problem stringlengths 16 1.47k | solution stringlengths 29 4.47k | type stringclasses 7
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What is the coefficient of $x^4$ in the expansion of $(1-2x^2)^5$? | Using the binomial theorem, we find that the $x^4=(x^2)^2$ term of the expansion is $\binom{5}{2}(1)^3(-2x^2)^2=10(4x^4)=40x^4$. Thus, the desired coefficient is $\boxed{40}$. | Counting & Probability | Level 4 | counting_and_probability |
A pizza parlor offers six toppings. What is the greatest number of four-topping pizzas that can be made such that no two pizzas have the same topping combination? | Because there are 6 choices of toppings, and each pizza must have 4 of them, there are ${6 \choose 4} = \boxed{15}$ four-topping pizzas. | Counting & Probability | Level 2 | counting_and_probability |
Suppose two distinct integers are chosen from between 5 and 17, inclusive. What is the probability that their product is odd? | There are 13 integers between 5 and 17 inclusive, so there are $\binom{13}{2} = 78$ ways to choose two of them without regard to order. In order for the product of two integers to be odd, both of the integers themselves must be odd. There are 7 odd integers between 5 and 17 inclusive, so there are $\binom72 = 21$ ways... | Counting & Probability | Level 3 | counting_and_probability |
A school's math club has 6 boys and 8 girls. A team needs to be selected to send to the state math competition. The team needs 6 people. In how many ways can the team be selected to have 3 boys and 3 girls? | We are picking 3 boys out of 6, so there are $\binom{6}{3} = 20$ options for the boys on the team. We are picking 3 girls out of 8, so there are $\binom{8}{3} = 56$ options for the girls on the team. This gives a total of $20 \times 56 = \boxed{1120}$ choices. | Counting & Probability | Level 2 | counting_and_probability |
The lottery in a state consists of two drawings. First, a MegaBall is picked from among 27 numbered balls. Second, five WinnerBalls are picked from among 44 numbered balls. To win the lottery, one must pick the MegaBall number correctly and also pick the numbers on the five WinnerBalls (but the order does not need to b... | The probability that the MegaBall matches is $\dfrac{1}{27}$ . The probability that the 5 WinnerBalls match is $\dfrac{1}{\binom{44}{5}}$. So my chances of winning are $\left(\dfrac{1}{27}\right)\times\left(\dfrac{1}{\binom{44}{5}}\right) = \boxed{\dfrac{1}{29,\!322,\!216}}$. | Counting & Probability | Level 4 | counting_and_probability |
How many ways are there to put 4 distinguishable balls into 2 distinguishable boxes? | For each ball, there are 2 choices of which box to place it in. Since this choice is independent for each of the 4 balls, we multiply the number of choices together. Hence there are $2^4 = \boxed{16}$ ways to place 4 distinguishable balls into 2 distinguishable boxes. | Counting & Probability | Level 3 | counting_and_probability |
Someone has two 20-sided dice that each have 4 maroon sides, 7 teal sides, 8 cyan sides, and one sparkly side. If both dice are rolled, what is the probability they come up the same? | This problem requires a little bit of casework. There are four ways in which the dice can both show the same thing: if they both show maroon, both show teal, both show cyan or both show sparkly. The probability of getting maroon is $\dfrac{4}{20}$, so the probability that they will both show maroon is $\left(\dfrac{4}... | Counting & Probability | Level 3 | counting_and_probability |
Ten 6-sided dice are rolled. What is the probability that exactly three of the dice show a 1? Express your answer as a decimal rounded to the nearest thousandth. | There are $\binom{10}{3}=120$ ways to select exactly three dice to roll 1's ones out of ten dice total. The probability of any one of these outcomes occurring is $\left(\dfrac{1}{6}\right)^3\left(\dfrac{5}{6}\right)^7$ and they're all mutually exclusive cases, so the probability that one of them will occur (which is th... | Counting & Probability | Level 4 | counting_and_probability |
A club has 20 members, 10 boys and 10 girls. In how many ways can a president and a vice-president be chosen if they must be of different gender? | The president can be any one of the 20 members, and the vice-president can be any one of the 10 members of the opposite sex. The answer is $20\times 10=\boxed{200}$. | Counting & Probability | Level 2 | counting_and_probability |
What is the second smallest four-digit number in Pascal's triangle? | Every positive integer appears in Pascal's triangle! The number 1000 appears in the row that starts 1, 1000. Then 1001 appears in the next row. So, the answer is $\boxed{1001}$. | Counting & Probability | Level 3 | counting_and_probability |
A person guesses randomly on a 15-question multiple choice test in which each problem has two choices, a correct one and an incorrect one. What is the probability that they get at least half of the questions correct? Express your answer as a common fraction. | Steve cannot get exactly half the questions right since there are an odd number of questions. So he can get either more than half or less than half correct, with equal probability, since he has a $1/2$ chance of getting any individual question correct. This means that Steve has a probability of $\boxed{\frac{1}{2}}$ of... | Counting & Probability | Level 3 | counting_and_probability |
Suppose four coins are flipped simultaneously: a penny, a nickel, a dime, and a quarter. What is the probability that at least 15 cents worth of coins come up heads? | There are $2^4=16$ possible outcomes, since each of the 4 coins can land 2 different ways (heads or tails). If the quarter is heads, there are 8 possibilities, since each of the other three coins may come up heads or tails. If the quarter is tails, then the nickel and dime must be heads, so there are 2 possibilities, ... | Counting & Probability | Level 5 | counting_and_probability |
Two cards are chosen at random from a standard 52-card deck. What is the probability that both cards are numbers (2 through 10) totaling to 12? | There are two cases that we have to consider.
$\bullet~$ Case 1: The first card is one of 2, 3, 4, 5, 7, 8, 9, 10.
