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vKUBJgogUkTqMHWSIq45x | maths | permutations-and-combinations | divisibility-of-numbers | The number of numbers between 2,000 and 5,000 that can be formed with the digits 0, 1, 2, 3, 4 (repetition of digits is not allowed) and are multiple of 3 is : | [{"identifier": "A", "content": "24"}, {"identifier": "B", "content": "30"}, {"identifier": "C", "content": "36"}, {"identifier": "D", "content": "48"}] | ["B"] | null | Here number should be divisible by 3, that means sum of numbers should be divisible by 3.
<br><br>Possible 4 digits among 0, 1, 2, 3, 4 which are divisible by 3 are
<br><br>(1)$$\,\,\,\,$$ (0, 2, 3, 4) Sum of digits = 0 + 2 + 3 +4 = 9 (divisible by 3)
<br><br>(2) $$\,\,\,\,$$ (0, 1, 2, 3) Sum of digits = 0 + 1 + 2 + 3... | mcq | jee-main-2018-online-16th-april-morning-slot | 7,454 |
OjrcXgele0CMwAVPCy1kluvyg3b | maths | permutations-and-combinations | divisibility-of-numbers | A natural number has prime factorization given by n = 2<sup>x</sup>3<sup>y</sup>5<sup>z</sup>, where y and z are such <br/>that y + z = 5 and y<sup>$$-$$1</sup> + z<sup>$$-$$1</sup> = $${5 \over 6}$$, y > z. Then the number of odd divisions of n, including 1, is : | [{"identifier": "A", "content": "11"}, {"identifier": "B", "content": "6"}, {"identifier": "C", "content": "12"}, {"identifier": "D", "content": "6x"}] | ["C"] | null | y + z = 5 ....... (1)<br><br>$${1 \over y} + {1 \over z} = {5 \over 6}$$<br><br>$$ \Rightarrow {{y + z} \over {yz}} = {5 \over 6}$$<br><br>$$ \Rightarrow {5 \over {yz}} = {5 \over 6}$$<br><br>$$ \Rightarrow $$ yz = 6<br><br>Also, (y $$-$$ z)<sup>2</sup> = (y + z)<sup>2</sup> $$-$$ 4yz<br><br>$$ \Rightarrow $$ (y $$-$$ ... | mcq | jee-main-2021-online-26th-february-evening-slot | 7,455 |
1krygm1ew | maths | permutations-and-combinations | divisibility-of-numbers | Let n be a non-negative integer. Then the number of divisors of the form "4n + 1" of the number (10)<sup>10</sup> . (11)<sup>11</sup> . (13)<sup>13</sup> is equal to __________. | [] | null | 924 | N = 2<sup>10</sup> $$\times$$ 5<sup>10</sup> $$\times$$ 11<sup>11</sup> $$\times$$ 13<sup>13</sup><br><br>Now, power of 2 must be zero,<br><br>power of 5 can be anything,<br><br>power of 13 can be anything<br><br>But, power of 11 should be even.<br><br>So, required number of divisors is <br><br>1 $$\times$$ 11 $$\times... | integer | jee-main-2021-online-27th-july-evening-shift | 7,456 |
1ldsg0itd | maths | permutations-and-combinations | divisibility-of-numbers | <p>The total number of 4-digit numbers whose greatest common divisor with 54 is 2, is __________.</p> | [] | null | 3000 | <p>$$\gcd (a,54) = 2$$ when a is a 4 digit no.</p>
<p>And $$54 = 3 \times 3 \times 3 \times 2$$</p>
<p>So, $$a=$$ all even no. of 4 digits $$-$$ Even multiple of 3 (4 digits)</p>
<p>$$ = 4500 - 1500$$</p>
<p>$$ = 3000$$</p> | integer | jee-main-2023-online-29th-january-evening-shift | 7,458 |
uKpYZD2xI25YxVqSvGYd0 | maths | permutations-and-combinations | factorial | The sum $$\sum\limits_{r = 1}^{10} {\left( {{r^2} + 1} \right) \times \left( {r!} \right)} $$ is equal to : | [{"identifier": "A", "content": "(11)!"}, {"identifier": "B", "content": "10 $$ \\times $$ (11!)"}, {"identifier": "C", "content": "101 $$ \\times $$ (10!)"}, {"identifier": "D", "content": "11 $$ \\times $$ (11!)\n"}] | ["B"] | null | $$\sum\limits_{r = 1}^{10} {\left( {{r^2} + 1} \right)} \times r!$$
<br><br>$$ = \sum\limits_{r = 1}^{10} {\left[ {{{\left( {r + 1} \right)}^2} - 2r} \right]\,.\,r!} $$
<br><br>$$ = \sum\limits_{r = 1}^{10} {\left[ {{{\left( {r + 1} \right)}^2}\,.\,r!\,\, - \,\,2r\,.\,r!} \right]} $$
<br><br>$$ = \sum\limits_{r = 1}^{... | mcq | jee-main-2016-online-10th-april-morning-slot | 7,459 |
jaoe38c1lscnqca5 | maths | permutations-and-combinations | factorial | <p>Let $$\alpha=\frac{(4 !) !}{(4 !)^{3 !}}$$ and $$\beta=\frac{(5 !) !}{(5 !)^{4 !}}$$. Then :</p> | [{"identifier": "A", "content": "$$\\alpha \\in \\mathbf{N}$$ and $$\\beta \\in \\mathbf{N}$$\n"}, {"identifier": "B", "content": "$$\\alpha \\in \\mathbf{N}$$ and $$\\beta \\notin \\mathbf{N}$$\n"}, {"identifier": "C", "content": "$$\\alpha \\notin \\mathbf{N}$$ and $$\\beta \\in \\mathbf{N}$$\n"}, {"identifier": "D",... | ["A"] | null | <p>$$\begin{aligned}
& \alpha=\frac{(4 !) !}{(4 !)^{3 !}}, \beta=\frac{(5 !) !}{(5 !)^{4 !}} \\
& \alpha=\frac{(24) !}{(4 !)^6}, \beta=\frac{(120) !}{(5 !)^{24}}
\end{aligned}$$</p>
<p>Let 24 distinct objects are divided into 6 groups of 4 objects in each group.</p>
<p>No. of ways of formation of group $$=\frac{24 !}{(... | mcq | jee-main-2024-online-27th-january-evening-shift | 7,460 |
jaoe38c1lsd4qjvk | maths | permutations-and-combinations | factorial | <p>If for some $$m, n ;{ }^6 C_m+2\left({ }^6 C_{m+1}\right)+{ }^6 C_{m+2}>{ }^8 C_3$$ and $${ }^{n-1} P_3:{ }^n P_4=1: 8$$, then $${ }^n P_{m+1}+{ }^{\mathrm{n}+1} C_m$$ is equal to</p> | [{"identifier": "A", "content": "380"}, {"identifier": "B", "content": "376"}, {"identifier": "C", "content": "372"}, {"identifier": "D", "content": "384"}] | ["C"] | null | <p>$$\begin{aligned}
& { }^6 \mathrm{C}_{\mathrm{m}}+2\left({ }^6 \mathrm{C}_{\mathrm{m}+1}\right)+{ }^6 \mathrm{C}_{\mathrm{m}+2}>{ }^8 \mathrm{C}_3 \\
& { }^7 \mathrm{C}_{\mathrm{m}+1}+{ }^7 \mathrm{C}_{\mathrm{m}+2}>{ }^8 \mathrm{C}_3 \\
& { }^8 \mathrm{C}_{\mathrm{m}+2}>{ }^8 \mathrm{C}_3 \\
& \therefore \mathrm{m}... | mcq | jee-main-2024-online-31st-january-evening-shift | 7,461 |
txW7MxHrrC595ct0 | maths | permutations-and-combinations | number-of-combinations | If $${}^n{C_r}$$ denotes the number of combination of n things taken r at a time, then the expression $$\,{}^n{C_{r + 1}} + {}^n{C_{r - 1}} + 2\, \times \,{}^n{C_r}$$ equals | [{"identifier": "A", "content": "$$\\,{}^{n + 1}{C_{r + 1}}$$ "}, {"identifier": "B", "content": "$${}^{n + 2}{C_r}$$ "}, {"identifier": "C", "content": "$${}^{n + 2}{C_{r + 1}}$$ "}, {"identifier": "D", "content": "$$\\,{}^{n + 1}{C_r}$$ "}] | ["C"] | null | Arrange it this way,
<br><br>$$^n{C_{r + 1}} + 2.{}^n{C_r} + {}^n{C_{r - 1}}$$
<br><br>$$ = {}^n{C_{r + 1}} + {}^n{C_r} + {}^n{C_r} + {}^n{C_{r - 1}}$$
<br><br>$$\left[ \, \right.$$ Now use the rule,
<br><br> $$\left. {{}^n{C_r} + {}^n{C_{r - 1}} = {}^{n + 1}{C_r}} \right... | mcq | aieee-2003 | 7,462 |
cxhMyG1Oo7EJnOxq | maths | permutations-and-combinations | number-of-combinations | The number of ways of distributing 8 identical balls in 3 distinct boxes so that none of the boxes is empty is | [{"identifier": "A", "content": "$${}^8{C_3}$$ "}, {"identifier": "B", "content": "21"}, {"identifier": "C", "content": "$${3^8}$$ "}, {"identifier": "D", "content": "5 "}] | ["B"] | null | To distribute n objects among p people where everyone should get atleast one object, then number of ways to distribute those n objects
<br><br>= $${}^{n - 1}{C_{p - 1}}$$
<br><br>For this question, n = 8 and p = 3
<br><br>$$ \therefore $$ Number of ways = $${}^{8 - 1}{C_{3 - 1}}$$ = $${}^7{C_2}$$ = 21 | mcq | aieee-2004 | 7,463 |
WAd9iWoQGJdqmxc0 | maths | permutations-and-combinations | number-of-combinations | At an election, a voter may vote for any number of candidates, not greater than the number to be elected. There are 10 candidates and 4 are of be selected, if a voter votes for at least one candidate, then the number of ways in which he can vote is | [{"identifier": "A", "content": "5040"}, {"identifier": "B", "content": "6210"}, {"identifier": "C", "content": "385"}, {"identifier": "D", "content": "1110"}] | ["C"] | null | A voter can give vote to either 1 candidate or 2 candidates or 3 candidates or 4 candidates.
<br><br>Case 1 : When he give vote to only 1 candidate then no ways = $${}^{10}{C_1}$$
<br><br>Case 2 : When he give vote to 2 candidates then no ways = $${}^{10}{C_2}$$
<br><br>Case 3 : When he give vote to 3 candidates then n... | mcq | aieee-2006 | 7,464 |
wZ6fkRvOqG6GRUaw | maths | permutations-and-combinations | number-of-combinations | The set S = {1, 2, 3, ........., 12} is to be partitioned into three sets A, B, C of equal size. Thus $$A \cup B \cup C = S,\,A \cap B = B \cap C = A \cap C = \phi $$. The number of ways to partition S is | [{"identifier": "A", "content": "$${{12!} \\over {{{(4!)}^3}}}\\,\\,$$ "}, {"identifier": "B", "content": "$${{12!} \\over {{{(4!)}^4}}}\\,\\,$$ "}, {"identifier": "C", "content": "$${{12!} \\over {3!\\,\\,{{(4!)}^3}}}$$ "}, {"identifier": "D", "content": "$${{12!} \\over {3!\\,\\,{{(4!)}^4}}}$$ "}] | ["A"] | null | <p>The total number of ways is</p>
<p>$${}^{12}{C_4} \times {}^{12 - 4}{C_4} \times {}^{12 - 4 - 4}{C_4} = {}^{12}{C_4} \times {}^8{C_4} \times {}^4{C_4} = {{12!} \over {{{(4!)}^3}}}$$</p> | mcq | aieee-2007 | 7,465 |
pW3FNSb5ohyzwnDR | maths | permutations-and-combinations | number-of-combinations | In a shop there are five types of ice-cream available. A child buys six ice-cream.
<br/> <b> Statement - 1: </b> The number of different ways the child can buy the six ice-cream is $${}^{10}{C_5}$$.
