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

Q 31 :

From all the English alphabets, five letters are chosen and are arranged in alphabetical order. The total number of ways, in which the middle letter is 'M', is : [2025]

14950
5148
4356
6084

1

2

3

4

(2)

Number of English alphabets before M = 12

Number of English alphabets after M = 13

$∴$ Required number of ways

= ${}^{12}C_{2} \times {}^{13}C_{2} = 66 \times 78$ = 5148

Q 32 :

Group A consists of 7 boys and 3 girls, while group B consists of 6 boys and 5 girls. The number of ways, 4 boys and 4 girls can be invited for a picnic if 5 of them must be from group A and the remaining 3 from group B, is equal to : [2025]

9100
8925
8750
8575

1

2

3

4

(2)

Group A : 7B and 3G

Group B : 6B and 5G.

The number of ways of inviting 4 boys and 4 girls (5 from Group A and 3 from Group B) is gven by

= (3G + 2B, 1G + 2B) + (2G + 3B, 2G + 1B) + (1G + 4B, 3G + 0B)

$= {}^{3}C_{3} \cdot {}^{7}C_{2} \cdot {}^{5}C_{1} \cdot {}^{6}C_{2} + {}^{3}C_{2} \cdot {}^{7}C_{3} \cdot {}^{5}C_{2} \cdot {}^{6}C_{1} + {}^{3}C_{1} \cdot {}^{7}C_{4} \cdot {}^{5}C_{3} \cdot {}^{6}C_{0}$

= 1575 + 6300 + 1050 = 8925.

Q 33 :

Let S be the set of all the words that can be formed by arranging all the letters of the word GARDEN. From the set S, one word is selected at random. The probability that the selected word will NOT have vowels in alphabetical order is : [2025]

$\frac{1}{3}$
$\frac{2}{3}$
$\frac{1}{4}$
$\frac{1}{2}$

1

2

3

4

(4)

We have, A E G R D N

Total cases = 6!

Number of cases when A and E are in order = ${}^{6}C_{2} \times 4!$

Probability, when A and E are in order = $\frac{{}^{6}C_{2} \times 4!}{6!} = \frac{1}{2}$

Probability when A and E are not in order = $1 - \frac{1}{2} = \frac{1}{2}$

Q 34 :

For $n \geq 2$ , let $S_{n}$ denote the set of all subsets of {1, 2, ..., n} with no two consecutive numbers. For example ${1, 3, 5} \in S_{6}$ but ${1, 2, 4} \notin S_{6}$ . Then $n (S_{5})$ is equal to __________. [2025]

0%

(13)

Let A = {1, 2, 3, 4, 5, ..., n}

Number of subsets having $r$ elements such that no two are consecutive = ${}^{n - r + 1}C_{r}$

For n = 5, number of ways = ${}^{6 - r}C_{r}$

Subsets having no elements = 1 i.e., $ϕ$

Subsets having exactly 1 element = ${}^{5}C_{1}$ = 5 i.e., {1}, {2}, {3}, {4}, {5}

Subsets having exactly 2 elements = ${}^{4}C_{2}$ = 6 i.e., {1, 3}, {1, 4}, {1, 5}, {2, 4}, {2, 5}, {3, 5}

Subsets having exactly 3 elements = ${}^{3}C_{3}$ = 1 i.e., {1, 3, 5}

$∴ n (S_{5})$ = 1 + 5 + 6 + 1 = 13

Q 35 :

The number of singular matrices of order 2, whose elements are from the set {2, 3, 6, 9}, is __________. [2025]

0%

(36)

For any singular matrix $[\begin{matrix} a & c \\ b & d \end{matrix}]$ ; we have

$| \begin{matrix} a & c \\ b & d \end{matrix} |$ $= a d - b c = 0 \Rightarrow a d = b c$

Case I : Exactly one number is used

$\Rightarrow$ All like, then required number = ${}^{4}C_{1}$ = 4

Case II : Exactly two number i.e., (a, a, a, b or a, a, b, b) are used, then required number

$\Rightarrow {}^{4}C_{2} \times 2 \times 2 = 24$

Case III : Exactly three numbers i.e., (a, a, b, c) are used, then none will be singular.