There are 32 such cards, so this occurs with probability $\dfrac{32}{52}$. For any of these cards, there are 4 cards left in the deck such that the cards sum to 12, so the probability of drawing one is ... | Counting & Probability | Level 5 | counting_and_probability |
A person has 10 pieces of art, including 3 prints by Escher. If they hang the pieces of art in a row in a random order, what is the probability that all three pieces by Escher will be placed consecutively? | To count the number of ways of arranging the 10 pieces of art with the three Escher's consecutively, treat the three of them as one item. It is clear that we are then selecting the location of 1 item out of 8 total which can be done in $\binom{8}{1}=8$ ways. There are also a total of $\binom{10}{3}=120$ ways to place t... | Counting & Probability | Level 5 | counting_and_probability |
A person rolls 6 fair 6-sided dice. What is the probability that at least two dice show the same number? | The only way in which there won't be two dice of the same number is if for each number between 1 and 6 there is exactly one die displaying that number. If we line the dice up, there are a total of $6!$ ways in which we could order 6 dice all displaying different numbers and a total of $6^6$ possible outcomes since each... | Counting & Probability | Level 5 | counting_and_probability |
Four horizontal lines and four vertical lines are drawn in a plane. In how many ways can four lines be chosen such that a rectangular region is enclosed? | In order for the four lines to enclose a rectangular region, we must choose two horizontal and two vertical lines. If we were to choose more than two of one of these types of lines, we would not be able to enclose any region. We can count independently the number of ways to choose vertical and horizontal lines. Ther... | Counting & Probability | Level 4 | counting_and_probability |
Interior numbers begin in the third row of Pascal's Triangle. The sum of the interior numbers in the fourth row is 6. The sum of the interior numbers of the fifth row is 14. What is the sum of the interior numbers of the seventh row? | If Pascal's Triangle begins with row 1, then the sum of the elements in row $n$ is $2^{n-1}$. The interior numbers refer to all numbers in the row except the $1$ at each end, so the sum of the interior elements in row $n$ is $2^{n-1}-1-1=2^{n-1}-2$. For the fourth row, the sum is $2^3-2=6$. For the fifth row, the sum i... | Counting & Probability | Level 3 | counting_and_probability |
What is the 39th number in the row of Pascal's triangle that has 41 numbers? | The row 1, 1 has 2 numbers. The row 1, 2, 1 has 3 numbers. The row 1, 3, 3, 1 has 4 numbers. Each time we go down one row, we have one more number in the list. So, the row that starts 1, $k$ has $k+1$ numbers (namely, the numbers $\binom{k}{0}, \binom{k}{1}, \binom{k}{2}, \ldots, \binom{k}{k}$.) So the row with 41... | Counting & Probability | Level 4 | counting_and_probability |
How many ways are there to put 4 balls in 3 boxes if the balls are not distinguishable and neither are the boxes? | Since the balls and boxes are indistinguishable, we only need to consider the number of the balls in boxes without considering order. The arrangements are (4,0,0),(3,1,0),(2,2,0),(2,1,1), for a total of $\boxed{4}$ ways. | Counting & Probability | Level 3 | counting_and_probability |
In how many ways is it possible to arrange the digits of 1150 to get a four-digit multiple of 5? | A multiple of 5 has to end in 0 or 5. If it ends in 0, the three remaining digits can go anywhere. There are 3! ways to arrange 3 digits, but we must divide by 2! to correct for overcounting since the 1's are identical. If the number ends in 5, the digit 0 can go in either of 2 places. Then the two remaining digits can... | Counting & Probability | Level 5 | counting_and_probability |
After a gymnastics meet, each gymnast shook hands once with every gymnast on every team (except herself). Afterwards, a coach came down and only shook hands with each gymnast from her own team. There were a total of 281 handshakes. What is the fewest number of handshakes the coach could have participated in? | The number of gymnasts is some integer $n$, so that the number of gymnast-gymnast handshakes is ${n \choose 2}$ for some $n$. Also, the coach must participate in an integer $k<n$ number of handshakes. So, ${n \choose 2} + k = 281$. If we want to minimize $k$, we need the maximal $n$ such that ${n \choose 2} \le 281$, w... | Counting & Probability | Level 5 | counting_and_probability |
If two distinct members of the set $\{ 3, 7, 21, 27, 35, 42, 51 \}$ are randomly selected and multiplied, what is the probability that the product is a multiple of 63? Express your answer as a common fraction. | In order to produce a multiple of 63, we must choose at least two factors of 3 and one factor of 7 among the prime factorizations of the two numbers we choose. We count the number of ways in which we can do this by considering the four multiples of 7 in our list. There are two which are not multiples of 3 (7 and 35) ... | Counting & Probability | Level 5 | counting_and_probability |
A student must choose a program of four courses from a list of courses consisting of English, Algebra, Geometry, History, Art, and Latin. This program must contain English and at least one mathematics course. In how many ways can this program be chosen? | The program must contain exactly $3$ of the $5$ classes that are not English. Therefore, there are $\tbinom{5}{3} = 10$ valid programs if we ignore the mathematics requirement.
Since there are $2$ math classes (Algebra and Geometry), $3$ of the $5$ classes besides English are not math. Therefore, there is only one pro... | Counting & Probability | Level 5 | counting_and_probability |
A pet store has 15 puppies, 6 kittens, and 8 hamsters. Three people each want to buy a pet. For variety, they each want a different kind of pet. How many ways can they buy pets and leave the store satisfied? | Ignoring who gets which pet for now, we can see that there are $15 \cdot 6 \cdot 8$ ways to choose one puppy, one kitten, and one hamster. Now, Alice has three choices for which type of pet she wants, Bob has two, and Charlie has one, due to the constraint that they must each have a different type of pet. Thus, there a... | Counting & Probability | Level 4 | counting_and_probability |
How many four-digit numbers greater than 2999 can be formed such that the product of the middle two digits exceeds 5? | For the first digit, there are seven choices (3, 4, 5, 6, 7, 8, or 9). For the last digit, there are ten choices (0 through 9).
