(m+1)(m+2)(m+3) \ldots(m+n)
$$ | [
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aops_1699873 | [hide=diff (but worse) approach]If the first number chosen is $1$ or $n$, then there is only one number out of $n-1$ that cannot be chosen, so there are $2(n-2)$ possibilities with $1$ and $n$
If the first number chosen is not $1$ or $n$, then there are two numbers out of $n-1$ that cannot be chosen, so there are $(n-... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "How many ways are there to choose 2 numbers from the first n natural numbers so that the two numbers are not adjacent?",
"content_html": "How many ways are there to choose 2 numbers from the first n ... | How many ways are there to choose 2 numbers from the first \(n\) natural numbers so that the two numbers are not adjacent? | [
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42680,
5780,
12391,
198,
37711,
... | null | null |
numina_10168550 | We will prove the two statements (F. 2712. and F. 2717.) together. Let
$$
S_{n, k}=\sum_{i=1}^{n}(-1)^{i}\binom{n}{i} i^{k}
$$
We will show that if $n \geq 2$, then
$$
S_{n, 0}=-1, \quad S_{n, k}=0, \quad \text { if } \quad 0 < k < n, \quad \text { and } \quad S_{n, n}=(-1)^{n} n!
$$
If $1 \leq i \leq n$, then $\bi... | proof | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} | We will prove that
$$
\sum_{i=1}^{100}(-1)^{i}\binom{100}{i} i^{100}=100!
$$ | [
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5... | null | null |
aops_1565125 | Use the sticks and stones/stars and bars method.
We need to divide 12 cookies into three groups, with two dividers (sticks). Because order doesn't matter, WLOG (without loss of generality) we let the left most group be the first kind of cookies, the middle group the second kind, and the right group the third kind. Then... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "I am stuck on the math count warm-up 13 question. I would like some hints to how to solve this problem below.\n\nAlexander Clifton visits Sweet Dreams bakery, which sells three kinds of cookies. How many u... | Alexander Clifton visits Sweet Dreams bakery, which sells three kinds of cookies. How many unique assortments of a dozen cookies can Alexander buy? | [
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21716... | null | null |
aops_1163523 | [hide=Hint] Consider doing casework. How might you organize the cases? [/hide]
[hide=Sol] All that matters is the numbers of balls in each box.
So we do casework by the number of balls in the box with the most balls.
(5,0,0): 3 ways here
(4,1,0): 3 x 2 = 6 ways here
(3,2,0): 6 ways here
(3,1,1): only 3 ways here
... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "If I have 5 indistinguishable balls how many ways can I place them in 3 distinguishable boxes?\n",
"content_html": "If I have 5 indistinguishable balls how many ways can I place them in 3 distinguish... | If I have 5 indistinguishable balls, how many ways can I place them into 3 distinguishable boxes? | [
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aops_73814 | [hide]Is it ${11\choose4}=330$? I did several numbers (a+b+c=5, a+b=6, etc.) and found a pattern that I knew before. I forgot how to explain it...
Maybe if you took the last digit b away to leave x+y+z+a, since what ever that equals will determine b (If x+y+z+a=9, b must be 3). There are 11 numbers to choose from sin... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "$x+y+z+a+b=12$ and $x,y,z,a,b$ are all positive, and note that $(6,2,1,1,2)$ is different from $(2,1,1,2,6)$, how many integral solutions are there to $(x,y,z,a,b)$?",
"content_html": "<img src=\"//l... | Let \(x,y,z,a,b\) be positive integers satisfying
\[
x+y+z+a+b=12.
\]
Note that the ordered 5-tuples \((6,2,1,1,2)\) and \((2,1,1,2,6)\) are considered different. How many integral solutions \((x,y,z,a,b)\) are there? | [
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... | null | null |
aops_1139852 | [hide="Solution"]Since $f(1), f(2), f(3)$ must be $1, 2, 4$ and sum to $4$, none of them can be $4$. Additionally, we can't have all three of them be $2$, and we can't even have two of them be $2$. Nor can we have none of them be $2$, since then $f(1) = f(2) = f(3) = 1$, and the sum is $3$.