<br/> <b> Statement - 2: </b> The number of different ways the child can buy the six ice-cream is equal to the number of ... | [{"identifier": "A", "content": "Statement - 1 is false, Statement - 2 is true "}, {"identifier": "B", "content": "Statement - 1 is true, Statement - 2 is true, Statement - 2 is a correct explanation for Statement - 1"}, {"identifier": "C", "content": "Statement - 1 is true, Statement - 2 is true, Statement - 2 i... | ["A"] | null | <b>Note :</b> n items can be distribute among p persons are $${}^{n + p - 1}{C_{p - 1}}$$ ways.
<br><br>Here n = 6 ice-cream
<br><br>p = 5 types of ice-cream
<br><br>Each ice-cream belongs to one of the 5 ice-cream type. So chosen 6 ice-crean can be divide into 5 types of ice-cream.
<br><br>$$ \therefore $$ The number ... | mcq | aieee-2008 | 7,466 |
gu4xEM8HE3wz9OiX | maths | permutations-and-combinations | number-of-combinations | There are two urns. Urn A has 3 distinct red balls and urn B has 9 distinct blue balls. From each urn two balls are taken out at random and then transferred to the other. The number of ways in which this can be done is | [{"identifier": "A", "content": "36"}, {"identifier": "B", "content": "66"}, {"identifier": "C", "content": "108"}, {"identifier": "D", "content": "3"}] | ["C"] | null | <p>Thus number of ways $$ = ({}^3{C_2}) \times ({}^9{C_2}) = 3 \times {{9 \times 8} \over 2} = 108$$</p> | mcq | aieee-2010 | 7,467 |
hZ0WqGlvxb9cOBdE | maths | permutations-and-combinations | number-of-combinations | These are 10 points in a plane, out of these 6 are collinear, if N is the number of triangles formed by joining these points. then: | [{"identifier": "A", "content": "$$N \\le 100$$ "}, {"identifier": "B", "content": "$$100 < N \\le 140$$ "}, {"identifier": "C", "content": "$$140 < N \\le 190\\,$$ "}, {"identifier": "D", "content": "$$N > 190$$ "}] | ["A"] | null | <p>We need 3 points to create a triangle. With 10 points number of triangle possible $${}^{10}{C_3}$$</p>
<p>Here 6 points are on the same line so we can't make any triangle with those 6 points.</p>
<p> So subtract $${}^{6}{C_3}$$.</p>
<p>$$\therefore$$ $$N = {}^{10}{C_3} - {}^6{C_3}$$</p>
<p>$$ = {{10\,.\,9\,.\,8} \ov... | mcq | aieee-2011 | 7,468 |
lWJFreINJxGWYGy2 | maths | permutations-and-combinations | number-of-combinations | <br/> <b> Statement - 1: </b> The number of ways of distributing 10 identical balls in 4 distinct boxes such that no box is emply is $${}^9{C_3}$$.
<br/> <b> Statement - 2: </b> The number of ways of choosing any 3 places from 9 different places is $${}^9{C_3}$$. | [{"identifier": "A", "content": "Statement - 1 is true, Statement - 2 is true, Statement - 2 is not a correct explanation for Statement - 1."}, {"identifier": "B", "content": "Statement - 1 is true, Statement - 2 is false."}, {"identifier": "C", "content": "Statement - 1 is false, Statement - 2 is true."}, {"identi... | ["A"] | null | <p>Let XA<sub></sub>, X<sub>B</sub>, X<sub>C</sub> and X<sub>D</sub> represent number of balls present in box A, B, C and D respectively.</p>
<p>As no box can be empty so,</p>
<p>X<sub>A</sub> $$\ge$$ 1, X<sub>B</sub> $$\ge$$ 1, X<sub>C</sub> $$\ge$$ 1 and X<sub>D</sub> $$\ge$$ 1</p>
<p>$$\Rightarrow$$ X<sub>A</sub> $$... | mcq | aieee-2011 | 7,469 |
FrmiixXJ0JcQkToz | maths | permutations-and-combinations | number-of-combinations | Assuming the balls to be identical except for difference in colours, the number of ways in which one or more balls can be selected from 10 white, 9 green and 7 black balls is: | [{"identifier": "A", "content": "880"}, {"identifier": "B", "content": "629"}, {"identifier": "C", "content": "630"}, {"identifier": "D", "content": "879"}] | ["D"] | null | <p>For alike n objects, number of ways we can select zero or more objects = n + 1 and number of ways we can select at least one object = n</p>
<p>Given 10 identical white balls, 9 identical green balls and 7 identical black balls.</p>
<p>To find number of ways for selecting atleast one ball.</p>
<p>Number of ways to c... | mcq | aieee-2012 | 7,470 |
foj68NgT56Ij10bv | maths | permutations-and-combinations | number-of-combinations | Let $${T_n}$$ be the number of all possible triangles formed by joining vertices of an n-sided regular polygon. If $${T_{n + 1}} - {T_n}$$ = 10, then the value of n is : | [{"identifier": "A", "content": "7 "}, {"identifier": "B", "content": "5 "}, {"identifier": "C", "content": "10 "}, {"identifier": "D", "content": "8"}] | ["B"] | null | Number of possible triangle using n vertices = <sup>n</sup>C<sub>3</sub>
<br><br>$$ \therefore $$ T<sub>n</sub> = <sup>n</sup>C<sub>3</sub>
<br><br>then T<sub>n + 1</sub> = <sup>n + 1</sup>C<sub>3</sub>
<br><br>Given, $${T_{n + 1}} - {T_n}$$ = 10
<br><br>$$ \Rightarrow $$ <sup>n + 1</sup>C<sub>3</sub> - <sup>n</sup>C<s... | mcq | jee-main-2013-offline | 7,472 |
IYJ08gkoyHdlZ5bWWXi8X | maths | permutations-and-combinations | number-of-combinations | If $${{{}^{n + 2}C{}_6} \over {{}^{n - 2}{P_2}}}$$ = 11, then n satisfies the
equation : | [{"identifier": "A", "content": "n<sup>2</sup> + 3n \u2212 108 = 0"}, {"identifier": "B", "content": "n<sup>2</sup> + 5n \u2212 84 = 0\n"}, {"identifier": "C", "content": "n<sup>2</sup> + 2n \u2212 80 = 0"}, {"identifier": "D", "content": "n<sup>2</sup> + n \u2212 110 = 0"}] | ["A"] | null | $${{{}^{n + 2}{C_6}} \over {{}^{n - 2}{P_2}}} = 11$$
<br><br>$$ \Rightarrow $$ $${{\left( {n + 2} \right)!} \over {6!\,\left( {n - 4} \right)!}} = 11\,.\,{{\left( {n - 2} \right)!} \over {\left( {n - 4} \right)!}}$$
<br><br>$$ \Rightarrow $$ (n + 2)! = 11.6! (n $$... | mcq | jee-main-2016-online-10th-april-morning-slot | 7,474 |
hPoq3OSkej5kavXb | maths | permutations-and-combinations | number-of-combinations | A man X has 7 friends, 4 of them are ladies and 3 are men. His wife Y also has 7 friends, 3 of them are
ladies and 4 are men. Assume X and Y have no common friends. Then the total number of ways in which X
and Y together can throw a party inviting 3 ladies and 3 men, so that 3 friends of each of X and Y are in
this par... | [{"identifier": "A", "content": "468"}, {"identifier": "B", "content": "469"}, {"identifier": "C", "content": "484"}, {"identifier": "D", "content": "485"}] | ["D"] | null | <table class="tg">
<tbody><tr>
<th class="tg-0lax"></th>
<th class="tg-s6z2" colspan="2"><span style="font-weight:bold">X(7 Friends)</span></th>
<th class="tg-s6z2" colspan="2"><span style="font-weight:bold">Y(7 Friends)</span></th>
</tr>
<tr>
<td class="tg-0lax"></td>
<td class="tg-s6z2">4 La... | mcq | jee-main-2017-offline | 7,475 |
nopfCkMOXIUlIZhNggMvD | maths | permutations-and-combinations | number-of-combinations | If $$\sum\limits_{r = 0}^{25} {\left\{ {{}^{50}{C_r}.{}^{50 - r}{C_{25 - r}}} \right\} = K\left( {^{50}{C_{25}}} \right)} ,\,\,$$ then K is equal to : | [{"identifier": "A", "content": "2<sup>24</sup> "}, {"identifier": "B", "content": "2<sup>25</sup>$$-$$ 1"}, {"identifier": "C", "content": "2<sup>25</sup>"}, {"identifier": "D", "content": "(25)<sup>2</sup>"}] | ["C"] | null | $$\sum\limits_{r = 0}^{25} {^{50}} {C_r}.{}^{50 - r}{C_{25 - r}}$$
<br><br>$$ = \sum\limits_{r = 0}^{25} {{{50!} \over {r!\left( {50 - r} \right)!}}} \times {{\left( {50 - r} \right)!} \over {\left( {25} \right)!\left( {25 - r} \right)!}}$$
<br><br>$$ = \sum\limits_{r = 0}^{25} {{{50!} \over {25!25!}} \times {{25!} \o... | mcq | jee-main-2019-online-10th-january-evening-slot | 7,476 |
d9rYjwxlxiAXsKNTjN8AV | maths | permutations-and-combinations | number-of-combinations | Consider three boxes, each containing, 10 balls labelled 1, 2, … , 10. Suppose one ball is randomly drawn from each of the boxes. Denote by n<sub>i</sub>, the label of the ball drawn from the i<sup>th</sup> box, (i = 1, 2, 3). Then, the number of ways in which the balls can be chosen such that n<sub>1</sub> < n<sub>... | [{"identifier": "A", "content": "164"}, {"identifier": "B", "content": "240"}, {"identifier": "C", "content": "82"}, {"identifier": "D", "content": "120"}] | ["D"] | null | Number of ways = <sup>10</sup>C<sub>3</sub> = 120 | mcq | jee-main-2019-online-12th-january-morning-slot | 7,477 |
wyrLSbxVQxkAIJHo7FrTT | maths | permutations-and-combinations | number-of-combinations | There are m men and two women participating in a chess tournament. Each participant plays two games with every other participant. If the number of games played by the men between themselves exceeds the number of games played between the men and the women by 84, then the value of m is : | [{"identifier": "A", "content": "12"}, {"identifier": "B", "content": "9"}, {"identifier": "C", "content": "7"}, {"identifier": "D", "content": "11"}] | ["A"] | null | Let m-men, 2-women
<br><br><sup>m</sup>C<sub>2</sub> $$ \times $$ 2 = <sup>m</sup><sup></sup>C<sub>1</sub> <sup>2</sup>C<sub>1</sub> . 2 + 84
<br><br>m<sup>2</sup> $$-$$ 5m $$-$$ 84 = 0 $$ \Rightarrow $$ (m $$-$$ 12) (m + 7) = 0
<br><br>m = 12 | mcq | jee-main-2019-online-12th-january-evening-slot | 7,478 |
IiAhsoFZAqNRGsldWG3rsa0w2w9jx5e78h8 | maths | permutations-and-combinations | number-of-combinations | The number of ways of choosing 10 objects out of 31 objects of which 10 are identical and the remaining 21
are distinct, is : | [{"identifier": "A", "content": "2<sup>20</sup> - 1"}, {"identifier": "B", "content": "2<sup>20</sup>"}, {"identifier": "C", "content": "2<sup>20</sup> + 1"}, {"identifier": "D", "content": "2<sup>21</sup> "}] | ["B"] | null | <sup>21</sup>C<sub>0</sub> + <sup>21</sup>C<sub>1</sub> + <sup>21</sup>C<sub>2</sub> + ....... + <sup>21</sup>C<sub>10</sub> = $${{{2^{21}}} \over 2} = {2^{20}}$$ | mcq | jee-main-2019-online-12th-april-morning-slot | 7,479 |
Df2NYHYq0uAHcIQV3H7k9k2k5gjegt5 | maths | permutations-and-combinations | number-of-combinations | If a, b and c are the greatest value of <sup>19</sup>C<sub>p</sub>, <sup>20</sup>C<sub>q</sub>
and <sup>21</sup>C<sub>r</sub> respectively, then : | [{"identifier": "A", "content": "$${a \\over {11}} = {b \\over {22}} = {c \\over {21}}$$"}, {"identifier": "B", "content": "$${a \\over {10}} = {b \\over {22}} = {c \\over {21}}$$"}, {"identifier": "C", "content": "$${a \\over {10}} = {b \\over {11}} = {c \\over {42}}$$"}, {"identifier": "D", "content": "$${a \\over {1... | ["D"] | null | (
<sup>19</sup>C<sub>p</sub>)<sub>max</sub> =
<sup>19</sup>C<sub>9</sub> or
<sup>19</sup>C<sub>10</sub> = a
<br><br>(
<sup>20</sup>C<sub>p</sub>)<sub>max</sub> =
<sup>20</sup>C<sub>10</sub> = b
<br><br>(
<sup>21</sup>C<sub>r</sub>)<sub>max</sub> =
<sup>21</sup>C<sub>10</sub> or
<sup>21</sup>C<sub>11</sub> = c
<br><br>1... | mcq | jee-main-2020-online-8th-january-morning-slot | 7,481 |
TgQzv9xzFjU3klQ0D27k9k2k5h0eczd | maths | permutations-and-combinations | number-of-combinations | An urn contains 5 red marbles, 4 black marbles
and 3 white marbles. Then the number of ways
in which 4 marbles can be drawn so that at the
most three of them are red is ___________. | [] | null | 490 | Here 5 red marbels and 7 non red marbels presents.