We know that if either of the middle digits is 0, their product will not exceed 5. So, only consider pairs of middle digits formed from choosing two numbers between 1 and 9, inclusive. There ... | Counting & Probability | Level 4 | counting_and_probability |
What is the nearest integer to $(2+\sqrt3)^4$? | If we expand this using the binomial theorem, we get a bunch of terms with $\sqrt3$ in them. To avoid painful estimation, we use the following trick: Add $(2-\sqrt3)^4$ to this expression. We know that $(2-\sqrt3)^4$ is small, since $2-\sqrt3<1$. When we add these together, the $\sqrt3$ terms magically cancel out. By t... | Counting & Probability | Level 3 | counting_and_probability |
A person has 5 books, two of which are identical copies of the same math book (and all of the rest of the books are different). In how many ways can they arrange them on a shelf? | There are $5!$ ways to arrange the books if they are unique, but two are identical so we must divide by $2!$ for an answer of $\dfrac{5!}{2!} = \boxed{60}$. | Counting & Probability | Level 2 | counting_and_probability |
How many 4-letter words with at least one vowel can be constructed from the letters A, B, C, D, and E? (Note that A and E are vowels, any word is valid, not just English language words, and letters may be used more than once.) | First we count the number of all 4-letter words with no restrictions on the word. Then we count the number of 4-letter words with no vowels. We then subtract to get the answer.
Each letter of a word must be one of A, B, C, D, or E, so the number of 4-letter words with no restrictions on the word is $5\times 5\times 5\... | Counting & Probability | Level 3 | counting_and_probability |
A club has 20 members: 10 boys and 10 girls. A 4-person committee is chosen at random. What is the probability that the committee has at least 1 boy and at least 1 girl? | The number of ways to choose a committee of all boys or all girls is $2\times\binom{10}{4}=420$. The total number of committees is $\binom{20}{4}=4845$. Thus the answer is $1-\dfrac{420}{4845} = \dfrac{4425}{4845} = \boxed{\dfrac{295}{323}}$. | Counting & Probability | Level 4 | counting_and_probability |
Find the greatest common divisor of $7!$ and $(5!)^2.$ | $$ \begin{array}{rcrcr} 7! &=& 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 &=& 2^4 \cdot 3^2 \cdot 5^1 \cdot 7^1 \\ (5!)^2 &=& (5 \cdot 4 \cdot 3 \cdot 2 \cdot 1)^2 &=& 2^6 \cdot 3^2 \cdot 5^2 \\ \text{gcd}(7!, (5!)^2) &=& 2^4 \cdot 3^2 \cdot 5^1 &=& \boxed{720} \end{array} $$ | Counting & Probability | Level 2 | counting_and_probability |
A student is taking the 25-question, multiple choice American Mathematics Competition. Each question has five answer choices. The student guesses random answers for the last four questions. What is the probability that they will get at least one of these final four questions right? | We find the probability that William misses the last four questions, and subtract from 1. The probability of William missing a question is $\frac{4}{5}$, so the probability that he misses all four is $\left(\frac{4}{5}\right)^4 = \frac{256}{625}$. The probability that he gets at least one right is $1-\frac{256}{625} =\... | Counting & Probability | Level 4 | counting_and_probability |
How many ways are there to put 5 balls in 2 boxes if the balls are distinguishable but the boxes are not? | Since the boxes are indistinguishable, there are 3 possibilities for arrangements of the number of balls in each box.
Case 1: 5 balls in one box, 0 in the other box. We must choose 5 balls to go in one box, which can be done in $\binom{5}{5} = 1$ way.
Case 2: 4 balls in one box, 1 in the other box. We must choose 4 b... | Counting & Probability | Level 4 | counting_and_probability |
At breakfast, lunch, and dinner, a person randomly chooses with equal probabilities either an apple, an orange, or a banana to eat. On a given day, what is the probability that they will eat at least two different kinds of fruit? | The only way for Joe not to eat at least two different kinds of fruit is if he eats all apples, all oranges, or all bananas. The probability that he eats all apples is $\left( \frac{1}{3} \right) ^3=\frac{1}{27}$. The probability that he eats all oranges and the probability that he eats all bananas are the same. So, th... | Counting & Probability | Level 4 | counting_and_probability |
Forty cards are placed into a box, each bearing a number 1, 2, 3, 4, 5, 6, 7, 8, 9, or 10, with each number entered on four cards. Four cards are drawn from the box at random and without replacement. Let $p$ be the probability that all four cards bear the same number. Let $q$ be the probability that three of the cards ... | The total number of ways that the numbers can be chosen is $\binom{40}{4}.$ Exactly 10 of these possibilities result in the four cards having the same number.
Now we need to determine the number of ways that three cards can have a number $a$ and the other card have a number $b$, with $b\ne a$. There are $10\cdot 9 = 9... | Counting & Probability | Level 5 | counting_and_probability |
A weather forecast predicts that for each day from Monday to Friday this week, there is a $40\%$ chance of sun, a $25\%$ chance of getting 4 inches of rain, and a $35\%$ chance of getting 10 inches of rain. What is the expected value of the total number of inches of rain which will fall from Monday to Friday? Give your... | For each day, the expected amount of rain is $(.40)(0)+(.25)(4)+(.35)(10)=0+1+3.5=4.5$. To find the total expected amount of rain for the days from Monday to Friday, we can add the amounts for each day, getting $5\cdot4.5= \boxed{22.5}$ inches total. | Counting & Probability | Level 4 | counting_and_probability |
In how many distinct ways can one arrange five keys on a keychain, if one wants to put a house key next to a car key? Two arrangements are not considered different if the keys are in the same order (or can be made to be in the same order without taking the keys off the chain--that is, by reflection or rotation). | First, put the house key and car key next to each other on the keychain. It doesn't matter where they are put, because the keychain can be rotated and reflected to move them to any other pair of adjacent locations. The remaining three keys can be put on the keychain in $3!=\boxed{6}$ ways. | Counting & Probability | Level 4 | counting_and_probability |
Compute $\frac{6! + 7!}{5!}$ | This simplifies nicely because $5!$ divides both $6!$ and $7!$ evenly:
\begin{align*}
\frac{6! + 7!}{5!} &=\frac{6!}{5!} + \frac{7!}{5!}\\&= 6 + 6 \cdot 7\\
&= 6 \cdot 8 \\
&= \boxed{48}
\end{align*} | Counting & Probability | Level 1 | counting_and_probability |
How many positive integers less than $500$ can be written as the sum of two positive perfect cubes? | Note that $7^3 < 500 < 8^3,$ so any positive integer that can be written as the sum of two positive perfect cubes must be written as the sum of two cubes $a^3 + b^3$ where $1 \le a \le 7$ and $1 \le b \le 7.$ We can make a chart of the sum of two such cubes: $$
\begin{array}{c|ccccccc}
& 1^3 & 2^3 & 3^3 & 4^3 & 5^3 & 6... | Counting & Probability | Level 5 | counting_and_probability |
A basketball team must decide on a new uniform. The seventh-graders will pick the color of the shorts (black or gold) and the eighth-graders will pick the color of the jersey (black, white, or gold), but the two groups of players will not confer together. If, for both garments, each possible color is equally likely to ... | We will count the number of garment configurations in which the garments do not match color and divide by the total number of garment configurations in order to find the probability that the garments do not match. If the seventh-graders choose a black garment, there are two garments the eighth-graders can choose such t... | Counting & Probability | Level 2 | counting_and_probability |
Compute $\dbinom{8}{2}$. | $\dbinom{8}{2} = \dfrac{8!}{2!6!}=\dfrac{8\times 7}{2\times 1}=\boxed{28}.$ | Counting & Probability | Level 2 | counting_and_probability |
Three tiles are marked X and two other tiles are marked O. The five tiles are randomly arranged in a row. What is the probability that the arrangement reads XOXOX? | There are three X's and two O's, and the tiles are selected without replacement, so the probability is \[
\frac{3}{5}\cdot\frac{2}{4}\cdot\frac{2}{3}\cdot\frac{1}{2}\cdot\frac{1}{1}= \frac{1}{10}.