Therefore, two of $f(1), ... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Find the number of functions $f\\colon \\mathbb(1,2,3)\\to\\mathbb(1,2,4)$ satisfying the condition $f(1)+f(2)+f(3)=4$\n",
"content_html": "Find the number of functions <img src=\"//latex.artofproble... | Find the number of functions \(f\colon \{1,2,3\}\to\{1,2,4\}\) satisfying the condition
\[
f(1)+f(2)+f(3)=4.
\] | [
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6654,... | null | null |
aops_480607 | Let $k_i$ be the number of sets that $i$ is contained in. Then $P_i \cap P_j = \{a\}$ for exactly $\binom{k_a}{2}$ pairs of subsets. Note that if $P_i \cap P_j$ is nonempty then it is a one element set, so summing over all $a$ we get that $P_i \cap P_j$ is nonempty for number of pairs equal to
\[ \binom{k_1}{2} + \bino... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Let $P_1, P_2, \\ldots, P_n$ be distinct $2$-element subsets of $\\{1, 2, \\ldots, n\\}$. Suppose that for every $1 \\le i < j \\le n$, if $P_i \\cap P_j \\neq \\emptyset$, then there is some $k$ such that... | Let \(P_1, P_2, \ldots, P_n\) be distinct 2-element subsets of \(\{1,2,\ldots,n\}\). Suppose that for every \(1\le i<j\le n\), if \(P_i\cap P_j\neq\varnothing\), then there is some \(k\) such that \(P_k=\{i,j\}\). Prove that if \(a\in P_i\) for some \(i\), then \(a\in P_j\) for exactly one value of \(j\ne i\). | [
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... | null | null |
ours_12260 | There are \(2^{10} = 1024\) subsets of \(\{1,2, \ldots, 10\}\) altogether. Any subset without the specified property must be either the empty set or a block of consecutive integers. To specify a block of consecutive integers, we either have just one element (10 choices) or a pair of distinct endpoints \(\left(\binom{10... | 968 | {
"competition": "hmmt",
"dataset": "Ours",
"posts": null,
"source": "adv_feb_2002.md"
} | Determine the number of subsets \( S \) of \(\{1,2,3, \ldots, 10\}\) with the following property: there exist integers \( a<b<c \) with \( a \in S, b \notin S, c \in S \). | [
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50... | null | null |
aops_374446 | There are $6$ ways to choose a European country, $4$ ways to choose an Asian country, $3$ ways to choose a North American country, and $7$ ways to choose an African country. So $6 \cdot 4 \cdot 3 \cdot 7 = \boxed{504}$ ways. | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "In how many ways can we form an international commission if we must choose one European country from among 6 European countries, one Asian country from among 4, one North American country from among 3, and... | In how many ways can we form an international commission if we must choose one European country from among 6 European countries, one Asian country from among 4, one North American country from among 3, and one African country from among 7? | [
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... | null | null |
numina_10102673 | None
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly.
Note: The provided instruction is a meta-instruction and not part of the text to be translated. Since the text to be translated is "None", the translation is also "None". ... | notfound | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} | 2.76 In selecting $k$ different and non-adjacent numbers from the natural numbers from 1 to $n$, let the number of such selection schemes be $f(n, k)$.
（1）Find the recurrence relation for $f(n, k)$.
（2）Use induction to find $f(n, k)$.
（3）If 1 and $n$ are considered adjacent numbers, and if under this assumption the num... | [
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17... | null | null |
numina_10103924 | 6. The maximum value sought is $N=45$, an example is shown in Figure 2.
If we let $n=5, m=\left\lceil\frac{n}{2}\right\rceil=3$, we will next prove: if $n$ is odd, in an $n \times n$ grid, fill in $1 \sim n^{2}$, then the sum of the numbers in some $2 \times 2$ subgrid is at least $8 m^{2}-11 m+6$.
