Permutations and Combinations jee mains questions | JEE Main Maths Permutations and Combinations Previous Year Question

Q 1 :
Let the sets $S = {2, 4, 8, 16, \dots, 512}$ be partitioned into 3 sets $A, B, C$ with equal number of elements such that $A \cup B \cup C = S$ and $A \cap B = B \cap C = A \cap C = ϕ .$ The maximum number of such possible partitions of $S$ is equal to [2024]

1520
1640
1710
1680

1

2

3

4

(D)

We have, $S = {2, 4, 8, 16, \dots, 512}$

$= {2^{1}, 2^{2}, 2^{3}, 2^{4}, \dots, 2^{9}}$ so, $| S | = 9$

$| A | = | B | = | C |$ and $A \cup B \cup C = S$

and $A \cap B = B \cap C = A \cap C = ϕ$

Since, $| S | = 9$ so $| A | = | B | = | C | = 3$

So, number of ways of making partition of $S = {}^{9}C_{3} \times {}^{6}{C_{3}} \times {}^{3}{C_{3}}$

$= \frac{9!}{6! 3!} \times \frac{6!}{3! \times 3!} \times 1$

$= \frac{9!}{3! \times 3! \times 3!} = 1680$

Q 2 :
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 [2024]

55
50
60
62

1

2

3

4

(B)

We have, ${}^{n + 1}{C_{r + 1} : {}^{n}{C_{r}}} = 55 : 35$

$\Rightarrow \frac{(n + 1)!}{(r + 1)! (n - r)!} \times \frac{(n - r)! r!}{n!} = \frac{55}{35}$

$\Rightarrow \frac{n + 1}{r + 1} = \frac{11}{7}$

$\Rightarrow 7 n = 11 r + 4$ ...(i)

Also, ${}^{n}{C_{r}} : {}^{n - 1}{C_{r - 1}} = 35 : 21$

$\Rightarrow \frac{n!}{r! (n - r)!} \times \frac{(r - 1)! (n - r)!}{(n - 1)!} = \frac{35}{21}$

$\Rightarrow \frac{n}{r} = \frac{5}{3}$

$\Rightarrow 3 n = 5 r$ ...(ii)

Solving (i) and (ii), we get

$\Rightarrow 7 (\frac{5 r}{3}) = 11 r + 4$

$\Rightarrow 35 r - 33 r = 12$

$\Rightarrow r = 6$ and $n = 10$

So, $2 n + 5 r = 20 + 30 = 50$

Q 3 :
The number of ways five alphabets can be chosen from the alphabets of the word MATHEMATICS, where the chosen alphabets are not necessarily distinct, is equal to [2024]

175
181
179
177

1

2

3

4

(C)

We have, MMAATTHEICS

Case-I : All distinct $= {}^{8}{C_{5}} = 56$

Case-II : 2 identical + 3 distinct $= {}^{3}{C_{1} \times {}^{7}{C_{3} = 105}}$

Case-III : 2 identical + 2 identical + 1 distinct $= {}^{3}{C_{2} \times {}^{6}{C_{1}} = 18}$

Hence, total number of ways = 56 + 105 + 18 = 179

Q 4 :
${}^{n - 1}{C_{r}} = (k^{2} - 8) {}^{n}{C_{r + 1}}$ if and only if [2024]

$2 \sqrt{2} < k \leq 3$
$2 \sqrt{3} < k \leq 3 \sqrt{2}$
$2 \sqrt{2} < k < 2 \sqrt{3}$
$2 \sqrt{3} < k < 3 \sqrt{3}$

1

2

3

4

(A)

We have, ${}^{n - 1}{C_{r}} = (k^{2} - 8) {}^{n}{C_{r + 1}}$

$\Rightarrow \frac{(n - 1)!}{(n - (1 + r))! r!} = (k^{2} - 8) \frac{n!}{(n - (r + 1))! (r + 1)!}$

$\Rightarrow 1 = \frac{(k^{2} - 8) n}{r + 1} \Rightarrow k^{2} - 8 = \frac{r + 1}{n} \leq 1$ $[∵ n \geq r + 1]$

$\Rightarrow k^{2} - 8 \leq 1 \Rightarrow k^{2} - 9 \leq 0 \Rightarrow (k - 3) (k + 3) \leq 0$

$\Rightarrow - 3 \leq k \leq 3$ ...(i)

But $k^{2} > 8,$ as ${}^{n - 1}{C_{r}}$ can't be negative and 0.