\]OR
The three tiles marked X are equally likely to lie in any of $\binom{5}{3}=10$ positions, so the probability of this ... | Counting & Probability | Level 3 | counting_and_probability |
How many two-digit numbers are there in which the tens digit is greater than the ones digit? | In addition to two-digit numbers, consider the one-digit numbers $01$ through $09.$ Of these $99$ numbers, $9$ have repeated digits $01,$ $02,$ $03,$ $\ldots,$ $99,$ namely: $11,$ $22,$ $33,$ $\ldots,$ and $99.$ Of the remaining $90$ numbers, each one has a unique counterpart formed by swapping their tens and units dig... | Counting & Probability | Level 2 | counting_and_probability |
There are 5 blue chips and 3 yellow chips in a bag. One chip is drawn from the bag. That chip is placed back into the bag. A second chip is then drawn. What is the probability that the two selected chips are of different colors? Express your answer as a common fraction. | If we draw a blue chip and then a yellow chip, or if we draw a yellow chip and then a blue chip, then our draws will be different colors. So, the probability is $\frac{5}{8} \cdot \frac{3}{8} + \frac{3}{8} \cdot \frac{5}{8} = 2 \cdot \frac{15}{64} = \boxed{\frac{15}{32}}$. | Counting & Probability | Level 3 | counting_and_probability |
A person has $3$ apples, $2$ oranges and $2$ bananas. If the person eats one piece of their fruit per day for a week and the pieces of fruit within each category are indistinguishable, in how many orders can the fruit be eaten? One such order is $AAAOOBB.$ | There are a total of $7$ fruits, so there are $7!$ ways to arrange them. However, since the fruits in each category are indistinguishable, we must divide out the repeats: $$\frac{7!}{3!2!2!}=7\times6\times5=\boxed{210}.$$ | Counting & Probability | Level 3 | counting_and_probability |
Compute $\dbinom{7}{4}$. | $\dbinom{7}{4} = \dfrac{7!}{4!3!}=\dfrac{7\times 6\times 5\times 4}{4\times 3\times 2\times 1}=\boxed{35}.$ | Counting & Probability | Level 2 | counting_and_probability |
Determine the number of ways to arrange the letters of the word TOOT. | First we count the arrangements if all the letters are unique, which is $4!$. Then since the T's and the O's are not unique, we divide by $2!$ twice for the arrangements of T's and the arrangement of O's, for an answer of $\dfrac{4!}{2! \times 2!} = \boxed{6}$. | Counting & Probability | Level 2 | counting_and_probability |
A play has two different male roles, two different female roles and two different roles that can be either gender. Only a man can be assigned to a male role, and only a woman can be assigned to a female role. If five men and six women audition, in how many ways can the six roles be assigned? | We start with the most restrictive conditions: the roles that are not open to both men and women. First, we will fill the two male roles. There are 5 men, so there are $5 \cdot 4= 20$ ways to assign 2 of the 5 men to the distinct male roles. Doing the same for the female roles, there are $6 \cdot 5 = 30$ ways. Finally... | Counting & Probability | Level 4 | counting_and_probability |
There are five unmarked envelopes on a table, each with a letter for a different person. If the mail is randomly distributed to these five people, with each person getting one letter, what is the probability that exactly four people get the right letter? | If 4 people get the right letter, then there is only one letter left, and only one person left. So, the last person will get the right letter, too. Therefore, it is impossible for exactly four people to get the right letter. So, the probability is $\boxed{0}$. | Counting & Probability | Level 4 | counting_and_probability |
A palindrome is a number that reads the same forwards and backwards, such as 3003. How many positive four-digit integers are palindromes? | Constructing palindromes requires that we choose the thousands digit (which defines the units digit) and the hundreds digit (which defines the tens digit). Since there are 9 choices for the thousands digit, and 10 choices for the hundreds digit, creating $9 \cdot 10 = \boxed{90}$ palindromes. | Counting & Probability | Level 2 | counting_and_probability |
In the row of Pascal's triangle that starts with 1 and then 12, what is the fourth number? | The row that starts 1, 12 is the row $\binom{12}{0}, \binom{12}{1}, \binom{12}{2},\binom{12}{3}$, so the fourth number in this row is $\binom{12}{3} = \frac{12\cdot 11\cdot 10}{3\cdot 2 \cdot 1} = \boxed{220}$. | Counting & Probability | Level 3 | counting_and_probability |
The product of the digits of 3214 is 24. How many distinct four-digit positive integers are such that the product of their digits equals 12? | We first have to figure out the different groups of 4 one-digit numbers whose product is 12. We obviously can't use 12 as one of the numbers, nor can we use 9, 8, or 7 (none divides 12). We can use 6, in which case one of the other numbers is 2 and the other two are 1s. So, we can have the number 6211, or any number... | Counting & Probability | Level 3 | counting_and_probability |
A club has 20 members, 10 boys and 10 girls. In how many ways can a president and a vice-president be chosen if they must be of the same gender? Assume no one can hold both offices. | The president can be any one of the 20 members, and the vice-president can be any one of the 9 remaining members of the same sex. The answer is $20\times 9=\boxed{180}$. | Counting & Probability | Level 2 | counting_and_probability |
How many integers are between $(10.2)^3$ and $(10.3)^3$? | We want to count the integers from the first one greater than $10.2^3$ to the last one less than $10.3^3$. We note that $(a+b)^3=a^3+3a^2b+3ab^2+b^3$ for any $a,b$. Using this expansion for $(10+.2)^3$ and $(10+.3)^3$, we have: \begin{align*}