Let the number fil... | 45 | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} | 6. Determine the maximum value of the integer $N$ that satisfies the following conditions: In a $5 \times 5$ grid, fill in the numbers from $1 \sim 25$ such that each cell contains a different number, and the sum of the numbers in some $2 \times 2$ subgrid is at least $N$. | [
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numina_10083168 | Consider a vertex $X$. It has 5 edges. Therefore, there are at least 3 edges of the same color. Without loss of generality, we assume that there are at least 3 red edges. We consider $A, B$ and $C$ as the 3 vertices connected to $X$ by these 3 edges. If one of the edges $A B, B C$ or $A C$ is red, we have a monochromat... | proof | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} | A complete graph with $n$ vertices, by definition, is a graph that contains all possible edges.
In a complete graph with 6 vertices, the edges are colored either yellow or red. Show that it is always possible to find a red triangle or a yellow triangle. | [
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22870... | null | null |
aops_2366053 | [hide]Since $\{a,b\}$ is the same as $\{b,a\}$, we can assume WLOG that $a>b$.
$a=b+i$ where $i\leq n$ has solutions $b\in [0,2n-i]$. So the number of solutions is $2n-i+1$
We want to find $\sum_{i=1}^n (2n+1)-i$
This is $(2n+1)(n)-\frac{n(n+1)}{2}=\boxed{\frac{n(3n+1)}{2}}$[/hide] | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "How many subsets $\\left\\{ a,b \\right\\}$ are there of the set $\\left\\{ 0,1,2,3,...,2n \\right\\} $ such that $\\left| a-b \\right|\\le n$?\n\n\n",
"content_html": "How many subsets <img src=\"//... | How many subsets \(\{a,b\}\) of the set \(\{0,1,2,\dots,2n\}\) are there such that \(|a-b|\le n\)? | [
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204... | null | null |
aops_329514 | \[ 1\minus{}\frac{\binom{40\minus{}n}6}{\binom{40}6}\equal{}1\minus{}\frac{(40\minus{}n)!34!}{(34\minus{}n)!40!}\]\[ \equal{}1\minus{}\frac{(40\minus{}n)(39\minus{}n)(38\minus{}n)(37\minus{}n)(36\minus{}n)(35\minus{}n)}{2763633600}\] | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "40 cards, $ n$ of them are green and the rest are red (of course $ n\\le 40$). If we shuffle them. What is the probability of finding a green card in the first 6 cards.",
"content_html": "40 cards, <... | 40 cards, \(n\) of them are green and the rest are red (with \(0\le n\le 40\)). The cards are shuffled. What is the probability that at least one of the first six cards is green? | [
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58998,
53167,... | null | null |
aops_480365 | [quote="GameBot"]How many four-character license plates consist of a consonant, followed by a vowel, followed by a consonant, and then a digit? (For this problem, consider Y a vowel.)[/quote]
I think it is below.
