Permutation & Combination – JEE Advanced Previous Year Questions with Solutions
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Q. The number of seven digit integers, with sum of the digits equal to 10 and formed by using the digits 1, 2 and 3 only, is (A) 55            (B) 66           (C) 77               (D) 88 [JEE 2009, 3]

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Sol. (C) For sum to be 10 Case-I $\quad 1,2,3,1,1,1,1,=\frac{7 !}{5 !}$ Case-II $\quad 2,2,2,1,1,1,1,=\frac{7 !}{3 ! 4 !}$ So total number $=77$

Q. Let S = {1,2,3,4}. The total number of unordered pairs of disjoint subsets of S is equal to – (A) 25           (B) 34           (C) 42             (D) 41 [JEE 10, 5M, –2M]

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Sol. (D) Case- $\mathrm{I}:$ The number of elements in the pairs $(\mathrm{A}, \mathrm{B})$ can be $(1,1) ;(1,2) ;(1,3) ;(2,2)$ $=^{4} \mathrm{C}_{2}+^{4} \mathrm{C}_{1} \times^{3} \mathrm{C}_{2}+^{4} \mathrm{C}_{1} \times^{3} \mathrm{C}_{3}+\frac{^{4} \mathrm{C}_{2} \cdot^{2} \mathrm{C}_{2}}{2}=25$ Case- $\mathrm{II}:$ Number of pairs with $\phi$ as one of subsets $=2^{4}=16$ $\therefore$ Total pairs $=25+16=41$

Q. The total number of ways in which 5 balls of different colours can be distributed among 3 persons so that each person gets at least one ball is – (A) 75           (B) 150           (C) 210              (D) 243 [JEE 2012, 3M, –1M]

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Sol. (B) Balls can be distributed as $1,1,3$ or $1,2,2$ to each person. When $1,1,3$ balls are distributed to each person, then total number of ways: $=\frac{5 !}{1 ! 1 ! 3 !} \cdot \frac{1}{2 !} 3 !=60$ When $1,2,2$ balls are distributed to each person, then total number of ways : $=\frac{5 !}{1 ! 2 ! 2 !} \cdot \frac{1}{2 !} 3 !=90$ $\therefore$ total $=60+90=150$

Let $\mathrm{a}_{\mathrm{n}}$ denotes the number of all n-digit positive integers formed by the digits 0, 1 or both such that no consecutive digits in them are 0. Let $b_{n}$ = the number of such n-digit integers ending with digit 1 and $\mathrm{C}_{\mathrm{n}}$ = the number of such n-digit integers ending with digit 0.
Q. The value of $\mathrm{b}_{6}$ is (A) 7            (B) 8          (C) 9             (D) 11 [JEE 2012, 3M, –1M]

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Sol. (B) $\mathrm{a}_{1}=1, \mathrm{a}_{2}=2$ So $a_{3}=3, \quad a_{4}=5 a_{5}=8$ $\Rightarrow b_{6}=a_{5}=8$

Q. Which of the following is correct ? (A) $a_{17}=a_{16}+a_{15}$ (B) $\mathrm{c}_{17} \neq \mathrm{c}_{16}+\mathrm{c}_{15}$ (C) $\mathrm{b}_{17} \neq \mathrm{b}_{16}+\mathrm{c}_{16}$ (D) $a_{17}=c_{17}+b_{16}$ [JEE 2012, 3M, –1M]

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Sol. (A) $a_{n}=a_{n-1}+a_{n-2}$ put $n=17$ $a_{17}=a_{16}+a_{15} \quad($ A) is correct $c_{n}=c_{n-1}+c_{n-2}$ $\mathrm{c}_{\mathrm{n}}=\mathrm{c}_{\mathrm{n}-1}+\mathrm{c}_{\mathrm{n}-2}$ $\mathrm{So}$ put $\mathrm{n}=17$ $\mathrm{c}_{17}=\mathrm{c}_{16}+\mathrm{c}_{15} \quad(\mathrm{B})$ is incorrect $\mathrm{b}_{\mathrm{n}}=\mathrm{b}_{\mathrm{n}-1}+\mathrm{b}_{\mathrm{n}-2}$ put $\mathrm{n}=17$ $\mathrm{b}_{17}=\mathrm{b}_{16}+\mathrm{b}_{15} \quad(\mathrm{C})$ is incorrect $\mathrm{a}_{17}=\mathrm{a}_{16}+\mathrm{a}_{15}$ while $(\mathrm{D})$ says $\mathrm{a}_{17}=\mathrm{a}_{15}+\mathrm{a}_{15} \quad$ (D) is incorrect