<br><br>No of ways 4 marbels can be chosen where atmost 3 red marbels can be present.
<br><br><b>Case 1:</b> When 3 red marbels present
<br><br>No of ways = <sup>5</sup>C<sub>3</sub> $$ \times $$ <sup>7</sup>C<sub>1</sub>
<br><br><b>Case 2:</b> When 2 red marbels pres... | integer | jee-main-2020-online-8th-january-morning-slot | 7,482 |
SLjj7ptO9uQcIkLlhTjgy2xukfakunkv | maths | permutations-and-combinations | number-of-combinations | A test consists of 6 multiple choice questions, each having 4 alternative answers of which only one is correct. The number of ways, in which a candidate answers all six questions such that exactly four of the answers are correct, is __________. | [] | null | 135 | Select any 4 questions in <sup>6</sup>C<sub>4</sub>
ways which are
correct.
<br><br>Answering right option for each question is possible in 1 way.
<br><br>So ways of choosing right option for 4 questions = 1.1.1.1 = (1)<sup>4</sup>
<br><br>Number of ways of choosing wrong option for each question = 3
<br><br>So ways o... | integer | jee-main-2020-online-4th-september-evening-slot | 7,483 |
FlAz2HmdDhqlvaBRdA1kmjbf0bo | maths | permutations-and-combinations | number-of-combinations | Team 'A' consists of 7 boys and n girls and Team 'B' has 4 boys and 6 girls. If a total of 52 single matches can be arranged between these two teams when a boy plays against a boy and a girl plays against a girl, then n is equal to : | [{"identifier": "A", "content": "5"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "4"}, {"identifier": "D", "content": "6"}] | ["C"] | null | Total matches between boys of both team = $${}^7{C_1} \times {}^4{C_1} = 28$$<br><br>Total matches between girls of both team = $${}^n{C_1}\,{}^6{C_1} = 6n$$<br><br>Now, 28 + 6n = 52<br><br>$$ \Rightarrow $$ n = 4 | mcq | jee-main-2021-online-17th-march-morning-shift | 7,484 |
1krzq59sa | maths | permutations-and-combinations | number-of-combinations | If $${}^n{P_r} = {}^n{P_{r + 1}}$$ and $${}^n{C_r} = {}^n{C_{r - 1}}$$, then the value of r is equal to : | [{"identifier": "A", "content": "1"}, {"identifier": "B", "content": "4"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "3"}] | ["C"] | null | $${}^n{P_r} = {}^n{P_{r + 1}} \Rightarrow {{n!} \over {(n - r)!}} = {{n!} \over {(n - r - 1)!}}$$<br><br>$$ \Rightarrow (n - r) = 1$$ .....(1)<br><br>$${}^n{C_r} = {}^n{C_{r - 1}}$$<br><br>$$ \Rightarrow {{n!} \over {r!(n - r)!}} = {{n!} \over {(r - 1)!(n - r + 1)!}}$$<br><br>$$ \Rightarrow {1 \over {r(n - r)!}} = {1 \... | mcq | jee-main-2021-online-25th-july-evening-shift | 7,485 |
1l6p3l4h6 | maths | permutations-and-combinations | number-of-combinations | <p>The number of matrices of order $$3 \times 3$$, whose entries are either 0 or 1 and the sum of all the entries is a prime number, is __________.</p> | [] | null | 282 | <p>In a $$3\times3$$ order matrix there are $$9$$ entries.</p>
<p>These nine entries are zero or one.</p>
<p>The sum of positive prime entries are $$2, 3, 5$$ or $$7$$.</p>
<p>Total possible matrices $$ = {{9!} \over {2!\,.\,7!}} + {{9!} \over {3!\,.\,6!}} + {{9!} \over {5!\,.\,4!}} + {{9!} \over {7!\,.\,2!}}$$</p>
<... | integer | jee-main-2022-online-29th-july-morning-shift | 7,487 |
ldoaldz7 | maths | permutations-and-combinations | number-of-combinations | Let $\mathrm{A}=\left[\mathrm{a}_{i j}\right], \mathrm{a}_{i j} \in \mathbb{Z} \cap[0,4], 1 \leq i, j \leq 2$.
<br/><br/>The number of matrices A such that the sum of all entries is a prime number $\mathrm{p} \in(2,13)$ is __________. | [] | null | 204 | $A=\left[\begin{array}{ll}a_{11} & a_{12} \\ a_{13} & a_{14}\end{array}\right]$
<br/><br/>Such that $\Sigma a_{i i}=3,5,7$ or 11
<br/><br/>Then for sum 3, the possible entries are $(0,0,0,3)$, $(0,0,1,2),(0,1,1,1)$.
<br/><br/>Then total number of possible matrices
$$
=4+12+4
$$
$=20$
<br/><br/>For sum 5 the poss... | integer | jee-main-2023-online-31st-january-evening-shift | 7,488 |
1ldu5w6ww | maths | permutations-and-combinations | number-of-combinations | <p>$$\sum\limits_{k = 0}^6 {{}^{51 - k}{C_3}} $$ is equal to :</p> | [{"identifier": "A", "content": "$$\\mathrm{{}^{51}{C_4} - {}^{45}{C_4}}$$"}, {"identifier": "B", "content": "$$\\mathrm{{}^{51}{C_3} - {}^{45}{C_3}}$$"}, {"identifier": "C", "content": "$$\\mathrm{{}^{52}{C_3} - {}^{45}{C_3}}$$"}, {"identifier": "D", "content": "$$\\mathrm{{}^{52}{C_4} - {}^{45}{C_4}}$$"}] | ["D"] | null | $$
\begin{aligned}
& \sum_{\mathrm{k}=0}^6{ }^{51-\mathrm{k}} \mathrm{C}_3 \\\\
& ={ }^{51} \mathrm{C}_3+{ }^{50} \mathrm{C}_3+{ }^{49} \mathrm{C}_3+\ldots+{ }^{45} \mathrm{C}_3 \\\\
& ={ }^{45} \mathrm{C}_3+{ }^{46} \mathrm{C}_3+\ldots \ldots+{ }^{51} \mathrm{C}_3 \\\\
& ={ }^{45} \mathrm{C}_4+{ }^{45} \mathrm{C}_3+{ ... | mcq | jee-main-2023-online-25th-january-evening-shift | 7,489 |
1lgzzp7q6 | maths | permutations-and-combinations | number-of-combinations | <p>Let the number of elements in sets $$A$$ and $$B$$ be five and two respectively. Then the number of subsets of $$A \times B$$ each having at least 3 and at most 6 elements is :</p> | [{"identifier": "A", "content": "782"}, {"identifier": "B", "content": "772"}, {"identifier": "C", "content": "752"}, {"identifier": "D", "content": "792"}] | ["D"] | null | <p>First, let's determine the number of elements in the Cartesian product $$A \times B$$. If set $$A$$ has 5 elements and set $$B$$ has 2 elements, then the number of elements in $$A \times B$$ is:</p>
<p>
<p>$$|A \times B| = |A| \times |B| = 5 \times 2 = 10$$</p>
</p>
<p>We need to find the number of subsets of $$... | mcq | jee-main-2023-online-8th-april-morning-shift | 7,492 |
lsaoelst | maths | permutations-and-combinations | number-of-combinations | If $\mathrm{n}$ is the number of ways five different employees can sit into four indistinguishable offices where any office may have any number of persons including zero, then $\mathrm{n}$ is equal to : | [{"identifier": "A", "content": "47"}, {"identifier": "B", "content": "53"}, {"identifier": "C", "content": "51"}, {"identifier": "D", "content": "43"}] | ["C"] | null | <p>To determine the number of ways in which five different employees can be seated in four indistinguishable offices, we will use the concept of partitioning of an integer.</p>
<p>The problem is equivalent to partitioning the number 5 (representing the 5 different employees) into at most 4 parts (representing the 4 in... | mcq | jee-main-2024-online-1st-february-morning-shift | 7,493 |
lsapvsz6 | maths | permutations-and-combinations | number-of-combinations | The number of elements in the set $\mathrm{S}=\{(x, y, z): x, y, z \in \mathbf{Z}, x+2 y+3 z=42, x, y, z \geqslant 0\}$ equals __________. | [] | null | 169 | $$
x+2 y+3 z=42
$$
<br/><br/>$$
x, y, z \geq 0
$$
<br/><br/>as
<br/><br/>$\begin{array}{ll}z=0 & x+2 y=42 \Rightarrow 22 \text { cases } \\\\ z=1 & x+2 y=39 \Rightarrow 20 \text { cases } \\\\ z=2 & x+2 y=36 \Rightarrow 19 \text { cases } \\\\ z=3 & x+2 y=33 \Rightarrow 17 \text { cases } \\\\ z=4 & x+2 y=30 \Rightarro... | integer | jee-main-2024-online-1st-february-morning-shift | 7,494 |
1lsg4z01p | maths | permutations-and-combinations | number-of-combinations | <p>In an examination of Mathematics paper, there are 20 questions of equal marks and the question paper is divided into three sections : $$A, B$$ and $$C$$. A student is required to attempt total 15 questions taking at least 4 questions from each section. If section $$A$$ has 8 questions, section $$B$$ has 6 questions ... | [] | null | 11376 | <p>The problem involves choosing 15 questions out of a total of 20 available questions, with the constraint that at least 4 questions must be chosen from each of the three sections A, B, and C. To evaluate the total number of ways a student can select these questions, we need to consider every possible combination of q... | integer | jee-main-2024-online-30th-january-evening-shift | 7,496 |
lv0vxc34 | maths | permutations-and-combinations | number-of-combinations | <p>There are 5 points $$P_1, P_2, P_3, P_4, P_5$$ on the side $$A B$$, excluding $$A$$ and $$B$$, of a triangle $$A B C$$. Similarly there are 6 points $$\mathrm{P}_6, \mathrm{P}_7, \ldots, \mathrm{P}_{11}$$ on the side $$\mathrm{BC}$$ and 7 points $$\mathrm{P}_{12}, \mathrm{P}_{13}, \ldots, \mathrm{P}_{18}$$ on the si... | [{"identifier": "A", "content": "751"}, {"identifier": "B", "content": "776"}, {"identifier": "C", "content": "796"}, {"identifier": "D", "content": "771"}] | ["A"] | null | <p>Number of points on side $$A B=5$$</p>
<p>Number of points on side $$B C=6$$</p>
<p>Number of points on side $$A C=7$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lwk6ko9r/1d69134a-fee4-473a-84cb-efe47fa9ec3c/ecc716f0-1985-11ef-a7bd-376696e028ce/file-1lwk6ko9s.png?format=png" data-orsrc="... | mcq | jee-main-2024-online-4th-april-morning-shift | 7,497 |
lv2er3yn | maths | permutations-and-combinations | number-of-combinations | <p>There are 4 men and 5 women in Group A, and 5 men and 4 women in Group B. If 4 persons are selected from each group, then the number of ways of selecting 4 men and 4 women is ________.</p> | [] | null | 5626 | <p><style type="text/css">
.tg {border-collapse:collapse;border-spacing:0;}
.tg td{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
overflow:hidden;padding:10px 5px;word-break:normal;}
.tg th{border-color:black;border-style:solid;border-width:1px;font-family:Arial,... | integer | jee-main-2024-online-4th-april-evening-shift | 7,498 |
lv9s208h | maths | permutations-and-combinations | number-of-combinations | <p>Let the set $$S=\{2,4,8,16, \ldots, 512\}$$ be partitioned into 3 sets $$A, B, C$$ with equal number of elements such that $$\mathrm{A} \cup \mathrm{B} \cup \mathrm{C}=\mathrm{S}$$ and $$\mathrm{A} \cap \mathrm{B}=\mathrm{B} \cap \mathrm{C}=\mathrm{A} \cap \mathrm{C}=\phi$$. The maximum number of such possible parti... | [{"identifier": "A", "content": "1640"}, {"identifier": "B", "content": "1520"}, {"identifier": "C", "content": "1710"}, {"identifier": "D", "content": "1680"}] | ["D"] | null | <p>Given set $$S=\left\{2^1, 2^2, \ldots 2^9\right\}$$ which consist of 9 elements.</p>
<p>Maximum number of possible partitions (in set $$A, B$$ and $$C$$)</p>