[hide="Solution"]$20\times6\times20\times10=\boxed{\text{24000}}$[/hide] | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "How many four-character license plates consist of a consonant, followed by a vowel, followed by a consonant, and then a digit? (For this problem, consider Y a vowel.)",
"content_html": "How many four... | How many four-character license plates consist of a consonant, followed by a vowel, followed by a consonant, and then a digit? (For this problem, consider \(Y\) a vowel.) | [
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12192,
2... | null | null |
numina_10081688 | We divide the chessboard into 16 small $2 \times 2$ squares. According to the pigeonhole principle, there exist 3 pawns in the same square. They necessarily form an 'L' as desired. | proof | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} | On place 33 pawns on a chessboard of 8 rows and 8 columns. Show that there exist 3 pawns that form an 'L' (or "corner"). | [
6631,
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12024,
63354,
69880,
40531,
1036,
8717,
211,
10749,
28060,
30398,
4052,
6190,
510... | null | null |
aops_3145333 | [hide]
If $x=0$, $y+z \le 17$. Set $w=17-(y+z)$. Then, we have $w+y+z=17$, so there are $\binom{19}{2}$ solutions
If $x=1$, $y+z \le 14$. Set $w=14-(y+z)$. Then, we have $w+y+z=14$, so there are $\binom{16}{2}$ solutions
If $x=2$, $y+z \le 11$. Set $w=11-(y+z)$. Then, we have $w+y+z=11$, so there are $\binom{13}{2}$ ... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "How many solutions are there to $3x+y+z\\le17$ where x, y, and z are nonnegative integers?",
"content_html": "How many solutions are there to <img src=\"//latex.artofproblemsolving.com/8/9/a/89a87663... | How many solutions are there to
\[
3x+y+z\le 17
\]
where \(x,y,z\) are nonnegative integers? | [
22596,
7334,
30822,
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10045,
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1169,
22784,
10860,
11662,
69447,
66919,
24215,
2041,
9292,
48036,
352... | null | null |
aops_3313174 | We will show that the maximum number of princes that can be saved is $\min(k,n)$. It is clear that $\min(k,n)$ princes can be saved if the princesses take $\min(k,n)$ frogs and each kisses all of them. Let us show that this is the maximum number. Look on it as on a game between the princesses and the witch. Assume ... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Once upon a time there are $n$ pairs of princes and princesses who are in love with each other. One day a witch comes along and turns all the princes into frogs; the frogs can be distinguished by sight but... | Once upon a time there are \(n\) pairs of princes and princesses who are in love with each other. One day a witch comes along and turns all the princes into frogs; the frogs can be distinguished by sight but the princesses cannot tell which frog corresponds to which prince. The witch tells the princesses that if any of... | [
31172,
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24969,
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13372,
39758,
14106,
15514,
26814,
458,
17355,
22617,
52342,
18885,
3582,
... | null | null |
aops_278625 | Let $ n$ be the pupils that don't play any sport. By the Principle of Inclusion-Exclusion, we have
$ 7 \plus{} 16 \plus{} 16 \minus{} 4 \minus{} 6 \minus{} 3 \plus{} 1 \plus{} n \equal{} 32 \implies 27 \plus{} n \equal{} 32 \implies n \equal{} \boxed{5}$. | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "The class has 32 pupils , 7 of them play football , 16 of them play volleyball , 16 of them play basketball . 4 of them play football and volleyball , 6 of them play football and basketball , 3 of play vol... | A class has 32 pupils. Seven of them play football, 16 play volleyball, and 16 play basketball. Four of them play football and volleyball, six play football and basketball, three play volleyball and basketball, and one pupil plays all three sports. How many pupils in the class do not play any sport? | [
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63887,
36641,
29495,
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37331,
14706,
13375,
40323,
44112,
30341,
... | null | null |
ours_18837 | All one-digit numbers have no repeating digits, so that gives us \(9\) numbers. For a two-digit number to have no repeating digits, the first digit must be between \(1\) and \(9\), while the second digit must not be equal to the first, giving us \(9 \cdot 9 = 81\) numbers. For a three-digit number to have no repeating ... | 1242 | {