Case IV : Exactly four number i.e., (a, b, c, d) are used, then ad = bc i.e.,

$2 \times 9 = 3 \times 6 \Rightarrow {}^{4}C_{1} \times 2! = 8$

Total number of matrices = 36.

Q 36 :

If $\sum_{r = 0}^{5} \frac{{}^{11}C_{2 r + 1}}{2 r + 2} = \frac{m}{n}$ , gcd (m, n) = 1, then m – n is equal to __________. [2025]

0%

(2035)

We know that,

$C_{0} + \frac{C_{1}}{2} + \frac{C_{2}}{3} + . . . . . . + \frac{C_{n}}{n + 1} = \frac{2^{n + 1} - 1}{(n + 1)}$ ... (i)

$C_{0} - \frac{C_{1}}{2} + \frac{C_{2}}{3} - . . . . . . + {(- 1)}^{n} \frac{C_{n}}{n + 1} = \frac{1}{n + 1}$ ... (ii)

Subtracting (ii) from (i), we get

$2 [\frac{C_{1}}{2} + \frac{C_{3}}{4} + . . . . . .] = \frac{2^{n + 1} - 1}{(n + 1)} - \frac{1}{(n + 1)}$

Put n = 11, we get

$2 [\frac{C_{1}}{2} + \frac{C_{3}}{4} + . . . . . . + \frac{C_{11}}{12}] = \frac{2^{12} - 1}{12} - \frac{1}{12}$

$\Rightarrow \frac{C_{1}}{2} + \frac{C_{3}}{4} + . . . . . . + \frac{C_{11}}{12} = \frac{1}{2} [\frac{2^{12} - 2}{12}]$ ... (iii)

Now, $\sum_{r = 0}^{5} \frac{{}^{11}C_{2 r + 1}}{2 r + 2} = \frac{m}{n}$

$\Rightarrow \frac{{}^{11}C_{1}}{2} + \frac{{}^{11}C_{3}}{4} + \frac{{}^{11}C_{5}}{6} + . . . . . . + \frac{{}^{11}C_{11}}{12} = \frac{m}{n}$

$\Rightarrow \frac{1}{2} [\frac{2^{12} - 2}{12}] = \frac{m}{n}$ [From (iii)]

$\Rightarrow \frac{2^{11} - 1}{12} = \frac{m}{n} \Rightarrow \frac{m}{n} = \frac{2047}{12}$

$∴$ m – n = 2047 – 12 = 2035.

Q 37 :

If $\sum_{r = 1}^{30} \frac{r^{2} {({}^{30}C_{r})}^{2}}{{}^{30}C_{r - 1}} = α \times 2^{29}$ , then $α$ is equal to __________. [2025]

0%

(465)

$\sum_{r = 1}^{30} \frac{r^{2} {({}^{30}C_{r})}^{2}}{{}^{30}C_{r - 1}} = \sum_{r = 1}^{30} r^{2} (\frac{31 - r}{r}) \cdot \frac{30!}{r! (30 - r)!}$

$= \sum_{r = 1}^{30} \frac{(31 - r) 30!}{(r - 1)! (30 - r)!} = 30 \sum_{r = 1}^{30} (30 - r + 1) \times {}^{29}C_{30 - r}$

$= 30 (\sum_{r = 1}^{30} (30 - r) \times {}^{29}C_{30 - r} + \sum_{r = 1}^{30} {}^{29}C_{30 - r})$

$= 30 (29 \times 2^{28} + 2^{29})$

$[∵ {}^{n}C_{1} + 2 {}^{n}C_{2} + 3 {}^{n}C_{3} + 4 {}^{n}C_{4} + . . . + n {}^{n}C_{n} = n 2^{n - 1} and {}^{n}C_{1} + {}^{n}C_{2} + {}^{n}C_{3} + . . . + {}^{n}C_{n} = 2^{n - 1}]$

$= 30 (29 + 2) 2^{28} = 15 \times 31 \times 2^{29} = 465 (2^{29})$

$∴ α = 465$ .