Q. Let $n_{1}<n_{2}<n_{3}<n_{4}<n_{5}$ be positive integers such that $n_{1}+n_{2}+n_{3}+n_{4}+n_{5}=20 .$ The the number of such distinct arrangements $\left(n_{1}, n_{2}, n_{3}, n_{4}, n_{5}\right)$ is [JEE(Advanced)-2014, 3]

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Sol. 5 as $n_{1} \geq 1, n_{2} \geq 2, n_{3} \geq 3, n_{4} \geq 4, n_{5} \geq 5$ Let $n_{1}-1=x_{1} \geq 0, n_{2}-2=x_{2} \geq 0 \ldots \ldots \ldots \ldots n_{5}-5=x_{5} \geq 0$ $\Rightarrow$ New equation will be $\quad x_{1}+1+x_{2}+2+\ldots \ldots x_{5}+5=20$ $\Rightarrow x_{1}+x_{2}+x_{3}+x_{4}+x_{5}=20-15=5$ Now $x_{1} \leq x_{2} \leq x_{3} \leq x_{4} \leq x_{5}$ AND Numbr of red line segments $=^{\mathrm{n}} \mathrm{C}_{2}-\mathrm{n}$ Number of blue line segments $=\mathrm{n}$ $\therefore^{\mathrm{n}} \mathrm{C}_{2}-\mathrm{n}=\mathrm{n}$ $\frac{\mathrm{n}(\mathrm{n}-1)}{2}=2 \mathrm{n} \Rightarrow \mathrm{n}=5$ Ans.

Q. Let n $\geq$ 2 b an integer. Take n distinct points on a circle and join each pair of points by a line segment. Colour the line segment joining every pair of adjacent points by blue and the rest by red. If the number of red and blue line segments are equal, then the value of n is [JEE(Advanced)-2014, 3]

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Sol. 7 Q. Six cards and six envelopes are numbered 1, 2, 3, 4, 5, 6 and cards are to be placed in envelopes so that each envelope contains exactly one card and no card is placed in the envelope bearing the same number and moreover the card numbered 1 in always placed in envelope numbered 2. Then the number of ways it can be done is – (A) 264         (B) 265         (C) 53             (D) 67 [JEE(Advanced)-2014, 3(–1)]

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Sol. (C) Total number of dearrangement 6! $\left[\frac{1}{2 !}-\frac{1}{3 !}+\frac{1}{4 !}-\frac{1}{5 !}+\frac{1}{6 !}\right]$ $=360-120+30-6+1$ $=240+25=265$ There are equal chances that card 1 goes into any envelope from 2 to $6 \quad \therefore \quad \frac{1}{5}(265)=53$

Q. Let n be the number of ways in which 5 boys and 5 girls can stand in a queue in such a way that all the girls stand consecutively in the queue. Let m be the number of ways in which 5 boys and 5 girls can stand in a queue in such a way that exactly four girls stand consecutively in the queue. Then the value of $\frac{\mathrm{m}}{\mathrm{n}}$ is [JEE (Advanced) 2015, 4M, –0M]

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Sol. 5 $n=6 ! \times 5 !$ $n=6 ! \times 4 ! \times 5 \times^{5} C_{4}$ $\frac{m}{n}=\frac{6 ! \times 4 ! \times 5 \times^{5} C_{4}}{6 ! \times 5 !}=5$

Q. A debate club consists of 6 girls and 4 body. A team of 4 members is to be selected from this club including the selection of a captain (from among these 4 member) for the team. If the team has to include at most one boy, then the number of ways of selecting the team is (A) 380            (B) 320              (C) 260              (D) 95 [JEE (Advanced) 2016]

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Sol. (A)

Q. Words of length 10 are formed using the letters A, B, C, D, E, F, G, H, I, J. Let x be the number of such words where no letter is repeated; and let y be the number of such words where exactly one letter is repeated twice and no other letter is repeated. Then, $\frac{\mathrm{y}}{9 \mathrm{x}}$ = [JEE (Advanced) 2017]

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Sol. 5 $\mathrm{x}=10 !$ $\mathrm{y}=^{10} \mathrm{C}_{1}^{9} \mathrm{C}_{8} \frac{10 !}{2 !}$