<p>$$={ }^9 C_3 \cdot{ }^6 C_3 \cdot{ }^3 C_3=1680$$</p> | mcq | jee-main-2024-online-5th-april-evening-shift | 7,500 |
lvb294ut | maths | permutations-and-combinations | number-of-combinations | <p>Let $$0 \leq r \leq n$$. If $${ }^{n+1} C_{r+1}:{ }^n C_r:{ }^{n-1} C_{r-1}=55: 35: 21$$, then $$2 n+5 r$$ is equal to :</p> | [{"identifier": "A", "content": "62"}, {"identifier": "B", "content": "60"}, {"identifier": "C", "content": "55"}, {"identifier": "D", "content": "50"}] | ["D"] | null | <p>Given $0 \leq r \leq n$. If $\binom{n+1}{r+1} : \binom{n}{r} : \binom{n-1}{r-1} = 55 : 35 : 21$, then we are to determine the value of $2n + 5r$.</p>
<h3>Step-by-Step Solution:</h3>
<ol>
<li>Write the given proportions involving binomial coefficients:</li>
</ol>
<p>$ \frac{n+1}{r+1} \times \binom{n}{r} : \binom... | mcq | jee-main-2024-online-6th-april-evening-shift | 7,501 |
MWbLRehjPiKaEgcg | maths | permutations-and-combinations | number-of-permutations | If the letter of the word SACHIN are arranged in all possible ways and these words are written out as in dictionary, then the word SACHIN appears at serial number | [{"identifier": "A", "content": "601"}, {"identifier": "B", "content": "600"}, {"identifier": "C", "content": "603"}, {"identifier": "D", "content": "602"}] | ["A"] | null | <picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263736/exam_images/ru0tccwaposwg6xx5sll.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264497/exam_images/bawohr8zdetwd8ozm1dc.webp"><source media="(max-wid... | mcq | aieee-2005 | 7,502 |
KscQ5CpVa6h3aZAR | maths | permutations-and-combinations | number-of-permutations | From 6 different novels and 3 different dictionaries, 4 novels and 1 dictionary are to be selected and arranged in a row on a shelf so that the dictionary is always in the middle. Then the number of such arrangement is : | [{"identifier": "A", "content": "at least 500 but less than 750"}, {"identifier": "B", "content": "at least 750 but less than 1000"}, {"identifier": "C", "content": "at least 1000"}, {"identifier": "D", "content": "less than 500"}] | ["C"] | null | From 6 different novels 4 novels can be chosen = $${}^6{C_4}$$ ways
<br><br>And from 4 different dictionaries 1 can be chosen = $${}^3{C_1}$$ ways
<br><br>$$\therefore$$ From 6 different novels and 3 different dictionaries, 4 novels and 1 dictionary can be chosen = $${}^6{C_4} \times {}^3{C_1}$$ ways
<br><br>Let 4 nove... | mcq | aieee-2009 | 7,503 |
cpWdidtToPeRE0b1 | maths | permutations-and-combinations | number-of-permutations | The number of integers greater than 6,000 that can be formed, using the digits 3, 5, 6, 7 and 8, without repetition, is: | [{"identifier": "A", "content": "120"}, {"identifier": "B", "content": "72"}, {"identifier": "C", "content": "216"}, {"identifier": "D", "content": "192"}] | ["D"] | null | <p>For a four digit number the first place can be filled in 3 ways with 6 or 7 or 8 and the remaining four places in 4! ways i.e., 3 $$\times$$ 4! = 72.</p>
<p>For a five digit number it can be arranged in 5! ways,</p>
<p>$$\therefore$$ total number of integers = (72 + 120) = 192.</p> | mcq | jee-main-2015-offline | 7,504 |
IplDhMsmdCoRdapu | maths | permutations-and-combinations | number-of-permutations | If all the words (with or without meaning) having five letters,formed using the letters of the word SMALL and arranged as in a dictionary, then the position of the word SMALL is : | [{"identifier": "A", "content": "$${46^{th}}$$ "}, {"identifier": "B", "content": "$${59^{th}}$$"}, {"identifier": "C", "content": "$${52^{nd}}$$"}, {"identifier": "D", "content": "$${58^{th}}$$"}] | ["D"] | null | <p>Clearly, number of words start with $$A = {{4!} \over {2!}} = 12$$</p>
<p>Number of words start with $$L = 4! = 24$$</p>
<p>Number of words start with $$M = {{4!} \over {2!}} = 12$$</p>
<p>Number of words start with $$SA = {{3!} \over {2!}} = 3$$</p>
<p>Number of words start with $$SL = 3! = 6$$</p>
<p>Note that, ne... | mcq | jee-main-2016-offline | 7,505 |
4sIda7kghDbn5SBTSeZMx | maths | permutations-and-combinations | number-of-permutations | If all the words, with or without meaning, are written using the letters of the word QUEEN and are arranged as in English dictionary, then the position of the word QUEEN is : | [{"identifier": "A", "content": "44<sup>th</sup> "}, {"identifier": "B", "content": "45<sup>th</sup> "}, {"identifier": "C", "content": "46<sup>th</sup> "}, {"identifier": "D", "content": "47<sup>th</sup> "}] | ["C"] | null | <p>To find the position of the word QUEEN:</p>
<p>$$\bullet$$ The number of words starting with E is 4! = 24.</p>
<p>$$\bullet$$ The number of words starting with N is $${{4!} \over 2} = 12$$.</p>
<p>$$\bullet$$ The number of words starting with QE is 3! = 6.</p>
<p>$$\bullet$$ Number of words starting with QN is $${{3... | mcq | jee-main-2017-online-8th-april-morning-slot | 7,506 |
dE3mDjQalcUd2nH6 | maths | permutations-and-combinations | number-of-permutations | From 6 different novels and 3 different dictionaries, 4 novels and 1 dictionary are to be selected and
arranged in a row on a shelf so that the dictionary is always in the middle. The number of such
arrangements is : | [{"identifier": "A", "content": "at least 750 but less than 1000"}, {"identifier": "B", "content": "at least 1000"}, {"identifier": "C", "content": "less than 500"}, {"identifier": "D", "content": "at least 500 but less than 750"}] | ["B"] | null | From 6 different novels 4 novels can be chosen = $${}^6{C_4}$$ ways
<br><br>And from 3 different dictionaries 1 can be chosen = $${}^3{C_1}$$ ways
<br><br>$$\therefore$$ From 6 different novels and 3 different dictionaries, 4 novels and 1 dictionary can be chosen = $${}^6{C_4} \times {}^3{C_1}$$ ways
<br><br>Let 4 nove... | mcq | jee-main-2018-offline | 7,507 |
XLd4i0VsYtBkRaHaQmAcd | maths | permutations-and-combinations | number-of-permutations | n$$-$$digit numbers are formed using only three digits 2, 5 and 7. The smallest value of n for which 900 such distinct numbers can be formed, is : | [{"identifier": "A", "content": "6"}, {"identifier": "B", "content": "7"}, {"identifier": "C", "content": "8"}, {"identifier": "D", "content": "9"}] | ["B"] | null | In n digit number first place can be filled with any one of 2, 5, 7. So no of ways first digit can be filled = 3<br><br>
Similarly,<br>
no of ways 2nd digit can be filled = 3 ways<br>
.<br>
.<br>
.<br>
.<br>
- - - - - - nth - - - - - - - = 3 ways<br>
<br>
$$ \therefore $$ Total numbers = 3 $$ \times $$ 3 $$ \times $... | mcq | jee-main-2018-online-15th-april-morning-slot | 7,508 |
xe5ZfgVxYCdbpMKEGIuOF | maths | permutations-and-combinations | number-of-permutations | The number of four letter words that can be formed using the letters of the word <b>BARRACK</b> is : | [{"identifier": "A", "content": "120"}, {"identifier": "B", "content": "144"}, {"identifier": "C", "content": "264"}, {"identifier": "D", "content": "270"}] | ["D"] | null | <b>Case 1 :</b>
<br><br>When all the four letters different then no of words
<br>= <sup>5</sup>C<sub>4</sub> $$ \times $$4!
<br><br><b>Case 2 :</b>
<br><br>When out of four letters two letters are R and other two different letters are chosen from B, A, C, K then the no of words <br>= <sup>4</sup>C<sub>2</sub> $$ \times... | mcq | jee-main-2018-online-15th-april-evening-slot | 7,509 |
lrAD8xOCEl2vEKUBAe7k9k2k5e2s565 | maths | permutations-and-combinations | number-of-permutations | Total number of 6-digit numbers in which only and all the five digits 1, 3, 5, 7 and 9 appear, is : | [{"identifier": "A", "content": "$${5 \\over 2}\\left( {6!} \\right)$$"}, {"identifier": "B", "content": "$${6!}$$"}, {"identifier": "C", "content": "5<sup>6</sup>"}, {"identifier": "D", "content": "$${1 \\over 2}\\left( {6!} \\right)$$"}] | ["A"] | null | Here none number repeats more than once.
<br><br>We can choose the number which repeats more than once among 1, 3, 5, 7, 9 in <sup>5</sup>C<sub>1</sub> ways.
<br><br>Let number 3 repeats more than once. So six digits are 1, 3, 3, 5, 7, 9.
<br><br>We can arrange those six digits in $${{6!} \over {2!}}$$ ways.
<br><br>$$... | mcq | jee-main-2020-online-7th-january-morning-slot | 7,510 |
qbRvgJn6hv5comatCj7k9k2k5hketjb | maths | permutations-and-combinations | number-of-permutations | The number of 4 letter words (with or without
meaning) that can be formed from the eleven
letters of the word 'EXAMINATION' is
_______. | [] | null | 2454 | 2A, 2I, 2N, E, X, M, T, O
<br><br>To form four letter words
<br><br><b>Case 1 :</b> All same ( not possible)
<br><br><b>Case 2 :</b> 1 different, 3 same (not possible)
<br><br><b>Case 3 :</b> 2 different, 2 same
<br><br>= <sup>3</sup>C<sub>1</sub> $$ \times $$ <sup>7</sup>C<sub>2</sub> $$ \times $$ $${{4!} \over {2!}}$... | integer | jee-main-2020-online-8th-january-evening-slot | 7,511 |
k3m0N9G0xSwL4wln7Cjgy2xukewtoyf9 | maths | permutations-and-combinations | number-of-permutations | If the letters of the word 'MOTHER' be permuted
and all the words so formed (with or without
meaning) be listed as in a dictionary, then the
position of the word 'MOTHER' is ______. | [] | null | 309 | <picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267803/exam_images/xbqcgksmphluv6sci9ir.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265109/exam_images/kdpljpefm6kve0qcetlt.webp"><source media="(max-wid... | integer | jee-main-2020-online-2nd-september-morning-slot | 7,512 |
KcZ71werwgGvco0DDojgy2xukf0wa8ut | maths | permutations-and-combinations | number-of-permutations | The value of (2.<sup>1</sup>P<sub>0</sub>
– 3.<sup>2</sup>P<sub>1</sub> + 4.<sup>3</sup>P<sub>2</sub> .... up to
51<sup>th</sup> term)<br/> + (1! – 2! + 3! – ..... up to 51<sup>th</sup> term)
is equal to : | [{"identifier": "A", "content": "1"}, {"identifier": "B", "content": "1 + (51)!"}, {"identifier": "C", "content": "1 \u2013 51(51)!"}, {"identifier": "D", "content": "1 + (52)!"}] | ["D"] | null | (2.<sup>1</sup>P<sub>0</sub>
– 3.<sup>2</sup>P<sub>1</sub> + 4.<sup>3</sup>P<sub>2</sub> .... up to
51<sup>th</sup> term)<br> + (1! – 2! + 3! – ..... up to 51<sup>th</sup> term)
<br><br>= ( 2.1! - 3.2! + 4.3! - ....+ 52.51!)