(10+.2)^3&=10^3+300\cdot .2+30\cdot .04+.008\\
(10+.3)^3&=10^3+300\cdot .3+30... | Counting & Probability | Level 4 | counting_and_probability |
A fair coin is flipped 10 times. What is the probability that heads are obtained in at least 8 of the 10 flips? | There are $2^{10} = 1024$ possible outcomes of the 10 coin flips. The number of ways to get 8, 9, or 10 heads is $\binom{10}{8}+\binom{10}{9}+\binom{10}{10}=45+10+1=56$. So the probability is $\dfrac{56}{1024}=\boxed{\dfrac{7}{128}}$. | Counting & Probability | Level 4 | counting_and_probability |
The digits $1,2,3,4$ and $5$ can be arranged to form many different $5$-digit positive integers with five distinct digits. In how many such integers is the digit $1$ to the left of the digit $2$? (The digits 1 and 2 do not have to be next to each other.) | For the first digit, we have 5 choices, then we have 4 choices left for the second digit, then 3 choices for the third digit, etc. So there are $5!=120$ arrangements of the digits. Notice that for each arrangement with 1 to the left of 2, we can reverse the arrangement so that 2 is to the left of 1. For instance, flipp... | Counting & Probability | Level 4 | counting_and_probability |
If the six digits 1, 2, 3, 5, 5 and 8 are randomly arranged into a six-digit positive integer, what is the probability that the integer is divisible by 15? Express your answer as a common fraction. | By the fact that the sum of the digits is a multiple of 3, any arrangement of the digits will be a multiple of 3. To be a multiple of 5, the number must end in one of the fives, which will happen with probability $\frac{2}{6} = \frac{1}{3}$. Since any number that is a multiple of 3 and a multiple of 5 is a multiple of ... | Counting & Probability | Level 4 | counting_and_probability |
Given that $8!=40320$, what is the value of $8!\div3!$? | Since $3!=6$, we need to divide 40320 by 6, which is $40320/6=\frac{36000+4200+120}{6}=6000+700+20=\boxed{6720}$. | Counting & Probability | Level 1 | counting_and_probability |
Each of five, standard, six-sided dice is rolled once. What is the probability that there is at least one pair but not a three-of-a-kind (that is, there are two dice showing the same value, but no three dice show the same value)? | There are a total of $6^5=7776$ possible sets of dice rolls. To get a pair without a three-of-a-kind, we can either have one pair and the other three dice all showing different numbers, or we have two pairs and the fifth die showing something different.
In the first case, there are $6$ ways to pick which number makes ... | Counting & Probability | Level 5 | counting_and_probability |
No two students in a 26-student mathematics class have the same two initials. Each student's first name and last name begin with the same letter. If the letter ``Y'' is considered a vowel, what is the probability of randomly picking a student whose initials are vowels? Express your answer as a common fraction. | The students' initials are AA, BB, CC, $\cdots$, ZZ, representing all 26 letters. The vowels are A, E, I, O, U, and Y, which are 6 letters out of the possible 26. So the probability of picking a student whose initials are vowels is $\frac{6}{26}=\boxed{\frac{3}{13}}$. | Counting & Probability | Level 2 | counting_and_probability |
A person has 6 daughters and no sons. Some of their daughters have 6 daughters, and the rest have none. The person has a total of 30 daughters and granddaughters, and no great-granddaughters. How many of the person's daughters and granddaughters have no daughters? | Bertha has $30 - 6 = 24$ granddaughters, none of whom have any daughters. The granddaughters are the children of $24/6 = 4$ of Bertha's daughters, so the number of women having no daughters is $30 - 4 = \boxed{26}$. | Counting & Probability | Level 3 | counting_and_probability |
How many positive multiples of $7$ that are less than $1000$ end with the digit $3$? | Listing the first 10 multiples of 7, which are 7, 14, 21, 28, 35, 42, 49, 56, 63, 70, we see that the only positive multiples of 7 that end in 3 are those that are the product of 7 and a number that ends in 9. Therefore, the positive multiples of 7 that are less than 1000 and end with the digit 3 are $7\cdot 9 = 63$, ... | Counting & Probability | Level 3 | counting_and_probability |
In how many ways can a person choose 3 captains from a team of 11 people? | Since the order that we choose the captains in doesn't matter, we can choose 3 of them out of 11 players in $\binom{11}{3}=\boxed{165}$ ways. | Counting & Probability | Level 2 | counting_and_probability |
A person uses a spinner to select a number from 1 through 5, each with equal probability. Another person uses a different spinner to select a number from 1 through 10, each with equal probability. What is the probability that the product of the second person's number and the first person's number is less than 30? Expre... | A little casework seems like the simplest approach. First, if Paco spins a 1 or 2, it does not matter what Manu spins; the product is less than 30 regardless. If Paco spins a 3, the product will be 30 or greater only if Manu spins a 10, and both of these will be true with probability $\frac{1}{5} \cdot \frac{1}{10} = \... | Counting & Probability | Level 3 | counting_and_probability |
How many positive three-digit integers less than 500 have at least two digits that are the same? | Case 1: The last two digits of our integer are equal. There are 10 possibilities for these last two digits and 4 choices for the hundred's digit, of a total of 40 possibilities. (Note that this case includes 111, 222, 333, and 444.)