Q 38 :

Let $S = {p_{1}, p_{2}, . . . . . ., p_{10}}$ be the set of first ten prime numbers, Let $A = S \cup P$ , where P is the set of all possible products of distinct elements of S. Then the number of all ordered pairs $(x, y), x \in S, y \in A$ , such that x divides y, is __________. [2025]

0%

(5120)

S = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29}

$P = {2 \times 3, 2 \times 3 \times 5, . . .}$

$A = S \cup P = {2, 2 \times 3, . . ., 3; 3 \times 5, . . .}$

For x = 2, the value of y can be

$1 + {}^{9}C_{1} + {}^{9}C_{2} + {}^{9}C_{3} + . . . + {}^{9}C_{9} = 2^{9}$

Similarly, for x = 3, 5, 7, 11, ...; y can be $2^{9}$

$∴$ Required number of ordered pair = $10 \times (2^{9})$

= $10 \times 512 = 5120$

Q 39 :

The number of 6-letter words, with or without meaning, that can be formed using the letters of the word MATHS such that any letter that appears in the word must appear at least twice, is __________. [2025]

0%

(1405)

(i) Single letter is used, then number of words = 5

(ii) Two distinct letters are used, then number of words

$= {}^{5}C_{2} \times (\frac{6!}{2! 4!} \times 2 + \frac{6!}{3! 3!}) = 10 (30 + 20) = 500$

(iii) Three distinct letters are used, then number of words

$= {}^{5}C_{3} \times \frac{6!}{2! 2! 2!} = 900$

$∴$ Total number of words = 1405.

Topic Question Set

Q 31 :

From all the English alphabets, five letters are chosen and are arranged in alphabetical order. The total number of ways, in which the middle letter is 'M', is : [2025]

Q 32 :

Group A consists of 7 boys and 3 girls, while group B consists of 6 boys and 5 girls. The number of ways, 4 boys and 4 girls can be invited for a picnic if 5 of them must be from group A and the remaining 3 from group B, is equal to : [2025]

Q 33 :

Let S be the set of all the words that can be formed by arranging all the letters of the word GARDEN. From the set S, one word is selected at random. The probability that the selected word will NOT have vowels in alphabetical order is : [2025]

Q 34 :

For $n \geq 2$ , let $S_{n}$ denote the set of all subsets of {1, 2, ..., n} with no two consecutive numbers. For example ${1, 3, 5} \in S_{6}$ but ${1, 2, 4} \notin S_{6}$ . Then $n (S_{5})$ is equal to __________. [2025]

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Q 35 :

The number of singular matrices of order 2, whose elements are from the set {2, 3, 6, 9}, is __________. [2025]

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Q 36 :

If $\sum_{r = 0}^{5} \frac{{}^{11}C_{2 r + 1}}{2 r + 2} = \frac{m}{n}$ , gcd (m, n) = 1, then m – n is equal to __________. [2025]

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Q 37 :

If $\sum_{r = 1}^{30} \frac{r^{2} {({}^{30}C_{r})}^{2}}{{}^{30}C_{r - 1}} = α \times 2^{29}$ , then $α$ is equal to __________. [2025]

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Q 38 :

Let $S = {p_{1}, p_{2}, . . . . . ., p_{10}}$ be the set of first ten prime numbers, Let $A = S \cup P$ , where P is the set of all possible products of distinct elements of S. Then the number of all ordered pairs $(x, y), x \in S, y \in A$ , such that x divides y, is __________. [2025]

0%

Q 39 :

The number of 6-letter words, with or without meaning, that can be formed using the letters of the word MATHS such that any letter that appears in the word must appear at least twice, is __________. [2025]

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Q 37 : If ∑r=130r2(Cr30)2Cr–130=α×229, then α is equal to __________. [2025] .question-row { display: flex; align-items: flex-start; gap: 8px; } .q-no { font-size: 17px; white-space: nowrap; margin-left: -15px; } .q-text { font-size: 18px; flex: 1; } .q-no, .q-text p { line-height: 30px; }

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Q 37 :

If $\sum_{r = 1}^{30} \frac{r^{2} {({}^{30}C_{r})}^{2}}{{}^{30}C_{r - 1}} = α \times 2^{29}$ , then $α$ is equal to __________. [2025]