<br>+ ( 1! – 2! + 3! – ..... + (51)! )
<br><br>= (2! - 3! + 4! .....+ 52! )
<br>+ ( 1! - 2! + ... | mcq | jee-main-2020-online-3rd-september-morning-slot | 7,513 |
t2DoGkNqDTClpiz752jgy2xukf49ludb | maths | permutations-and-combinations | number-of-permutations | The total number of 3-digit numbers, whose
sum of digits is 10, is __________. | [] | null | 54 | Let xyz is 3 digits number.<br><br>Given that sum of digits = 10<br><br>$$ \therefore $$ x + y + z = 10 ......(1)<br><br>Also x can't be 0 as if x = 0 then it will become 2 digits number.<br><br>So, x $$ \ge $$ 1, y $$ \ge $$ 0, z $$ \ge $$ 0<br><br>As x $$ \ge $$ 1<br><br>$$ \Rightarrow $$ x $$-$$ 1 $$ \ge $$ 0<br><br... | integer | jee-main-2020-online-3rd-september-evening-slot | 7,514 |
HzI0OnXimst1pZw1d21kmllyfaz | maths | permutations-and-combinations | number-of-permutations | The missing value in the following figure is <br/><br/><img src="data:image/png;base64,UklGRqYYAABXRUJQVlA4IJoYAACQiACdASqcAVsBPm00lkikIqKhITJLIIANiWlu/HRYhs4NYEDwnk/Ff8+/GTwA/tn9O/bD+u9kB4u/T/2s/uHsA/038e9AnPv+1/j3qp8z/wv8O/qv+5/wf7s/fb89/yP8b/nf+0/sfp78RvkD2CPTf+M/hv9Y/0f9O/df2+f7j+AfyLwM7GfoB7BHbv/L/xX+ffr58Qnpf+h/i... | [] | null | 4 | Inside number $=\left(\right.$ difference)${ }^{(\text {difference}) !}$
<br/><br/>$=($Greater number - Smaller number)<sup>(Greater number - Smaller number)!</sup>
<br/><br/>i.e. $1=(2-1)^{(2-1) !}, $
<br/><br/>$4^{24}=(12-8)^{(12-8) !}, $
<br/><br/>$3^6=(7-4)^{(7-4) !}$
<br/><br/>$\therefore \quad ?=(5-3)^{(5-3) !}$
... | integer | jee-main-2021-online-18th-march-morning-shift | 7,515 |
1ktbicc13 | maths | permutations-and-combinations | number-of-permutations | If $${}^1{P_1} + 2.{}^2{P_2} + 3.{}^3{P_3} + .... + 15.{}^{15}{P_{15}} = {}^q{P_r} - s,0 \le s \le 1$$, then $${}^{q + s}{C_{r - s}}$$ is equal to ______________. | [] | null | 136 | $${}^1{P_1} + 2.{}^2{P_2} + 3.{}^3{P_3} + .... + 15.{}^{15}{P_{15}}$$<br><br>= 1! + 2 . 2! + 3 . 3! + ..... 15 $$\times$$ 15!<br><br>$$ = \sum\limits_{r = 1}^{15} {(r + 1)! - (r)!} $$<br><br>= 16! $$-$$ 1<br><br>= $${}^{16}{P_{16}}$$ $$-$$ 1<br><br>$$\Rightarrow$$ q = r = 16, s = 1<br><br>$${}^{q + s}{C_{r - s}} = {}^{... | integer | jee-main-2021-online-26th-august-morning-shift | 7,517 |
1l6dx41xx | maths | permutations-and-combinations | number-of-permutations | <p>The letters of the word 'MANKIND' are written in all possible orders and arranged in serial order as in an English dictionary. Then the serial number of the word 'MANKIND' is _____________.</p> | [] | null | 1492 | <p>Step 1 :</p>
<p>Write all the alphabets in alphabetical order.</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l79egidm/a6690954-7aeb-421e-841b-1c4509718aab/ba67f6a0-24a6-11ed-8d2e-5f0df5271c2d/file-1l79egidn.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l79egidm/a66... | integer | jee-main-2022-online-25th-july-morning-shift | 7,518 |
1l6m6g453 | maths | permutations-and-combinations | number-of-permutations | <p>Let $$S$$ be the set of all passwords which are six to eight characters long, where each character is either an alphabet from $$\{A, B, C, D, E\}$$ or a number from $$\{1,2,3,4,5\}$$ with the repetition of characters allowed. If the number of passwords in $$S$$ whose at least one character is a number from $$\{1,2,3... | [] | null | 7073 | <p>If password is 6 character long, then</p>
<p>Total number of ways having atleast one number $$ = {10^6} - {5^6}$$</p>
<p>Similarly, if 7 character long $$ = {10^7} - {5^7}$$</p>
<p>and if 8-character long $$ = {10^8} - {5^8}$$</p>
<p>Number of password $$ = ({10^6} + {10^7} + {10^8}) - ({5^6} + {5^7} + {5^8})$$</p>
... | integer | jee-main-2022-online-28th-july-morning-shift | 7,519 |
ldoan5f0 | maths | permutations-and-combinations | number-of-permutations | If ${ }^{2 n+1} \mathrm{P}_{n-1}:{ }^{2 n-1} \mathrm{P}_{n}=11: 21$,
<br/><br/>then $n^{2}+n+15$ is equal to : | [] | null | 45 | $\frac{\frac{(2 n+1) !}{(n+2) !}}{\frac{(2 n-1) !}{(n-1) !}}=\frac{11}{21}$
<br/><br/>$\frac{(2 n+1) 2 n}{(n+2)(n+1) n}=\frac{11}{21}$
<br/><br/>$84 n+42=11\left(n^{2}+3 n+2\right)$
<br/><br/>$11 n^{2}-51 n-20=0$
<br/><br/>$(n-5)(11 n+4)=0$
<br/><br/>$n=5, \frac{-4}{11}$ (Rejected $)$
<br/><br/>$n^{2}+n+15=45$ | integer | jee-main-2023-online-31st-january-evening-shift | 7,520 |
1ldom13ou | maths | permutations-and-combinations | number-of-permutations | <p>The value of $$\frac{1}{1 ! 50 !}+\frac{1}{3 ! 48 !}+\frac{1}{5 ! 46 !}+\ldots .+\frac{1}{49 ! 2 !}+\frac{1}{51 ! 1 !}$$ is :</p> | [{"identifier": "A", "content": "$$\\frac{2^{51}}{50 !}$$"}, {"identifier": "B", "content": "$$\\frac{2^{51}}{51 !}$$"}, {"identifier": "C", "content": "$$\\frac{2^{50}}{50 !}$$"}, {"identifier": "D", "content": "$$\\frac{2^{50}}{51 !}$$"}] | ["D"] | null | $$
\begin{aligned}
& \mathrm{S}=\frac{1}{1 ! 50 !}+\frac{1}{3 ! 48 !}+\frac{1}{5 ! 46 !}+\ldots \ldots+\frac{1}{49 ! 2 !}+\frac{1}{51 ! 1 !} \\\\
& =\frac{1}{51 !}\left(\frac{51 !}{1 ! 50 !}+\frac{51 !}{3 ! 48 !}+\frac{51 !}{5 ! 46 !}+\ldots . .+\frac{51 !}{49 ! 2 !}+\frac{51 !}{51 ! 0 !}\right) \\\\
& =\frac{1}{51 !}\... | mcq | jee-main-2023-online-1st-february-morning-shift | 7,521 |
1ldoo8du5 | maths | permutations-and-combinations | number-of-permutations | <p>The number of 3-digit numbers, that are divisible by either 2 or 3 but not divisible by 7, is ____________.</p> | [] | null | 514 | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1le9xbqbh/6a03e2bb-e7c5-4885-abb8-b77d1c53361f/0699bad0-af86-11ed-8a80-c79017866e24/file-1le9xbqbi.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1le9xbqbh/6a03e2bb-e7c5-4885-abb8-b77d1c53361f/0699bad0-af86-11ed-8a80-c79017866e24/fi... | integer | jee-main-2023-online-1st-february-morning-shift | 7,522 |
1ldseml3w | maths | permutations-and-combinations | number-of-permutations | <p>The letters of the word OUGHT are written in all possible ways and these words are arranged as in a dictionary, in a series. Then the serial number of the word TOUGH is :</p> | [{"identifier": "A", "content": "79"}, {"identifier": "B", "content": "84"}, {"identifier": "C", "content": "89"}, {"identifier": "D", "content": "86"}] | ["C"] | null | <p>Lets arrange the letters of OUGHT in alphabetical
order.
<br/><br/>G, H, O, T, U</p>
<br/>Words starting with
<br/><br/>$$
\begin{aligned}
& \mathrm{G}----\rightarrow 4 ! \\\\
& \mathrm{H}----\rightarrow 4 ! \\\\
& \mathrm{O}----\rightarrow 4 !
\end{aligned}
$$
<br/><br/>$\mathrm{T} \,\mathrm{G}---\rightarrow 3$ !
... | mcq | jee-main-2023-online-29th-january-evening-shift | 7,524 |
1ldswzrar | maths | permutations-and-combinations | number-of-permutations | <p>Five digit numbers are formed using the digits 1, 2, 3, 5, 7 with repetitions and are written in descending order with serial numbers. For example, the number 77777 has serial number 1. Then the serial number of 35337 is ____________.</p> | [] | null | 1436 | <p>Descending order means from highest to lowest.</p>
<p>$$\therefore$$ Descending order of digits 1, 2, 3, 5, 7 is = 7, 5, 3, 2, 1.</p>
<p>(1) Number starts with 7 :</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1le2g13sg/30377b8c-aded-4085-8900-acce8af1dc3b/e75d6700-ab68-11ed-b566-111c81fc645... | integer | jee-main-2023-online-29th-january-morning-shift | 7,525 |
1ldww37nt | maths | permutations-and-combinations | number-of-permutations | <p>The number of integers, greater than 7000 that can be formed, using the digits 3, 5, 6, 7, 8 without repetition is :</p> | [{"identifier": "A", "content": "48"}, {"identifier": "B", "content": "120"}, {"identifier": "C", "content": "168"}, {"identifier": "D", "content": "220"}] | ["C"] | null | Four digit numbers greater than 7000 $=2 \times 4 \times 3 \times 2=48$<br/><br/>
Five digit number $=5 !=120$<br/><br/>
Total number greater than 7000 $=120+48=168$ | mcq | jee-main-2023-online-24th-january-evening-shift | 7,526 |
1lgow5dwx | maths | permutations-and-combinations | number-of-permutations | <p>All words, with or without meaning, are made using all the letters of the word MONDAY. These words are written as in a dictionary with serial numbers. The serial number of the word MONDAY is :</p> | [{"identifier": "A", "content": "324"}, {"identifier": "B", "content": "328"}, {"identifier": "C", "content": "326"}, {"identifier": "D", "content": "327"}] | ["D"] | null |
<p>The letters of "MONDAY" arranged in alphabetical order are: A, D, M, N, O, Y.</p>
<ol>
<li>Fix A as the first letter, we can arrange the remaining 5 letters in 5! ways = 120 ways.</li>
<li>Fix D as the first letter, we can arrange the remaining 5 letters in 5! ways = 120 ways.</li>
</ol>
<p>So far, we hav... | mcq | jee-main-2023-online-13th-april-evening-shift | 7,527 |
1lgsugvol | maths | permutations-and-combinations | number-of-permutations | <p>If the letters of the word MATHS are permuted and all possible words so formed are arranged as in a dictionary with serial numbers, then the serial number of the word THAMS is :</p> | [{"identifier": "A", "content": "103"}, {"identifier": "B", "content": "104"}, {"identifier": "C", "content": "102"}, {"identifier": "D", "content": "101"}] | ["A"] | null | To solve this problem, we start by finding the number of permutations before we reach a word that begins with T.
<br/><br/>We have 5 letters in the word MATHS. If we fix the first letter, there are 4! (=4$$ \times $$3$$ \times $$2$$ \times $$1=24) ways to arrange the remaining letters.
<br/><br/>The letters before T... | mcq | jee-main-2023-online-11th-april-evening-shift | 7,528 |
1lgvqlzes | maths | permutations-and-combinations | number-of-permutations | <p>The sum of all the four-digit numbers that can be formed using all the digits 2, 1, 2, 3 is equal to __________.</p> | [] | null | 26664 | We have, four digits are $2,1,2,3$.