"competition": "jhmt",
"dataset": "Ours",
"posts": null,
"source": "general2-solutions.md"
} | How many positive numbers up to and including \(2012\) have no repeating digits? | [
19682,
21783,
61268,
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222,
25958,
40651,
60373,
60541,
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12655,
3286,
12412,
9998,
69040,
54104,
17941,
67364,
49607,
30994,
45154... | null | null |
numina_10111763 | 【Analysis】(1) When all five greeting cards are given incorrectly, and there are no mutual wrongs between any two, then student No. 1 has four wrong ways. For example, if student No. 1 gets card No. 2, then student No. 2 cannot get card No. 1 or card No. 2, leaving only 3 wrong ways. Student No. 3, besides not getting c... | 44 | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} | 12. (5 points) Before New Year's Day, Xiaofang made greeting cards for her five classmates. When putting the cards into envelopes, she made a mistake, and none of the five classmates received the card Xiaofang made for them; instead, they received cards Xiaofang made for others. In total, there are $\qquad$ possible sc... | [
51659,
21658,
18120,
11665,
22155,
64306,
20060,
11508,
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61470,
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56502,
6199,
38566,
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66122,
50381,
40773,
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9394,
28984,
5760,
45534,
8906,
15311,
10346,
6855,
12609,
49120,
... | null | null |
aops_1248553 | [hide = Stars and Bars]
First we look at how many are possible, without caring about the maximum. There are 21 stars and 3 bars, so there are ${21+3 \choose 3} = 2024$ ways. However in some of these cases the student has more than 10 points in a subject . Remove the 11 stars that the subject has and use stars and bars... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "In an examination, the score in each of the four languages -- English,French,Spanish,Latin--- can be integers between $0$ and $10$.Then find the number of ways in which a student can secure a total score o... | In an examination, the score in each of the four languages—English, French, Spanish, Latin—can be integers between \(0\) and \(10\). Find the number of ways in which a student can secure a total score of \(21\). | [
30467,
52421,
2863,
17062,
41634,
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65581,
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51963,
36638,
13375,
5229,
20991,
7214,
12911,
46162,
53140,
68550,
4... | null | null |
numina_10253594 | To find the number of sequences \(a_1, a_2, \dots, a_{100}\) such that:
1. There exists \(i \in \{1, 2, \dots, 100\}\) such that \(a_i = 3\).
2. \(|a_i - a_{i+1}| \leq 1\) for all \(1 \leq i < 100\).
We will assume that \(a_i\) are positive integers, as the problem seems to imply this.
1. **Determine the possible val... | null | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "aops_forum"
} | Find the number of sequences $a_1,a_2,\dots,a_{100}$ such that
$\text{(i)}$ There exists $i\in\{1,2,\dots,100\}$ such that $a_i=3$, and
$\text{(ii)}$ $|a_i-a_{i+1}|\leq 1$ for all $1\leq i<100$. | [
15160,
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44767,
41015,
33925,
55090,
12939,
11086,
1742,
13177,
46490,
24121,
3... | null | null |
aops_2319310 | Nice , in a more general formulation :
Let $x_1,x_2,...,x_n\le m$ and $y_1,...,y_m\le n$ be positive integers then there are two sums one consisting of $x's$ and the other of $y's$ that are equal.
Assume that $y_1+y_2+...+y_m>x_1+x_2+...+x_n $ Consider the following for every $k=1,2,3,...,n$ choose $c_k$ such that
$... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Given a sequence of $19$ positive (not necessarily distinct) integers not greater than $93$, and a set of $93$ positive (not necessarily distinct) integers not greater than $19$. Show that we can find non-... | Given a sequence of \(19\) positive integers, each at most \(93\), and a multiset of \(93\) positive integers, each at most \(19\). Show that there exist nonempty subsequences (submultisets) of the two sequences whose sums are equal. | [
31099,
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29403,
60099,
43237,
522... | null | null |