$\Rightarrow k^{2} - 8 > 0 \Rightarrow (k - 2 \sqrt{2}) (k + 2 \sqrt{2}) > 0$

$\Rightarrow k < - 2 \sqrt{2} or k > 2 \sqrt{2}$ ...(ii)

From (i) and (ii), $2 \sqrt{2} < k \leq 3 or - 3 \leq k < - 2 \sqrt{2}$

Q 5 :
The number of ways in which 21 identical apples can be distributed among three children such that each child gets at least 2 apples, is [2024]

136
406
142
130

1

2

3

4

(A)

We have, 21 identical apples.

Let $x_{1}, x_{2}$ and $x_{3}$ be the number of apples received by the three children.

$∴$ $x_{1} + x_{2} + x_{3} = 21$

Now, let $x_{1}^{'} = x_{1} - 2, x_{2}^{'} = x_{2} - 2$ and $x_{3}^{'} = x_{3} - 2$

$∴$ $x_{1}^{'} + x_{2}^{'} + x_{3}^{'} = 21 - 6 = 15$

$∴$ Required number of ways $= {}^{15 + 3 - 1}{C_{3 - 1}}$

$= {}^{17}{C_{2}} = 136$

Q 6 :
Let $a = 1 + \frac{{}^{2}{C_{2}}}{3!} + \frac{{}^{3}{C_{2}}}{4!} + \frac{{}^{4}{C_{2}}}{5!} + \dots,$ $b = 1 + \frac{{}^{1}{C_{0}} +^{1} C_{1}}{1!} + \frac{{}^{2}{C_{0}} +^{2} C_{1} +^{2} C_{2}}{2!} + \frac{{}^{3}{C_{0}} +^{3} C_{1} +^{3} C_{2} +^{3} C_{3}}{3!} + \dots$ Then $\frac{2 b}{a^{2}}$ is equal to ______ . [2024]

0%

(8)

Given, $a = 1 + \frac{{}^{2}C_{2}}{3!} + \frac{{}^{3}C_{2}}{4!} + \frac{{}^{4}C_{2}}{5!} + \dots,$ $b = 1 + \frac{{}^{1}C_{0} + {}^{1}C_{1}}{1!} + \frac{{}^{2}C_{0} + {}^{2}C_{1} + {}^{2}C_{2}}{2!} + . . .,$

$\Rightarrow b = 1 + \frac{2}{1!} + \frac{2^{2}}{2!} + \frac{2^{3}}{3!} + \dots = e^{2}$ $[∵ e^{x} = 1 + \frac{x}{1!} + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + \dots]$

Now, $a = 1 + \sum_{r = 2}^{\infty} \frac{{}^{r}C_{2}}{(r + 1)!} = 1 + \sum_{r = 2}^{\infty} \frac{r (r - 1)}{2 (r + 1)!}$

$= 1 + \frac{1}{2} \sum_{r = 2}^{\infty} \frac{(r + 1) r - 2 r}{(r - 1)!} = 1 + \frac{1}{2} \sum_{r = 2}^{\infty} \frac{1}{(r - 1)!} - \frac{1}{2} \sum_{r = 2}^{\infty} \frac{2 r}{(r + 1)!}$

$= 1 + \frac{1}{2} (\frac{1}{1!} + \frac{1}{2!} + . . .) - \sum_{r = 2}^{\infty} \frac{(r + 1) - 1}{(r + 1)!}$