<br/><br/>Total numbers when 1 is at unit digit $=\frac{3 !}{2 !}=3$
<br/><br/>Total number when 2 is at unit digit $=3 !=6$
<br/><br/>Total numbers when 3 is at unit digit $=\frac{3 !}{2 !}=3$
<br/><br/>Sum of digits at unit place $=3 \times 1+6 \times 2+3 \times 3=24$
<br/><br/>$\t... | integer | jee-main-2023-online-10th-april-evening-shift | 7,529 |
1lh2xtxhh | maths | permutations-and-combinations | number-of-permutations | <p> All the letters of the word PUBLIC are written in all possible orders and these words are written as in a dictionary with serial numbers. Then the serial number of the word PUBLIC is :</p> | [{"identifier": "A", "content": "578"}, {"identifier": "B", "content": "576"}, {"identifier": "C", "content": "580"}, {"identifier": "D", "content": "582"}] | ["D"] | null | Given word is PUBLIC.
<br/><br/>Alphabetical order of letters is BCILPU. So, number of words start with letter.
<br/><br/>B - - - - is $5 !=120$
<br/><br/>C ---- is $5 !=120$
<br/><br/>I ----- is $5 !=120$
<br/><br/>$\mathrm{L}-----$ is $5 !=120$
<br/><br/>$\mathrm{PB}----$ is $4 !=24$
<br/><br/>$\mathrm{PC}----$ is ... | mcq | jee-main-2023-online-6th-april-evening-shift | 7,530 |
jaoe38c1lse5y7d4 | maths | permutations-and-combinations | number-of-permutations | <p>The total number of words (with or without meaning) that can be formed out of the letters of the word '<b>DISTRIBUTION</b>' taken four at a time, is equal to __________.</p> | [] | null | 3734 | <p>We have III, TT, D, S, R, B, U, O, N</p>
<p>Number of words with selection (a, a, a, b)</p>
<p>$$={ }^8 \mathrm{C}_1 \times \frac{4 !}{3 !}=32$$</p>
<p>Number of words with selection (a, a, b, b)</p>
<p>$$=\frac{4 !}{2 ! 2 !}=6$$</p>
<p>Number of words with selection (a, a, b, c)</p>
<p>$$={ }^2 \mathrm{C}_1 \times{... | integer | jee-main-2024-online-31st-january-morning-shift | 7,531 |
jaoe38c1lsf0oh50 | maths | permutations-and-combinations | number-of-permutations | <p>All the letters of the word "GTWENTY" are written in all possible ways with or without meaning and these words are written as in a dictionary. The serial number of the word "GTWENTY" is _________.</p> | [] | null | 553 | <p>Words starting with $$\mathrm{E}=360$$</p>
<p>Words starting with $$\mathrm{GE=60}$$</p>
<p>Words starting with $$\mathrm{GN=60}$$</p>
<p>Words starting with $$\mathrm{GTE=24}$$</p>
<p>Words starting with $$\mathrm{GTN=24}$$</p>
<p>Words starting with $$\mathrm{GTT=24}$$</p>
<p>GTWENTY $$=1$$</p>
<p>Total $$=553$$</... | integer | jee-main-2024-online-29th-january-morning-shift | 7,532 |
lv5gsk30 | maths | permutations-and-combinations | number-of-permutations | <p>The number of 3-digit numbers, formed using the digits 2, 3, 4, 5 and 7, when the repetition of digits is not allowed, and which are not divisible by 3 , is equal to ________.</p> | [] | null | 36 | <p>To solve this problem, we need to find the number of 3-digit numbers formed using the digits 2, 3, 4, 5, and 7, with no repetition of digits allowed, and these numbers should not be divisible by 3. Let's break down the solution step-by-step:</p>
<p>1. <strong>Calculating the total number of 3-digit numbers without ... | integer | jee-main-2024-online-8th-april-morning-shift | 7,534 |
lv9s20f0 | maths | permutations-and-combinations | number-of-permutations | <p>60 words can be made using all the letters of the word $$\mathrm{BHBJO}$$, with or without meaning. If these words are written as in a dictionary, then the $$50^{\text {th }}$$ word is:</p> | [{"identifier": "A", "content": "OBBJH"}, {"identifier": "B", "content": "HBBJO"}, {"identifier": "C", "content": "OBBHJ"}, {"identifier": "D", "content": "JBBOH"}] | ["A"] | null | <p>To find the $$50^{\text{th}}$$ word formed by the letters of "BHBJO" as if listed in a dictionary, let's analyze the arrangement methodically. Given the letters are B, H, B, J, O, there are some repetitions with the letter B appearing twice.</p>
<p>First, calculate the total number of permutations of these letters:... | mcq | jee-main-2024-online-5th-april-evening-shift | 7,535 |
lvb294sf | maths | permutations-and-combinations | number-of-permutations | <p>If all the words with or without meaning made using all the letters of the word "NAGPUR" are arranged as in a dictionary, then the word at $$315^{\text {th }}$$ position in this arrangement is :</p> | [{"identifier": "A", "content": "NRAPUG\n"}, {"identifier": "B", "content": "NRAGUP\n"}, {"identifier": "C", "content": "NRAPGU\n"}, {"identifier": "D", "content": "NRAGPU"}] | ["C"] | null | <p>NAGPUR</p>
<p>Word at $$315^{\text {th }}$$ position</p>
<p>$$\begin{aligned}
& \text { A...... }=5!=120 \\
& \text { G....... }=5!=120 \\
& \text { NA..... }=4!=24 \\
& \text { NG..... }=4!=24 \\
& \text { NP..... }=4!=24
\end{aligned}$$</p>
<p>..... Till 312 words</p>
<p>$$313^{\text {th }} \text { word }=\text { ... | mcq | jee-main-2024-online-6th-april-evening-shift | 7,536 |
ys2abeQH99za49tLBzsHL | maths | probability | bayes-theorem | A box 'A' contains $$2$$ white, $$3$$ red and $$2$$ black balls. Another box 'B' contains $$4$$ white, $$2$$ red and $$3$$ black balls. If two balls are drawn at random, without eplacement, from a randomly selected box and one ball turns out to be white while the other ball turns out to be red, then the probability th... | [{"identifier": "A", "content": "$${9 \\over {16}}$$ "}, {"identifier": "B", "content": "$${7 \\over {16}}$$"}, {"identifier": "C", "content": "$${9 \\over {32}}$$"}, {"identifier": "D", "content": "$${7 \\over {8}}$$"}] | ["B"] | null | Probability of drawing a white ball and then a red ball
<br><br>from bag B is given by
$${{{}^4{C_1} \times {}^2{C_1}} \over {{}^9{C_2}}}$$ = $${2 \over 9}$$
<br><br>Probability of drawing a white ball and then a red ball
<br><br>from bag A is given by $${{{}^2{C_1} \times {}^3{C_1}} \over {{}^7{C_2}}}$$ = $${2 \ov... | mcq | jee-main-2018-online-15th-april-morning-slot | 7,537 |
ddcEu6sJsoUFV2yTjBjgy2xukewmw2ol | maths | probability | bayes-theorem | Box I contains 30 cards numbered 1 to 30 and
Box II contains 20 cards numbered 31 to 50. A
box is selected at random and a card is drawn
from it. The number on the card is found to be
a non-prime number. The probability that the
card was drawn from Box I is : | [{"identifier": "A", "content": "$${8 \\over {17}}$$"}, {"identifier": "B", "content": "$${2 \\over 3}$$"}, {"identifier": "C", "content": "$${2 \\over 5}$$"}, {"identifier": "D", "content": "$${4 \\over {17}}$$"}] | ["A"] | null | Let B<sub>1</sub> be the event where Box-I is selected.
<br><br>And B<sub>2</sub> be the event where Box-II is selected.
<br><br>P(B<sub>1</sub>) = P(B<sub>2</sub>) = $${1 \over 2}$$
<br><br>Let E be the event where selected card is non prime.
<br><br>For B<sub>1</sub> : Prime numbers: {2, 3, 5, 7, 11, 13, 17, 19, 23, ... | mcq | jee-main-2020-online-2nd-september-morning-slot | 7,538 |
RTeG7N1NMdtZ3ITnFI1klt88cer | maths | probability | bayes-theorem | In a group of 400 people, 160 are smokers and non-vegetarian; 100 are smokers and vegetarian and the remaining 140 are non-smokers and vegetarian. Their chances of getting a particular chest disorder are 35%, 20% and 10% respectively. A person is chosen from the group at random and is found to be suffering from the che... | [{"identifier": "A", "content": "$${{14} \\over {45}}$$"}, {"identifier": "B", "content": "$${{8} \\over {45}}$$"}, {"identifier": "C", "content": "$${{7} \\over {45}}$$"}, {"identifier": "D", "content": "$${{28} \\over {45}}$$"}] | ["D"] | null | Consider following events<br><br>A : Person chosen is a smoker and non vegetarian.<br><br>B : Person chosen in a smoker and vegetarian.<br><br>C : Person chosen is a non-smoker and vegetarian.<br><br>E : Person chosen has a chest disorder<br><br>Given <br><br>$$P(A) = {{160} \over {400}}$$
<br><br>$$P(B) = {{100} \over... | mcq | jee-main-2021-online-25th-february-evening-slot | 7,539 |
1l5c0v06l | maths | probability | bayes-theorem | <p>Bag A contains 2 white, 1 black and 3 red balls and bag B contains 3 black, 2 red and n white balls. One bag is chosen at random and 2 balls drawn from it at random, are found to be 1 red and 1 black. If the probability that both balls come from Bag A is $${6 \over {11}}$$, then n is equal to __________.</p> | [{"identifier": "A", "content": "13"}, {"identifier": "B", "content": "6"}, {"identifier": "C", "content": "4"}, {"identifier": "D", "content": "3"}] | ["C"] | null | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l8la6rik/8b350f4d-948a-4095-a004-5c8116776062/a60740c0-3efb-11ed-8e30-11b9f1cb84fb/file-1l8la6ril.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l8la6rik/8b350f4d-948a-4095-a004-5c8116776062/a60740c0-3efb-11ed-8e30-11b9f1cb84fb/fi... | mcq | jee-main-2022-online-24th-june-morning-shift | 7,540 |
1l6rev2p3 | maths | probability | bayes-theorem | <p>Bag I contains 3 red, 4 black and 3 white balls and Bag II contains 2 red, 5 black and 2 white balls. One ball is transferred from Bag I to Bag II and then a ball is drawn from Bag II. The ball so drawn is found to be black in colour. Then the probability, that the transferred ball is red, is :</p> | [{"identifier": "A", "content": "$$\\frac{4}{9}$$"}, {"identifier": "B", "content": "$$\\frac{5}{18}$$"}, {"identifier": "C", "content": "$$\\frac{1}{6}$$"}, {"identifier": "D", "content": "$$\\frac{3}{10}$$"}] | ["B"] | null | Let $E \rightarrow$ Ball drawn from Bag II is black.
<br/><br/>
$E_{R} \rightarrow$ Bag I to Bag II red ball transferred.
<br/><br/>
$E_{B} \rightarrow$ Bag I to Bag II black ball transferred.
<br/><br/>
$E_{w} \rightarrow$ Bag I to Bag II white ball transferred.
<br/><br/>
$P\left(E_{R} / E\right)=\frac{P\left(E / E_{... | mcq | jee-main-2022-online-29th-july-evening-shift | 7,541 |
1ldprftih | maths | probability | bayes-theorem | <p>A bag contains 6 balls. Two balls are drawn from it at random and both are found to be black. The probability that the bag contains at least 5 black balls is :</p> | [{"identifier": "A", "content": "$$\\frac{3}{7}$$"}, {"identifier": "B", "content": "$$\\frac{5}{6}$$"}, {"identifier": "C", "content": "$$\\frac{5}{7}$$"}, {"identifier": "D", "content": "$$\\frac{2}{7}$$"}] | ["C"] | null | Let $E_{i} \rightarrow$ Bag have at least $i$ black balls
<br/><br/>$E \rightarrow 2$ balls are drawn & both black
<br/><br/>$$
\begin{aligned}
& \therefore P\left(\frac{E_{5} \text { or } E_{6}}{E}\right)=\frac{P\left(\frac{E}{E_{5}}\right)+P\left(\frac{E}{E_{6}}\right)}{\sum\limits_{i=1}^{6} P\left(\frac{E}{E_{... | mcq | jee-main-2023-online-31st-january-morning-shift | 7,542 |
1ldwxvdwq | maths | probability | bayes-theorem | <p>Three urns A, B and C contain 4 red, 6 black; 5 red, 5 black; and $$\lambda$$ red, 4 black balls respectively. One of the urns is selected at random and a ball is drawn. If the ball drawn is red and the probability that it is drawn from urn C is 0.4 then the square of the length of the side of the largest equilatera... | [] | null | 432 | $E_{1}$ : Ball is drawn from urn $A(4 R+6 B)$
<br><br>
$E_{2}:$ Ball is drawn from urn $B(5 R+5 B)$
<br><br>
$E_{3}$ : Ball is drawn from urn $C(\lambda R+4 B)$
<br><br>
$A \rightarrow$ Ball drawn is red.