numina_10085296 | Solution. By replacing 2013 with 1, 2, 3, and 4 and then calculating the sum, you get a suspicion of which number squared this sum yields. The factorials suggest that you should look in the binomial coefficients. We prove something more general:
$$
\sum_{n=0}^{m} \frac{(2 m) !}{(n !(m-n) !)^{2}}=\left(\begin{array}{c}... | proof | {
"competition": "Numina-1.5",
"dataset": "NuminaMath-1.5",
"posts": null,
"source": "olympiads"
} | Task 1. Prove that
$$
\sum_{n=0}^{2013} \frac{4026 !}{(n !(2013-n) !)^{2}}
$$
is the square of an integer. | [
28983,
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37085,
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61581,
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50335,
38341,
9720,
9130,
26540,
26193,
35013,
34909,
56050,
34966,... | null | null |
aops_1312364 | [quote=Jyzhang12]When they say "HW", it means AoPS HW. And it's not really related to bases. There are $4$ primes below $10$. So, every digit of the $3$ has $4$ possibilities. Thus, $4^3 = 64$ numbers?[/quote]
Hi AoPS HW is not allowed either. | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "How many three-digit numbers can be written using the digits that are prime numbers less than 10?",
"content_html": "How many three-digit numbers can be written using the digits that are prime number... | How many three-digit numbers can be written using the digits that are prime numbers less than 10? | [
7047,
10500,
57230,
53513,
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21247,
2235,
2420,
30895,
5593,
7974,
53868,
5546,
47088,
38203,
3600,
1732... | null | null |
aops_580036 | [hide="My thoughts"]
There are $\sum_{i=1}^{5}\binom{20}{2i-1}\binom{32}{11-2i}$ ways to choose an odd number of black cards,
where each $\binom{20}{2i-1}$ is the number of ways to choose $2i-1$ black cards,
and each $\binom{32}{11-2i}$ is the number of ways to choose $10-(2i-1)=11-2i$ red cards.
Also, there are a t... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Deck of 52 cards is given. 20 out of them are black and 32 are red. We choose 10 cards at random. Prove that probability of choosing odd number of black cards lies in interval $(\\frac{49}{100}, \\frac{51}... | A standard deck of 52 cards contains 20 black cards and 32 red cards. Ten cards are chosen at random. Prove that the probability of choosing an odd number of black cards lies in the interval \(\left(\tfrac{49}{100},\tfrac{51}{100}\right)\). | [
44610,
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47622,
44820,
52352,
49422,
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42702,
2305,
10357,... | null | null |
aops_1800159 | [quote=motorfinn]Is the answer [hide=this]$(n-1)!$[/hide]?
If not I may have misinterpreted...[/quote]
No, the answer is defined by $!n$ (the number of [url=https://en.wikipedia.org/wiki/Derangement]derangements[/url] of $n$ objects). The formula for $!n$ can be derived by PIE. | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "[quote]A party was attended by $n$ guests. When the guests arrived, they left their hats in the same coatroom. After the party ended, there was an electrical power failure, so each guest took a hat from th... | A party was attended by \(n\) guests. When the guests arrived, they left their hats in the same coatroom. After the party ended, there was an electrical power failure, so each guest took a hat from the coatroom at random. When the guests were back on the street, they were amused to find that none of them got his hat ba... | [
60417,
69421,
46053,
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9097,
65625,
12727,
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16466,
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61283,
6543,
53867,
29661,
41655,
26644,
18923,
12327,
17326,
9711,
54165,
15283,
24443... | null | null |
aops_3119549 | First we give two lemma, I'll skip the proof.
[b]Lemma 1 (Vandermonde): [/b] For $a,b,c\in\mathbb N_+,\dbinom{a+b}{c}=\sum\limits_{k=0}^a\dbinom{a}{k}\dbinom{b}{c-k}.$
[b]Lemma 2: [/b] For $a,b,c\in\mathbb N_+,a\geq c,$ $\dbinom{a}{b}\dbinom bc=\dbinom ac\dbinom{a-c}{b-c}.$
Then
$$\begin{aligned}\sum\limits_{k=0}^n\... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Let $n,p,q\\in\\mathbb N_+.$ Prove that\n$$\\dbinom{n}{p}\\dbinom{n}{q}=\\sum\\limits_{k=0}^n\\dbinom{p}{k}\\dbinom{q}{k}\\dbinom{n+k}{p+q}.$$",
"content_html": "Let <img src=\"//latex.artofproblemso... | Let \(n,p,q\in\mathbb{N}_+\). Prove that
\[
\binom{n}{p}\binom{n}{q}=\sum_{k=0}^{n}\binom{p}{k}\binom{q}{k}\binom{n+k}{p+q}.