$= 1 + \frac{1}{2} (e - 1) - \sum_{r = 2}^{\infty} \frac{1}{r!} + \sum_{r = 2}^{\infty} \frac{1}{(r + 1)!}$

$= 1 + \frac{1}{2} (e - 1) - (e - \frac{1}{1!} - \frac{1}{0!}) + (e - \frac{1}{1!} - \frac{1}{0!} - \frac{1}{2!})$

$= 1 + \frac{e}{2} - \frac{1}{2} - e + 2 + e - 2 - \frac{1}{2} = \frac{e}{2} \Rightarrow \frac{2 b}{a^{2}} = \frac{2 e^{2}}{\frac{e^{2}}{4}} = 8$

Q 7 :
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 _______. [2024]

0%

(5626)

Group A	Group B	Ways
$4 m$	$4 w$	${}^{4}{C_{4}} \cdot^{4} C_{4} = 1$
$3 m + 1 w$	$1 m + 3 w$	${}^{4}{C_{3}} \cdot^{5} C_{1} \cdot^{5} {C_{1}}^{4} C_{3} = 400$
$2 m + 2 w$	$2 m + 2 w$	${}^{4}{C_{2}} \cdot^{5} {C_{2}}^{5} {C_{2}}^{4} C_{2} = 3600$
$1 m + 3 w$	$3 m + w$	${}^{4}{C_{1}}^{5} {C_{3}}^{5} {C_{3}}^{4} C_{1} = 1600$
$4 w$	$4 m$	${}^{5}{C_{4}}^{5} C_{4} = 25$

Total number of ways = 1 + 400 + 3600 + 1600 + 25 = 5626

Q 8 :
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 _______. [2024]

0%

(36)

Possible triplets for which number is divisible by 3 are (2, 3, 4), (2, 3, 7), (3, 5, 4), (3, 5, 7)

These 4 triplets can be arrange in $4 \times 3! .$

Total number of 3-digit numbers made by using the digits 2, 3, 4, 5, and 7 $= {}^{5}{C_{3} \times 3!}$

Required number which are not divisible by 3 $= {}^{5}{C_{3} \times 3! - 4 \times 3! = 6 \times 3! = 36}$

Q 9 :
The lines $L_{1}, L_{2}, . . . L_{20}$ are distinct. For $n = 1, 2, 3, . . ., 10$ all the lines $L_{2 n - 1}$ are parallel to each other and all the lines $L_{2 n}$ pass through a given point P. The maximum number of points of intersection of pairs of lines from the set ${L_{1}, L_{2}, . . ., L_{20}}$ is equal to ______. [2024]

0%

(101)

Let $L_{1}, L_{3}, . . ., L_{19}$ are parallel to each other and $L_{2}, L_{4}, L_{6}, \dots, L_{20}$ are passing through a point P.

$∴$ Point of intersection of pairs of lines from the set ${L_{1}, L_{2}, \dots, L_{20}} = {}^{20}{C_{2} - {}^{10}{C_{2}} - {}^{10}{C_{2}}} + 1 = 101$

Q 10 :
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 and section C has 6 questions, then the total number of ways a student can select 15 questions is ______. [2024]

0%

(11376)

A B C Number of ways

5 6 4 ${}^{8}{C_{5}} \times {}^{6}{C_{6}} \times {}^{6}{C_{4}} = 56 \times 1 \times 15 = 840$

6 5 4 ${}^{8}{C_{6}} \times {}^{6}{C_{5}} \times {}^{6}{C_{4}} = 28 \times 6 \times 15 = 2520$

6 4 5 ${}^{8}{C_{6}} \times {}^{6}{C_{4}} \times {}^{6}{C_{5}} = 28 \times 15 \times 6 = 2520$

5 5 5 ${}^{8}{C_{5}} \times {}^{6}{C_{5}} \times {}^{6}{C_{5}} = 56 \times 6 \times 6 = 2016$

4 6 5 ${}^{8}{C_{4}} \times {}^{6}{C_{6}} \times {}^{6}{C_{5}} = 70 \times 1 \times 6 = 420$

