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

Q 11 :
Suppose Anil's mother wants to give 5 whole fruits to Anil from a basket of 7 red apples, 5 white apples and 8 oranges. If in the selected 5 fruits, at least 2 oranges, at least one red apple and at least one white apple must be given, then the number of ways, Anil's mother can offer 5 fruits to Anil is _______ . [2023]

0%

(6860)

$Let 7 red apples be denoted by (A), 5 white apples by (B), 8 oranges by (C) .$

$Now, 5 fruits are to be selected (Note: fruits taken different).$

$Possible selections: (2 C, 1 A, 2 B) or (2 C, 2 A, 1 B) or (3 C, 1 A, 1 B)$

$= {}^{8}{C_{2}} {}^{7}{C_{1}} {}^{5}{C_{2}} + {}^{8}{C_{2}} {}^{7}{C_{2}} {}^{5}{C_{1}} + {}^{8}{C_{3}} {}^{7}{C_{1}} {}^{5}{C_{1}}$

$= 28 \times 7 \times 10 + 28 \times 21 \times 5 + 56 \times 7 \times 5$

$= 1960 + 2940 + 1960 = 6860$

Q 12 :
If all the six digit numbers $x_{1} x_{2} x_{3} x_{4} x_{5} x_{6}$ with $0 < x_{1} < x_{2} < x_{3} < x_{4} < x_{5} < x_{6}$ are arranged in the increasing order, then the sum of the digits in the 72th number is ___________ . [2023]

0%

(32)

[IMAGE 23]

2 4 5 6 7 8 → 72th word. So, 2 + 4 + 5 + 6 + 7 + 8 = 32

Q 13 :
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

(4)

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 14 :
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

(2)

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 15 :
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

(3)

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 16 :
${}^{n - 1}{C_{r}} = (k^{2} - 8) {}^{n}{C_{r + 1}}$ 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

(1)

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 17 :
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

(1)

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 18 :
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 19 :
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 20 :
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 $= {}^{5}{C_{3} \times 3! - 4 \times 3! = 6 \times 3! = 36}$

Topic Question Set

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Q 12 :
If all the six digit numbers $x_{1} x_{2} x_{3} x_{4} x_{5} x_{6}$ with $0 < x_{1} < x_{2} < x_{3} < x_{4} < x_{5} < x_{6}$ are arranged in the increasing order, then the sum of the digits in the 72th number is ___________ . [2023]

0%

Q 13 :
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 14 :
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 15 :
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 16 :
${}^{n - 1}{C_{r}} = (k^{2} - 8) {}^{n}{C_{r + 1}}$ if and only if [2024]

Q 17 :
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 19 :
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 20 :
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

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Q 12 : If all the six digit numbers x1 x2 x3 x4 x5 x6 with 0<x1<x2<x3<x4<x5<x6 are arranged in the increasing order, then the sum of the digits in the 72th number is ___________ . [2023]

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Q 13 : 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 14 : Let 0≤r≤n. If Cr+1n+1:Cr:Cr-1n-1n=55:35:21, then 2n+5r is equal to [2024]

Q 15 : 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 16 : Crn-1=(k2-8)Cr+1n if and only if [2024]

Q 17 : 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 18 : 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 19 : 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 20 : 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%

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Q 12 :
If all the six digit numbers $x_{1} x_{2} x_{3} x_{4} x_{5} x_{6}$ with $0 < x_{1} < x_{2} < x_{3} < x_{4} < x_{5} < x_{6}$ are arranged in the increasing order, then the sum of the digits in the 72th number is ___________ . [2023]

Q 13 :
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 14 :
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 15 :
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 16 :
${}^{n - 1}{C_{r}} = (k^{2} - 8) {}^{n}{C_{r + 1}}$ if and only if [2024]

Q 17 :
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 19 :
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 20 :
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]