<br><br>
Required probability $=P\left(\frac{E_{3}}{A}\right)$
<br><br>
$=\frac{\frac{1}{3} \times \frac{\lambda}{\... | integer | jee-main-2023-online-24th-january-evening-shift | 7,543 |
1lsg4n319 | maths | probability | bayes-theorem | <p>Bag A contains 3 white, 7 red balls and Bag B contains 3 white, 2 red balls. One bag is selected at random and a ball is drawn from it. The probability of drawing the ball from the bag A, if the ball drawn is white, is</p> | [{"identifier": "A", "content": "1/4"}, {"identifier": "B", "content": "1/3"}, {"identifier": "C", "content": "3/10"}, {"identifier": "D", "content": "1/9"}] | ["B"] | null | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsoxmoxg/bddbd042-adc7-47f2-8492-6edfb8c4f19c/00793740-ccf2-11ee-a330-494dca5e9a63/file-6y3zli1lsoxmoxh.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lsoxmoxg/bddbd042-adc7-47f2-8492-6edfb8c4f19c/00793740-ccf2-11ee... | mcq | jee-main-2024-online-30th-january-evening-shift | 7,544 |
lv3ve61j | maths | probability | bayes-theorem | <p>There are three bags $$X, Y$$ and $$Z$$. Bag $$X$$ contains 5 one-rupee coins and 4 five-rupee coins; Bag $$Y$$ contains 4 one-rupee coins and 5 five-rupee coins and Bag $$Z$$ contains 3 one-rupee coins and 6 five-rupee coins. A bag is selected at random and a coin drawn from it at random is found to be a one-rupee ... | [{"identifier": "A", "content": "$$\\frac{1}{2}$$\n"}, {"identifier": "B", "content": "$$\\frac{1}{3}$$\n"}, {"identifier": "C", "content": "$$\\frac{5}{12}$$\n"}, {"identifier": "D", "content": "$$\\frac{1}{4}$$"}] | ["B"] | null | <p>To solve this problem, we use Bayes' theorem. Let's define the events:</p>
<ul>
<li>$$X$$: Selecting bag $$X$$</li>
<li>$$Y$$: Selecting bag $$Y$$</li>
<li>$$Z$$: Selecting bag $$Z$$</li>
<li>$$A$$: Drawing a one-rupee coin</li>
</ul>
<p>We are given that a bag is selected at random, so the probabilities for ... | mcq | jee-main-2024-online-8th-april-evening-shift | 7,546 |
lvc586ak | maths | probability | bayes-theorem | <p>A company has two plants $$A$$ and $$B$$ to manufacture motorcycles. $$60 \%$$ motorcycles are manufactured at plant $$A$$ and the remaining are manufactured at plant $$B .80 \%$$ of the motorcycles manufactured at plant $$A$$ are rated of the standard quality, while $$90 \%$$ of the motorcycles manufactured at plan... | [{"identifier": "A", "content": "54"}, {"identifier": "B", "content": "66"}, {"identifier": "C", "content": "56"}, {"identifier": "D", "content": "64"}] | ["A"] | null | <p>$$\begin{aligned}
& P(\text { standard automobile from } A)=\frac{6}{10} \times \frac{8}{10}=\frac{12}{25} \\
& P(\text { standard automobile from } B)=\frac{4}{10} \times \frac{9}{10}=\frac{9}{25}
\end{aligned}$$</p>
<p>$$\begin{aligned}
& \text { Required Probability } \frac{\frac{9}{25}}{\frac{12}{25}+\frac{9}{25... | mcq | jee-main-2024-online-6th-april-morning-shift | 7,547 |
byM0olMAPi99wdnA | maths | probability | binomial-distribution | A dice is tossed $$5$$ times. Getting an odd number is considered a success. Then the variance of distribution of success is : | [{"identifier": "A", "content": "$$8/3$$"}, {"identifier": "B", "content": "$$3/8$$"}, {"identifier": "C", "content": "$$4/5$$ "}, {"identifier": "D", "content": "$$5/4$$"}] | ["D"] | null | Total no of trials = 5
<br>$$\therefore$$ n = 5
<br>Odd no possibe are = {1, 3, 5} =3
<br>Sample space = {1, 2, 3, 4, 5, 6} = 6
<br><br>Probablity of getting odd number = $${3 \over 6}$$ = $${1 \over 2}$$
<br>$$\therefore$$ p = $${1 \over 2}$$
<br><br>Formula for variance = npq
<br>where q = 1 - p = 1 - $${1 \over 2}$$... | mcq | aieee-2002 | 7,548 |
KSCoYtIeFC6mdWAN | maths | probability | binomial-distribution | The mean and variance of a random variable $$X$$ having binomial distribution are $$4$$ and $$2$$ respectively, then $$P(X=1)$$ is : | [{"identifier": "A", "content": "$${1 \\over 4}$$ "}, {"identifier": "B", "content": "$${1 \\over 32}$$"}, {"identifier": "C", "content": "$${1 \\over 16}$$"}, {"identifier": "D", "content": "$${1 \\over 8}$$"}] | ["B"] | null | Mean = $$np$$ = 4
<br>Variance = $$npq$$ = 2
<br><br>$$ \Rightarrow 4 \times q$$ = 2
<br>$$ \Rightarrow$$ $$q$$ = $${1 \over 2}$$
<br>p = 1 - q = 1 - $${1 \over 2}$$ = $${1 \over 2}$$
<br>n = 8
<br><br>Now P(X = 1) = $${}^8{C_1}{\left( {{1 \over 2}} \right)^1}{\left( {{1 \over 2}} \right)^{8 - 1}}$$
<br> &nb... | mcq | aieee-2003 | 7,549 |
WwUXVwpKwqJWZr3N | maths | probability | binomial-distribution | The mean and the variance of a binomial distribution are $$4$$ and $$2$$ respectively. Then the probability of $$2$$ successes is : | [{"identifier": "A", "content": "$${28 \\over 256}$$"}, {"identifier": "B", "content": "$${219 \\over 256}$$"}, {"identifier": "C", "content": "$${128 \\over 256}$$"}, {"identifier": "D", "content": "$${37 \\over 256}$$"}] | ["A"] | null | <p>Given that mean = 4 = np = 4 and variance = 2</p>
<p>npq = 2 $$\Rightarrow$$ 4q = 2</p>
<p>$$ \Rightarrow q = {1 \over 2}$$</p>
<p>$$\therefore$$ $$p = 1 - q = 1 - {1 \over 2} = {1 \over 2}$$</p>
<p>Also, n = 8</p>
<p>Probability of 2 successes $$ = P(X = 2) = {}^8{C_2}{p^2}{q^6}$$</p>
<p>$$ = {{8!} \over {2! \times... | mcq | aieee-2004 | 7,550 |
uWKlDoLk0I45E7uG | maths | probability | binomial-distribution | In a binomial distribution $$B\left( {n,p = {1 \over 4}} \right),$$ if the probability of at least one success is greater than or equal to $${9 \over {10}},$$ then $$n$$ is greater than : | [{"identifier": "A", "content": "$${1 \\over {\\log _{10}^4 + \\log _{10}^3}}$$"}, {"identifier": "B", "content": "$${9 \\over {\\log _{10}^4 - \\log _{10}^3}}$$"}, {"identifier": "C", "content": "$${4 \\over {\\log _{10}^4 - \\log _{10}^3}}$$"}, {"identifier": "D", "content": "$${1 \\over {\\log _{10}^4 - \\log _{10}^... | ["D"] | null | Given that, no of trials = n
<br>Probability of success (p) = $${{1 \over 4}}$$
<br>$$\therefore$$ Probability of no success = 1 - $${{1 \over 4}}$$ = $${{3 \over 4}}$$
<br><br>As we know, probability of at least one success = 1 - probability of no success
<br><br>$$\therefore$$ According to the question,
<br><br>1 - (... | mcq | aieee-2009 | 7,552 |
OE21asjIebiZi6xf | maths | probability | binomial-distribution | Consider $$5$$ independent Bernoulli's trials each with probability of success $$p.$$ If the probability of at least one failure is greater than or equal to $${{31} \over 32},$$ then $$p$$ lies in the interval : | [{"identifier": "A", "content": "$$\\left( {{3 \\over 4},{{11} \\over {12}}} \\right]$$ "}, {"identifier": "B", "content": "$$\\left[ {0,{1 \\over 2}} \\right]$$ "}, {"identifier": "C", "content": "$$\\left( {{11 \\over 12},1} \\right]$$"}, {"identifier": "D", "content": "$$\\left( {{1 \\over 2},{{3} \\over {4}}} \\rig... | ["B"] | null | Here is 5 trials.
<br>So according to Bernoulli trial n = 5
<br><br>P( at least one failure) = 1 - P( no failure)
<br><br>According to question,
<br>1 - P( no failure) $$ \ge {{31} \over {32}}$$
<br><br>$$ \Rightarrow $$ 1 - P( x = 5 ) $$ \ge {{31} \over {32}}$$
<br><br>[<b>Note:</b> no failure = all success]
<br><br>$... | mcq | aieee-2011 | 7,553 |
6W0qksJnJVRp3xfN | maths | probability | binomial-distribution | A multiple choice examination has $$5$$ questions. Each question has three alternative answers of which exactly one is correct. The probability that a student will get $$4$$ or more correct answers just by guessing is : | [{"identifier": "A", "content": "$${{17} \\over {{3^5}}}$$ "}, {"identifier": "B", "content": "$${{13} \\over {{3^5}}}$$"}, {"identifier": "C", "content": "$${{11} \\over {{3^5}}}$$"}, {"identifier": "D", "content": "$${{10} \\over {{3^5}}}$$"}] | ["C"] | null | Each question has 3 alternative and exactly one is correct.
<br><br>$$\therefore$$ Probability of giving correct answer P(correct) = $${1 \over 3}$$
<br><br>$$\therefore$$ Probability of giving wrong answer P(wrong) = $${2 \over 3}$$
<br><br>Here student give 4 or more correct answer.