Case 2: The first two digits are equal and the third is different. This occurs in $... | Counting & Probability | Level 5 | counting_and_probability |
If two distinct numbers are selected at random from the first seven prime numbers, what is the probability that their sum is an even number? Express your answer as a common fraction. | The only way for the sum to not be even is if one of the primes chosen is 2. There are six pairs where one of the primes is 2, and there are $\binom{7}{2}=21$ total possible pairs, so the probability that the sum is NOT even is $\frac{6}{21}=\frac{2}{7}$. Therefore, the probability that the sum IS even is $1-\frac{2}{7... | Counting & Probability | Level 3 | counting_and_probability |
A person has forgotten their locker combination. It is a sequence of three numbers, each in the range from 1 to 30, inclusive. They know that the first number is odd, the second number is even, and the third number is a multiple of 3. How many combinations could possibly be theirs? | There are 15 odd numbers, 15 even numbers, and 10 multiples of 3, making for $15^2\cdot 10 = 225 \cdot 10 = \boxed{2250}$ combinations total. | Counting & Probability | Level 2 | counting_and_probability |
Three friends have a total of 6 identical pencils, and each one has at least one pencil. In how many ways can this happen? | The largest number of pencils that any friend can have is four. There are 3 ways that this can happen: $(4,1,1)$, $(1,4,1)$ and $(1,1,4)$. There are 6 ways one person can have 3 pencils: $(3,2,1)$, $(3,1,2)$, $(2,3,1)$, $(2,1,3)$, $(1,2,3)$ and $(1,3,2)$. There is only one way all three can have two pencils each: $(2,2... | Counting & Probability | Level 4 | counting_and_probability |
If a person flips 8 coins, what is the probability that they get more heads than tails? | We could solve this problem using casework, but using a little bit of symmetry and complementary probability gives us a more elegant solution. Since each coin flips heads and tails with equal probability, by the principle of symmetry the probability of getting more heads than tails is the equal to the probability of ge... | Counting & Probability | Level 4 | counting_and_probability |
A person rolls four fair 20-sided dice, and each of the dice has faces numbered from 1 to 20. What is the probability that exactly two of the dice show an even number? | There is a $\frac{1}{2}$ probability that a 20-sided die will show an even number and a $\frac{1}{2}$ probability that it will show an odd number. We can choose which dice will show the even numbers in $\binom{4}{2}=6$ ways. For each way, there is a $\left( \frac{1}{2} \right) ^4=\frac{1}{16}$ probability that the chos... | Counting & Probability | Level 4 | counting_and_probability |
How many three-digit whole numbers have no 7's and no 9's as digits? | In order to be a three-digit number, the first digit cannot be 0, so there are normally 9 choices for the hundreds digit. But it also cannot be 7 or 9, so there are actually only 7 choices for the hundreds digit, and then 8 digits each for the units and tens digits. So, there are $7 \cdot 8 \cdot 8 = \boxed{448}$ such ... | Counting & Probability | Level 3 | counting_and_probability |
What is the probability that when we roll four fair 6-sided dice, they won't all show the same number? | Counting the number of outcomes in which four 6-sided dice don't all show the same number would require some pretty delicate casework. However, counting all the outcomes in which four 6-sided dice do all show the same number is very easy: there are only 6 ways this can happen, namely all ones, all twos, all threes, al... | Counting & Probability | Level 4 | counting_and_probability |
The set $S = \{1, 2, 3, \ldots , 49, 50\}$ contains the first $50$ positive integers. After the multiples of 2 and the multiples of 3 are removed, how many integers remain in the set $S$? | The set $S$ contains $25$ multiples of 2 (that is, even numbers). When these are removed, the set $S$ is left with only the odd integers from 1 to 49. At this point, there are $50-25=25$ integers in $S$. We still need to remove the multiples of 3 from $S$.
Since $S$ only contains odd integers after the multiples of 2... | Counting & Probability | Level 2 | counting_and_probability |
Two distinct positive integers from 1 to 50 inclusive are chosen. Let the sum of the integers equal $S$ and the product equal $P$. What is the probability that $P+S$ is one less than a multiple of 5? | There are a total of $\binom{50}{2}=1225$ ways to choose the two positive integers. Call these integers $a$ and $b$. The problem asks what the probability is that: $$ab+a+b=n-1$$where $n$ is a multiple of 5. We can add one to each side of this equation and factor: $$ab+a+b+1=(a+1)(b+1)=n$$Now, we need to count the n... | Counting & Probability | Level 5 | counting_and_probability |
A person is entering a wildlife photo contest, and wishes to arrange seven snow leopards of different heights in a row. If the shortest two leopards have inferiority complexes and demand to be placed at the ends of the row, how many ways can the leopards be lined up? | There are two ways to arrange the shortest two leopards. For the five remaining leopards, there are $5!$ ways to arrange them.
Therefore, the answer is $2\times5!=\boxed{240\text{ ways.}}$ | Counting & Probability | Level 2 | counting_and_probability |
A very bizarre weighted coin comes up heads with probability $\frac12$, tails with probability $\frac13$, and rests on its edge with probability $\frac16$. If it comes up heads, a person wins 1 dollar. If it comes up tails, the person wins 3 dollars. But if it lands on its edge, the person loses 5 dollars. What is ... | The expected value is $E = \left(\dfrac{1}{2}\times\$1\right) + \left(\dfrac{1}{3}\times\$3\right) + \left(\dfrac{1}{6}\times(-\$5)\right) = \$\dfrac{4}{6} =\boxed{\$\dfrac23 \approx \$0.67}$. | Counting & Probability | Level 3 | counting_and_probability |
A person has entered a buffet line in which they choose one kind of meat, two different vegetables and one dessert. If the order of food items is not important, how many different meals might they choose? $\bullet$ Meat: beef, chicken, pork $\bullet$ Vegetables: baked beans, corn, potatoes, tomatoes $\bullet$ Desse... | There are $3$ choices for the meat and $4$ for dessert.