4 5 6 ${}^{8}{C_{4}} \times {}^{6}{C_{5}} \times {}^{6}{C_{6}} = 70 \times 6 \times 1 = 420$

5 4 6 ${}^{8}{C_{5}} \times {}^{6}{C_{4}} \times {}^{6}{C_{6}} = 56 \times 15 \times 1 = 840$

7 4 4 ${}^{8}{C_{7}} \times {}^{6}{C_{4}} \times {}^{6}{C_{4}} = 8 \times 15 \times 15 = 1800$

$∴$ Required number of ways = 840 + 2520 + 2520 + 2016 + 420 + 420 + 840 = 11376

Topic Question Set

Q 1 :
Let the sets $S = {2, 4, 8, 16, \dots, 512}$ be partitioned into 3 sets $A, B, C$ with equal number of elements such that $A \cup B \cup C = S$ and $A \cap B = B \cap C = A \cap C = ϕ .$ The maximum number of such possible partitions of $S$ is equal to [2024]

Q 2 :
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 [2024]

Q 3 :
The number of ways five alphabets can be chosen from the alphabets of the word MATHEMATICS, where the chosen alphabets are not necessarily distinct, is equal to [2024]

Q 4 :
${}^{n - 1}{C_{r}} = (k^{2} - 8) {}^{n}{C_{r + 1}}$ if and only if [2024]

Q 5 :
The number of ways in which 21 identical apples can be distributed among three children such that each child gets at least 2 apples, is [2024]

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Q 7 :
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 _______. [2024]

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Q 8 :
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 _______. [2024]

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Topic Question Set

Q 1 : Let the sets S={2,4,8,16,…,512} be partitioned into 3 sets A,B,C with equal number of elements such that A∪B∪C=S and A∩B=B∩C=A∩C=ϕ. The maximum number of such possible partitions of S is equal to [2024]

Q 2 : Let 0≤r≤n. If Cr+1n+1:Cr:Cr-1n-1n=55:35:21, then 2n+5r is equal to [2024]

Q 3 : The number of ways five alphabets can be chosen from the alphabets of the word MATHEMATICS, where the chosen alphabets are not necessarily distinct, is equal to [2024]

Q 4 : Crn-1=(k2-8)Cr+1n if and only if [2024]

Q 5 : The number of ways in which 21 identical apples can be distributed among three children such that each child gets at least 2 apples, is [2024]

Q 6 : Let a=1+C223!+C234!+C245!+…,b=1+C01+1C11!+C02+2C1+2C22!+C03+3C1+3C2+3C33!+… Then 2ba2 is equal to ______ . [2024]

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Q 7 : 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 _______. [2024]

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Q 8 : 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 _______. [2024]

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Q 9 : The lines L1,L2,...L20 are distinct. For n=1,2,3,...,10 all the lines L2n-1 are parallel to each other and all the lines L2n pass through a given point P. The maximum number of points of intersection of pairs of lines from the set {L1,L2,...,L20} is equal to ______. [2024]

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Q 1 :
Let the sets $S = {2, 4, 8, 16, \dots, 512}$ be partitioned into 3 sets $A, B, C$ with equal number of elements such that $A \cup B \cup C = S$ and $A \cap B = B \cap C = A \cap C = ϕ .$ The maximum number of such possible partitions of $S$ is equal to [2024]

Q 2 :
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 [2024]

Q 3 :
The number of ways five alphabets can be chosen from the alphabets of the word MATHEMATICS, where the chosen alphabets are not necessarily distinct, is equal to [2024]

Q 4 :
${}^{n - 1}{C_{r}} = (k^{2} - 8) {}^{n}{C_{r + 1}}$ if and only if [2024]

Q 5 :
The number of ways in which 21 identical apples can be distributed among three children such that each child gets at least 2 apples, is [2024]

Q 7 :
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 _______. [2024]

Q 8 :
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 _______. [2024]