<br><br>$$\therefore$$ Student giv... | mcq | jee-main-2013-offline | 7,554 |
fPr4yNo79AsCQj24OgUaE | maths | probability | binomial-distribution | An experiment succeeds twice as often as it fails. The probability of at least 5 successes in the six trials of this experiment is : | [{"identifier": "A", "content": "$${{240} \\over {729}}$$ "}, {"identifier": "B", "content": "$${{192} \\over {729}}$$"}, {"identifier": "C", "content": "$${{256} \\over {729}}$$"}, {"identifier": "D", "content": "$${{496} \\over {729}}$$"}] | ["C"] | null | Probability of fail, P(F) = p
<br><br>$$ \therefore $$ Probability of success, P(S) = 2p
<br><br>We know,
<br><br>p + 2Pp = 1
<br><br>$$ \Rightarrow $$ p = $${1 \over 3}$$
<br><br>$$ \therefore $$ Now, probability of at least 5 success in 6 trials,
<br><br>P(x $$ \ge ... | mcq | jee-main-2016-online-10th-april-morning-slot | 7,555 |
TNLBuSQYBYqruPcC | maths | probability | binomial-distribution | A box contains 15 green and 10 yellow balls. If 10 balls are randomly drawn, one-by-one, with
replacement, then the variance of the number of green balls drawn is : | [{"identifier": "A", "content": "6"}, {"identifier": "B", "content": "4"}, {"identifier": "C", "content": "$${6 \\over {25}}$$"}, {"identifier": "D", "content": "$${{12} \\over 5}$$"}] | ["D"] | null | We can apply binomial probability distribution
<br><br>n = 10
<br><br>p = Probability of drawing a green ball = $${{15} \over {25}}$$ = $${3 \over 5}$$
<br><br>Also q = 1 - $${3 \over 5}$$ = $${2 \over 5}$$
<br><br>Variance = npq
<br><br>= $$10 \times {3 \over 5} \times {2 \over 5}$$ = $${{12} \over 5}$$
| mcq | jee-main-2017-offline | 7,556 |
dNyf7SQZWaNvnOFNKi3rsa0w2w9jxb0q6m7 | maths | probability | binomial-distribution | For an initial screening of an admission test, a candidate is given fifty problems to solve. If the probability
that the candidate solve any problem is $${4 \over 5}$$
, then the probability that he is unable to solve less than two
problems is : | [{"identifier": "A", "content": "$${{164} \\over {25}}{\\left( {{1 \\over 5}} \\right)^{48}}$$"}, {"identifier": "B", "content": "$${{316} \\over {25}}{\\left( {{4 \\over 5}} \\right)^{48}}$$"}, {"identifier": "C", "content": "$${{201} \\over 5}{\\left( {{1 \\over 5}} \\right)^{49}}$$"}, {"identifier": "D", "content": ... | ["D"] | null | There are 50 questions in an exam.<br><br>
So Probability of each question to be correct is p = $${4 \over 5}$$<br><br>
and also probability of each question to be incorrect is q = $${1 \over 5}$$<br><br>
Let Y is the number of correct question in 50 questions, hence required probability is <br><br>
$${\left( {{4 \over... | mcq | jee-main-2019-online-12th-april-evening-slot | 7,558 |
y42Fg1dEZL6zPE14iO3rsa0w2w9jx66ep25 | maths | probability | binomial-distribution | Let a random variable X have a binomial distribution with mean 8 and variance 4. If $$P\left( {X \le 2} \right) = {k \over {{2^{16}}}}$$, then k
is equal to : | [{"identifier": "A", "content": "17"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "137"}, {"identifier": "D", "content": "121"}] | ["C"] | null | Let number of trials be n and probability of success = p, probability of failure = q<br><br>
Given np = 8, npq = 4<br><br>
$$ \Rightarrow $$ q = $${1 \over 2}$$, p = $${1 \over 2}$$, n = 16 (as p + q = 1)<br><br>
p$$\left( {x \le 2} \right)$$ = $${{{}^{16}{C_0} + {}^{16}{C_1} + {}^{16}{C_2}} \over {{2^{16}}}} = {{1 + 1... | mcq | jee-main-2019-online-12th-april-morning-slot | 7,559 |
8LM5BOab51jC57AsuYgDh | maths | probability | binomial-distribution | A bag contains 30 white balls and 10 red balls. 16 balls are drawn one by one randomly from the bag with replacement. If X be the number of white balls drawn, then $$\left( {{{mean\,\,of\,X} \over {s\tan dard\,\,deviation\,\,of\,X}}} \right)$$ is equal to : | [{"identifier": "A", "content": "4"}, {"identifier": "B", "content": "$$3\\sqrt 2 $$"}, {"identifier": "C", "content": "$${{4\\sqrt 3 } \\over 3}$$"}, {"identifier": "D", "content": "$$4\\sqrt 3 $$"}] | ["D"] | null | p (probability of getting white ball) = $${{30} \over {40}}$$
<br><br>q = $${1 \over 4}$$ and n = 16
<br><br>mean = np = 16.$${3 \over 4}$$ = 12
<br><br>and standard diviation
<br><br>= $$\sqrt {npq} $$ = $$\sqrt {16.{3 \over 4}.{1 \over 4}} = \sqrt 3 $$<br><br>
$$ \because $$ $$mean \over standard\,deviation$$ = $$12... | mcq | jee-main-2019-online-11th-january-evening-slot | 7,560 |
sLnGz2MXngSYJtUcmwWgT | maths | probability | binomial-distribution | Two cards are drawn successively with replacement from a well-shuffled deck of 52 cards. Let X denote the random variable of number of aces obtained in the two drawn cards. Then P(X = 1) + P (X = 2) equals : | [{"identifier": "A", "content": "$$25 \\over 169$$"}, {"identifier": "B", "content": "$$49\\over 169$$"}, {"identifier": "C", "content": "$$24 \\over 169$$"}, {"identifier": "D", "content": "$$52 \\over 169$$"}] | ["A"] | null | P (X = 1) means out of two drawn cards one card is ace.
<br><br>and P(X = 2) means both the drawn cards are ace.
<br><br>$$ \therefore $$ P(X = 1) = first card is ace or 2nd card is ace.
<br><br>= A _ $$+$$ _ A
<br><br>= $${4 \over {52}} \times {{48} \over {52}} + {{48} \over {52}} \times {4 \over {52}}$$
<... | mcq | jee-main-2019-online-9th-january-morning-slot | 7,561 |
2nkSMMiE7kPpEjjIP47k9k2k5filgzm | maths | probability | binomial-distribution | In a workshop, there are five machines and the probability of any one of them to be out of service on a day is $${{1 \over 4}}$$
. If the probability that at most two machines will be out of service on the same day is $${\left( {{3 \over 4}} \right)^3}k$$, then k is equal to : | [{"identifier": "A", "content": "$${{{17} \\over 4}}$$"}, {"identifier": "B", "content": "$${{{17} \\over 2}}$$"}, {"identifier": "C", "content": "$${{{17} \\over 8}}$$"}, {"identifier": "D", "content": "4"}] | ["C"] | null | Probablity of at most two machines will be out of service = $${\left( {{3 \over 4}} \right)^3}k$$
<br><br>$$ \Rightarrow $$ <sup>5</sup>C<sub>0</sub>$${\left( {{3 \over 4}} \right)^5}$$ + <sup>5</sup>C<sub>1</sub>$$\left( {{1 \over 4}} \right){\left( {{3 \over 4}} \right)^5}$$ + <sup>5</sup>C<sub>2</sub>$${\left( {{1 \... | mcq | jee-main-2020-online-7th-january-evening-slot | 7,562 |
ypdvxtIllDnLKrAoI3jgy2xukfqfwl3g | maths | probability | binomial-distribution | In a bombing attack, there is 50% chance that
a bomb will hit the target. Atleast two
independent hits are required to destroy the
target completely. Then the minimum number
of bombs, that must be dropped to ensure that
there is at least 99% chance of completely
destroying the target, is __________. | [] | null | 11 | Let n is total no. of bombs being dropped
<br><br>at least 2 bombs should hit.
<br><br>P(x > 2) $$ \ge $$ 0.99
<br><br>$$ \Rightarrow $$ 1 - p(x < 2) $$ \ge $$ 0.99
<br><br>$$ \Rightarrow $$ 1 - (p(x = 0) + p(x = 1)) $$ \ge $$ 0.99
<br><br>$$ \Rightarrow $$ 1 - <sup>n</sup>C<sub>0</sub>$${\left( {{1 \over 2}} \ri... | integer | jee-main-2020-online-5th-september-evening-slot | 7,563 |
wcKeZHloIi9ghYWsTA1klrhaah4 | maths | probability | binomial-distribution | An ordinary dice is rolled for a certain number of times. If the probability of getting an odd
number 2 times is equal to the probability of getting an even number 3 times, then the
probability of getting an odd number for odd number of times is : | [{"identifier": "A", "content": "$${5 \\over {36}}$$"}, {"identifier": "B", "content": "$${3 \\over {16}}$$"}, {"identifier": "C", "content": "$${1 \\over 2}$$"}, {"identifier": "D", "content": "$${1 \\over {32}}$$"}] | ["C"] | null | P(odd no. twice) = P(even no. thrice)<br><br>$$ \Rightarrow {}^n{C_2}{\left( {{1 \over 2}} \right)^n} = {}^n{C_3}{\left( {{1 \over 2}} \right)^n} \Rightarrow n = 5$$<br><br>Success is getting an odd number then P(odd successes) = P(1) + P(3) + P(5)<br><br>$$ = {}^5{C_1}{\left( {{1 \over 2}} \right)^5} + {}^5{C_3}{\left... | mcq | jee-main-2021-online-24th-february-morning-slot | 7,564 |
1krxij6cm | maths | probability | binomial-distribution | A student appeared in an examination consisting of 8 true-false type questions. The student guesses the answers with equal probability.
the smallest value of n, so that the probability of guessing at least 'n' correct answers is less than $${1 \over 2}$$, is : | [{"identifier": "A", "content": "5"}, {"identifier": "B", "content": "6"}, {"identifier": "C", "content": "3"}, {"identifier": "D", "content": "4"}] | ["A"] | null | $$P(E) < {1 \over 2}$$<br><br>$$ \Rightarrow \sum\limits_{r = n}^8 {{}^8{C_r}} {\left( {{1 \over 2}} \right)^{8 - r}}{\left( {{1 \over 2}} \right)^r} < {1 \over 2}$$<br><br>$$ \Rightarrow \sum\limits_{r = n}^8 {{}^8{C_r}} {\left( {{1 \over 2}} \right)^8} < {1 \over 2}$$<br><br>$$ \Rightarrow {}^8{C_n} + {}^8{C... | mcq | jee-main-2021-online-27th-july-evening-shift | 7,567 |
1ktfz7m08 | maths | probability | binomial-distribution | Each of the persons A and B independently tosses three fair coins. The probability that both of them get the same number of heads is : | [{"identifier": "A", "content": "$${1 \\over 8}$$"}, {"identifier": "B", "content": "$${5 \\over 8}$$"}, {"identifier": "C", "content": "$${5 \\over 16}$$"}, {"identifier": "D", "content": "1"}] | ["C"] | null | <p>Let x be the number of heads obtained by A, and y be the number of heads obtained by B.</p>
<p>Note that x and y are binomial variable with parameters n = 3 and p = $${1 \over 2}$$</p>
<p>$$\therefore$$ Probability that both A and B obtained the same number of heads is</p>
<p>$$ = P(x = 0)\,.\,P(y = 0) + P(x = 1)\,.... | mcq | jee-main-2021-online-27th-august-evening-shift | 7,568 |
1l57op7rj | maths | probability | binomial-distribution | <p>Let X be a random variable having binomial distribution B(7, p). If P(X = 3) = 5P(x = 4), then the sum of the mean and the variance of X is :</p> | [{"identifier": "A", "content": "$${105 \\over {16}}$$"}, {"identifier": "B", "content": "$${7\\over {16}}$$"}, {"identifier": "C", "content": "$${77\\over {36}}$$"}, {"identifier": "D", "content": "$${49\\over {16}}$$"}] | ["C"] | null | <p>Given P(X = 3) = 5P(X = 4) and n = 7</p>
<p>$$ \Rightarrow {}^7{C_3}{p^3}{q^4} = 5\,.\, \Rightarrow {}^7{C_4}{p^4}{q^3}$$</p>
<p>$$ \Rightarrow q = 5p$$ and also $$p + q = 1$$</p>
<p>$$ \Rightarrow p = {1 \over 6}$$ and $$q = {5 \over 6}$$</p>
<p>Mean $$ = {7 \over 6}$$ and variance $$ = {{35} \over {36}}$$</p>
<p>M... | mcq | jee-main-2022-online-27th-june-morning-shift | 7,569 |
1l58a4o04 | maths | probability | binomial-distribution | <p>Let a biased coin be tossed 5 times. If the probability of getting 4 heads is equal to the probability of getting 5 heads, then the probability of getting atmost two heads is :</p> | [{"identifier": "A", "content": "$${{275} \\over {{6^5}}}$$"}, {"identifier": "B", "content": "$${{36} \\over {{5^4}}}$$"}, {"identifier": "C", "content": "$${{181} \\over {{5^5}}}$$"}, {"identifier": "D", "content": "$${{46} \\over {{6^4}}}$$"}] | ["D"] | null | <p>Coin is tossed 5 times, so n = 5</p>
<p>Let, p = probability of getting heads</p>
<p>q = probability of getting tails.</p>
<p>$$\therefore$$ p + q = 1 ...... (1)</p>
<p>$$\therefore$$ Probability of getting 4 heads</p>
<p>= <sup>5</sup>C<sub>4</sub> . p<sup>4</sup> . q</p>
<p>And probability of getting 5 heads</p>
<... | mcq | jee-main-2022-online-26th-june-morning-shift | 7,570 |
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