\] | [
22300,
35723,
2369,
53870,
38186,
64711,
62455,
42790,
45779,
5922,
54058,
28652,
15098,
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70715,
25164,
61492,
15943,
59987,
14474,
12454,
29765,
17646,
64905,
51352,
44602,
35643,... | null | null |
aops_478472 | [hide="Outline"]Make a $6 \times 6$ subtraction chart with the numbers $1$ through $6$ on top and on the left, and put the difference in each box. Count how many squares have $2$ in them. (Don't worry about signs - just put the absolute value of the difference.) This will give you the answer to A when you put it ove... | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "You roll two six-sided dice.\nFind the Probability of:\na) Pips(numbers on die) differ by 2\nb) Rolling a 8 (when added together)\nc) Pips differ by 2 [b]AND[/b] rolling a 8 \nd) Pips differ by 2 [b]OR[/b]... | You roll two six-sided dice. Find the probability of each of the following:
a) The numbers on the two dice differ by 2.
b) The sum of the two dice is 8.
c) The numbers on the two dice differ by 2 AND the sum is 8.
d) The numbers on the two dice differ by 2 OR the sum is 8. | [
52514,
50447,
66592,
24133,
31241,
34319,
5604,
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55226,
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49676,
22900,
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51587,
20332,
21103,
63476,
42358,
50794,
37278,
50394,
12932,
22035,
59875,
47649,
... | null | null |
aops_417307 | [hide="similar answer but if you don't want to list them out, do this way"]There are two sums of one digit numbers, $9, 9, 7$ and $9,8,8$. You can order the first sum $\frac{3!}{2!}$ ways and the second one in $\frac{3!}{2!}$ ways. $3+3=\boxed{6}$[/hide] | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "For how many three-digit whole numbers does the sum of the digits equal $25$?\n\n$\\text{(A)}\\ 2 \\qquad \\text{(B)}\\ 4 \\qquad \\text{(C)}\\ 6 \\qquad \\text{(D)}\\ 8 \\qquad \\text{(E)}\\ 10$",
"... | For how many three-digit whole numbers does the sum of the digits equal \(25\)?
\(\text{(A)}\ 2 \qquad \text{(B)}\ 4 \qquad \text{(C)}\ 6 \qquad \text{(D)}\ 8 \qquad \text{(E)}\ 10\) | [
9099,
30004,
15671,
11662,
60089,
44192,
22490,
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63519,
36378,
71000,
21537,
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24164,
35253,
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26509,
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38362,
38781,
2289,
26754,
56968,
20838,
44376,
37460,
26027,
35942,
6440,
25021,... | null | null |
aops_1589404 | Sam can pull $20$ polka dot socks, and $1$ fuzzy and $1$ striped. This is the most she can pull w/o satisfying any requirement. So she must pull one more to fulfill it for sure. Therefore, the answer is $20+1+1+1=23$ | null | {
"competition": null,
"dataset": "AOPS",
"posts": [
{
"attachments": [],
"content_bbcode": "Samantha has a sock cupboard filled with different types of socks. She has 8 pairs of striped socks, 10 pairs of polka-dotted socks, and 4 pairs of fuzzy black socks. Only the black socks are fuzzy, and ea... | Samantha has a sock cupboard containing 8 pairs of striped socks, 10 pairs of polka-dotted socks, and 4 pairs of fuzzy black socks. Only the black socks are fuzzy, and each sock is separate (not attached to its mate).
With her eyes closed and with gloves on, Samantha wants to pick socks from the cupboard. What is the ... | [
4362,
25381,
37269,
11342,
13618,
13652,
14733,
14748,
40072,
54977,
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2,
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823,
27957,
7078,
16574,
10832,
5170,
18082,
14037,
36649,
49862,
3920,
210,
... | null | null |
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