Q. Let $S=\{1,2,3, \ldots \ldots, 9\} .$ For $\mathrm{k}=1,2, \ldots . ., 5,$ let $\mathrm{N}_{\mathrm{k}}$ be the number of subsets of $\mathrm{S},$ each containing five elements out of which exactly $\mathrm{k}$ are odd. Then $\mathrm{N}_{1}+\mathrm{N}_{2}+\mathrm{N}_{3}+\mathrm{N}_{4}+\mathrm{N}_{5}=$b (A) 125             (B) 252            (C) 210             (D) 126 [JEE (Advanced) 2017]

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Sol. (D) $\mathrm{N}_{1}+\mathrm{N}_{2}+\mathrm{N}_{3}+\mathrm{N}_{4}+\mathrm{N}_{5}=$ Total ways $-\{\text { when no odd }\}$ Total ways $=^{9} \mathrm{C}_{5}$ Number of ways when no odd, is zero $\quad(\because \text { only available even are } 2,4,6,8)$ $\therefore$ Ans $:^{9} \mathrm{C}_{5}-$ zero $=126$

Q. The number of 5 digit numbers which are divisible by 4, with digits from the set {1, 2, 3, 4, 5} and the repetition of digits is allowed, is —— [JEE (Advanced) 2018,3(0)]

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Sol. 625 Option for last two digits are $(12),(24),(32),(44)$ are $(52) .$ $\therefore \quad$ Total No. of digits $=5 \times 5 \times 5 \times 5=625$

Q. In a high school, a committee has to be formed from a group of 6 boys $\mathrm{M}_{1}, \mathrm{M}_{2}, \mathrm{M}_{3}, \mathrm{M}_{4}, \mathrm{M}_{5},$ $\mathrm{M}_{6}$ and 5 girls $\mathrm{G}_{1}, \mathrm{G}_{2}, \mathrm{G}_{3}, \mathrm{G}_{4}, \mathrm{G}_{5}$ (i) Let $\alpha_{1}$ be the total number of ways in which the committee can be formed such that the committee has 5 members, having exactly 3 body and 2 girls. (ii) Let $\alpha_{2}$ be the total number of ways in which the committe can be formed such that the committee has at least 2 members, and having an equal number of boys and girls. (iii) Let $\alpha_{3}$ be the total number of ways in which the committe can be formed such that the committee has 5 members, at least 2 of them being girls. (iv) Let $\alpha_{4}$ be the total number of ways in which the committee can be formed such that the commitee has 4 members, having at least 2 girls and such that both $\mathrm{M}_{1}$ and $\mathrm{G}_{1}$ are NOT in the committee together.  [JEE (Advanced) 2018,3(–1)]

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Sol. (C)  • January 14, 2021 at 1:47 pm

I THINK ANSWER TO QUESTION 3 FROM THE TOP IS WRONG… IT SHOULD BE 243-3=240 BECAUSE BALLS ARE DISTINCT SO IT SHOULD BE( 3^(5))-3

13
• January 14, 2021 at 1:48 pm

ESARAL PLZZ CLARIFY

3
• January 16, 2021 at 5:09 pm

It is correct 2cases are formed 1) when 5 distributed as (1,1,3) and 2) as(1,2,2 )and then add these cases 150 is answer
5!/1!1!1!3! *3!/2! + 5!/2!2!1!*3!/2!=150

2
• September 27, 2021 at 12:15 pm

Hey i also got the same ans .
First way to distribute would be in 1 1 3 so total Ways in which such groups formed will be 5C3 = 10 and no. Of ways to distribute them in 3 people would be 3!*10 = 60
Similarly for 1 2 2 no. Of ways to distribute then in 3 people will be = 5C2 * 3C2 * 1C1 * 3! = 180
So total Ways = 180+60 = 240 .

0
• December 31, 2020 at 9:05 am

which ans is wrong?

1
• December 19, 2020 at 8:27 pm

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15
• December 18, 2020 at 7:34 pm

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0
• October 15, 2020 at 11:21 pm

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4
• October 15, 2020 at 11:22 pm

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0
• September 15, 2020 at 2:52 pm

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0
• September 5, 2020 at 9:57 pm

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0
• September 5, 2020 at 9:57 pm

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0
• May 20, 2020 at 7:20 pm

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0