There are $\binom{4}{2} = 6$ ways to choose $2$ of the $4$ vegetables, since we don't care about the order in which the vegetables are chosen.
The answer therefore is $3\cdot 4\cdot 6=\boxed{72}.$ | Counting & Probability | Level 3 | counting_and_probability |
A person has 6 children. Assuming that the gender of each child is determined independently and with equal likelihood of male and female, what is the probability that they have more sons than daughters or more daughters than sons? | We could do this with a bit of casework, but that gets boring after a while. Instead, we can use complementary probability. Since each child can be male or female with equal likelihood, there are $2^6=64$ possible ways in which the genders of the children can be determined. The only way in which Mr. Jones won't have mo... | Counting & Probability | Level 4 | counting_and_probability |
When flipping a fair coin, what is the probability that the first two flips are both heads? Express your answer as a common fraction. | The probability that any single flip comes up heads is $1/2$. Since the flips are independent, the probability that the first two flips are both heads is $1/2\cdot1/2=\boxed{\frac{1}{4}}$. | Counting & Probability | Level 1 | counting_and_probability |
A Senate committee has 8 Republicans and 6 Democrats. In how many ways can a subcommittee with 3 Republicans and 2 Democrats be formed? | There are 8 Republicans and 3 spots for them, so there are $\binom{8}{3} = 56$ ways to choose the Republicans. There are 6 Democrats and 2 spots for them, so there are $\binom{6}{2} = 15$ ways to choose the Democrats. So there are $56 \times 15 = \boxed{840}$ ways to choose the subcommittee. | Counting & Probability | Level 2 | counting_and_probability |
We call a number a mountain number if its middle digit is larger than any other digit. For example, 284 is a mountain number. How many 3-digit mountain numbers are there? | We will break this into three cases.
Case 1: numbers of the form $xyx$ ($x \ne 0$).
Any pair of nonzero digits has a corresponding palindrome ($xyx$) mountain number, so the number of these is $\binom{9}{2} = 36$.
Case 2: numbers of the form $xyz$ ($z \ne 0, x \ne z$).
Any group of three nonzero digits ($y > x > z ... | Counting & Probability | Level 5 | counting_and_probability |
How many combinations of pennies (1 cent), nickels (5 cents) and/or dimes (10 cents) are there with a total value of 25 cents? | First, we count the number of combinations that include pennies: we could have all pennies, all pennies and one nickel, all pennies and two nickels, all pennies and three nickels, all pennies and four nickels, all pennies and one dime, all pennies and two dimes, all pennies and one dime and one nickel, all pennies and ... | Counting & Probability | Level 3 | counting_and_probability |
How many ways are there to put 5 balls in 3 boxes if the balls are not distinguishable but the boxes are? | Since the balls are indistinguishable, we must only count the number of balls in the different boxes.
There are $3$ ways to arrange the balls as $(5,0,0)$ (specifically, box 1 can have 5, box 2 can have 5, box 3 can have 5).
There are $3! = 6$ to arrange $(4,1,0)$ and $3! = 6$ ways to arrange $(3,2,0)$; in each case,... | Counting & Probability | Level 4 | counting_and_probability |
The planning committee at school has 10 members. Exactly four of these members are teachers. A four-person subcommittee with at least one member who is a teacher must be formed from the members of the planning committee. How many distinct subcommittees are possible? | Because there are 4 teachers on the committee, there are 6 non-teachers. Now, in total, we can form ${10 \choose 4} = 210$ subcomittees. The number of subcomittees with 0 teachers is the number of subcommittees formed by the 6 nonteachers, totaling ${6 \choose 4} = 15$. So, the number of subcomittees with at least 1 te... | Counting & Probability | Level 4 | counting_and_probability |
A business is eager to advertise how many flavors it has. But it really only has three basic flavors: chocolate, vanilla, and strawberry. However, they can make "new" flavors by taking four scoops of ice cream of those basic flavors and blending them together. Different proportions of the basic flavors give different n... | We can think of the $3$ basic flavors as $3$ distinguishable boxes, and the $4$ scoops as $4$ indistinguishable balls. For every ball we put in the chocolate box, for instance, we put a scoop of chocolate ice cream in the blending machine. In this way we can correlate each new flavor with an arrangement of balls in box... | Counting & Probability | Level 5 | counting_and_probability |
2 distinct vertices of an octagon are chosen at random. What is the probability that they are adjacent? | After we choose the first vertex, there are 7 ways to choose the second. There are only 2 of these that are adjacent to the first vertex, so the probability the two vertices are adjacent is $\boxed{\frac{2}{7}}$. | Counting & Probability | Level 3 | counting_and_probability |
How many 4-letter words with at least one consonant can be constructed from the letters $A$, $B$, $C$, $D$, and $E$? (Note that $B$, $C$, and $D$ are consonants, any word is valid, not just English language words, and letters may be used more than once.) | First we count the number of all 4-letter words with no restrictions on the word. Then we count the number of 4-letter words with no consonants. We then subtract to get the answer.
Each letter of a word must be one of $A$, $B$, $C$, $D$, or $E$, so the number of 4-letter words with no restrictions on the word is $5\ti... | Counting & Probability | Level 3 | counting_and_probability |
How many ways can we put 3 math books and 5 English books on a shelf if all the math books must stay together and all the English books must also stay together? (The math books are all different and so are the English books.) | First we arrange the 2 groups of books; there are $2!$ ways in which we can do this. Then we can arrange the 3 math books in $3!$ ways and the 5 English books in $5!$ ways. Therefore, there are $2! \times 3! \times 5!=\boxed{1440}$ ways to arrange the books. | Counting & Probability | Level 3 | counting_and_probability |
The chart below gives the air distance in miles between selected world cities. If two different cities from the chart are chosen at random, what is the probability that the distance between them is less than $7000$ miles? Express your answer as a common fraction. \begin{tabular}{|c|c|c|c|c|} \hline & Bangkok & Cape Tow... | Although there are 12 entries on the table, there are only 6 different ``pairs'' of cities. Of the 6 pairs, 4 of them are less than 7,000 miles apart, yielding a $\boxed{\frac{2}{3}}$ probability that you select a pair less than 7,000 miles apart. | Counting & Probability | Level 3 | counting_and_probability |
Compute $\dbinom{15}{3}$. | \begin{align*}
\dbinom{15}{3} &= \dfrac{15!}{12!3!} \\
&= \dfrac{15\times 14\times 13}{3\times 2\times 1} \\
&= \dfrac{15}{3} \times \dfrac{14}{2} \times \dfrac{13}{1} \\
&= 5\times 7\times 13 \\
&= \boxed{455}.
\end{align*} | Counting & Probability | Level 1 | counting_and_probability |
How many ways are there to arrange the letters of the word $\text{BA}_1\text{N}_1\text{A}_2\text{N}_2\text{A}_3$, in which the three A's and the two N's are considered different? | This is counting the number of ways that six distinct objects can be put in order, so there are $6! = \boxed{720}$ different arrangements. | Counting & Probability | Level 2 | counting_and_probability |
On the refrigerator, MATHEMATICS is spelled out with $11$ magnets, one letter per magnet. Two vowels and four consonants fall off and are put away in a bag. If the T's, M's, and A's are indistinguishable, how many distinct possible collections of letters could be put in the bag? | We count the number of ways to choose the vowels and the consonants separately. There are four vowels, of which two are As. If there are no As, then we must choose both the remaining vowels, so there is $1$ choice; if there is one A, then we can choose the remaining vowel in $2$ ways; and if there are two As, then ther... | Counting & Probability | Level 5 | counting_and_probability |
A bowl contains 10 jellybeans (four red, one blue and five white). If you pick three jellybeans from the bowl at random and without replacement, what is the probability that exactly two will be red? Express your answer as a common fraction. | First, we consider how many total sets of three jellybeans we can select, which is very simply ${10 \choose 3} = 120$, if we treat all 10 jellybeans as distinct. Now, if we have exactly 2 red jellybeans, there are ${4 \choose 2} = 6$ pairs of red jellybeans, and $5+1 = 6$ choices for the third non-red jellybean. So, th... | Counting & Probability | Level 5 | counting_and_probability |
The local weatherman says there is a 75 percent chance of rain for each day of the Labor Day four-day weekend. If it doesn't rain, then the weather will be sunny. Two people want it to be sunny one of those days for an event, but if it's sunny for more than one day they won't know what to do with themselves. What is th... | There are $\binom{4}{3}=4$ ways to choose which three of the four days it will be rainy so that the other day it will be sunny. For any of those 4 choices, there is a $\left( \frac{3}{4} \right) ^3 \left( \frac{1}{4} \right) ^1 = \frac{27}{256}$ chance for that choice to happen, because there is a $\frac{3}{4}$ chance ... | Counting & Probability | Level 4 | counting_and_probability |
A bag has 4 red marbles, 5 white marbles, and 6 blue marbles. Three marbles are drawn from the bag (without replacement). What is the probability that they are all the same color? | We can have all red, all white, or all blue. Thus the answer is \begin{align*}
&P(\text{all red}) + P(\text{all white}) + P(\text{all blue}) \\
&\qquad = \left(\frac{4}{15}\times\frac{3}{14}\times\frac{2}{13}\right) +\left(\frac{5}{15}\times\frac{4}{14}\times\frac{3}{13}\right) \\
&\qquad\qquad+\left(\frac{6}{15}\time... | Counting & Probability | Level 4 | counting_and_probability |
How many three-digit numbers are composed of three distinct digits such that one digit is the average of the other two? | The set of the three digits of such a number can be arranged to form an increasing arithmetic sequence. There are 8 possible sequences with a common difference of 1, since the first term can be any of the digits 0 through 7. There are 6 possible sequences with a common difference of 2, 4 with a common difference of 3... | Counting & Probability | Level 5 | counting_and_probability |
Compute $\dbinom{25}{2}$. | \begin{align*}
\dbinom{25}{2} &= \dfrac{25!}{23!2!} \\
&= \dfrac{25\times 24}{2\times 1} \\
&= 25 \times \dfrac{24}{2} \\
&= 25 \times 12 \\
&= \boxed{300}.
\end{align*} | Counting & Probability | Level 1 | counting_and_probability |
How many three-digit whole numbers have at least one 7 or at least one 9 as digits? | We know that there are a total of $999 - 100 + 1 = 900$ three digit numbers. If we try to count how many have at least one 7 or one 9 as digits directly, we'll run into a load of casework. So instead, we proceed by counting the complement, the number of three digit numbers having no 7s or 9s as digits. We can choose... | Counting & Probability | Level 3 | counting_and_probability |
One team plays another team 4 times. The first team is the much better team, and are $75\%$ likely to win any given game. What is the probability that the first team will win all 4 games? Express your answer as a common fraction. | Each of the 4 games is independent of the others, and in each game, the Grunters have probability $\frac34$ of winning. Therefore, to get the probability that the Grunters will win all 4 games, we multiply the probabilities that the Grunters win each individual game. This gives: \begin{align*}
&P(\text{Grunters win ... | Counting & Probability | Level 2 | counting_and_probability |
In how many ways can the digits of $45,\!520$ be arranged to form a 5-digit number? (Remember, numbers cannot begin with 0.) | First we place the $0$, which we only have four options for (everywhere but the first digit). Then we have 4 remaining places to put the last 4 digits, two of which are not unique (the fives), so there are $\dfrac{4!}{2!}$ options for arranging the other 4 digits. This gives a final answer of $\dfrac{4 \times 4!}{2!}... | Counting & Probability | Level 4 | counting_and_probability |
On each of the first three days of January, there is a $\frac{1}{3}$ chance that it will snow where someone lives. On each of the next four days, there is a $\frac{1}{4}$ chance that it will snow. What is the probability that it snows at least once during the first week of January? | On each of the first three days, there is a $\frac{2}{3}$ chance that there will be no snow. On each of the next four days, there is a $\frac{3}{4}$ chance that there will be no snow. The chance that there will be no snow at all during the first week of January is $\left( \frac{2}{3} \right) ^3 \left( \frac{3}{4} \righ... | Counting & Probability | Level 4 | counting_and_probability |
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