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

Q 11 :

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 12 :

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 13 :

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 14 :

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]

(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 15 :

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

(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 16 :

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]

(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 17 :

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]

(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 18 :

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]

(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 19 :

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]

(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.

Q 20 :

Let ABC be a triangle. Consider four points $p_{1}, p_{2}, p_{3}, p_{4}$ on the side AB, five points $p_{5}, p_{6}, p_{7}, p_{8}, p_{9}$ on the side BC, and four points $p_{10}, p_{11}, p_{12}, p_{13}$ on the side AC. None of these points is a vertex of triangle ABC. Then the total number of pentagons that can be formed by taking all the vertices from the points $p_{1}, p_{2}, \dots, p_{13}$ is _________ . [2026]

(660)

$Case 1 :$

$2 from A B, 2 from B C, 1 from A C$

$(\binom{4}{2}) . (\binom{5}{2}) . (\binom{4}{1}) = 6 \cdot 10 \cdot 4 = 240$

$Case 2:$

$2 from A B, 1 from B C, 2 from A C$

$(\binom{4}{2}) . (\binom{5}{1}) . (\binom{4}{2}) = 6 \cdot 5 \cdot 6 = 180$

$Case 3:$

$1 from A B, 2 from B C, 2 from A C$

$(\binom{4}{1}) . (\binom{5}{2}) . (\binom{4}{2}) = 4 \cdot 10 \cdot 6 = 240$

Q 21 :

The number of ways, in which 16 oranges can be distributed to four children such that each child gets at least one orange, is [2026]

403
429
455
384

1

2

3

4

(3)

$Let oranges be identical, then$

$x_{1} + x_{2} + x_{3} + x_{4} = 16, and x_{1}, x_{2}, x_{3}, x_{4} \geq 1$

$or x_{1}' + x_{2}' + x_{3}' + x_{4}' = 12$

$so total number of solutions are$

$= {}^{12 + 3}C_{3} = {}^{15}C_{3} = 455$

Q 22 :

Let S denote the set of 4-digit numbers abcd such that $a > b > c > d$ and P denote the set of 5-digit numbers having product of its digits equal to 20. Then $n (S) + n (P)$ is equal to ______ [2026]

(260)

$For n (s) = {}^{10}C_{4} = 210$

$(5, 4, 1, 1, 1), (5, 2, 2, 1, 1)$

$For n (p) = \frac{5!}{3!} + \frac{5!}{2! 2!} = 50$

$n (s) + n (p) = 210 + 50 = 260$

Q 23 :

Let $S = {x^{3} + a x^{2} + b x + c : a, b, c \in ℕ and a, b, c \leq 20}$ be a set of polynomials. Then the number of polynomials in S, which are divisible by $x^{2} + 2$ , is [2026]

10
20
120
6

1

2

3

4

(1)

$x^{3} + a x^{2} + b x + c = (x^{2} + 2) (x + \frac{c}{2})$

$x^{2} : a = \frac{c}{2}$

$x : b = 2$

$b = 2, a = \frac{c}{2}, c \in {2, 4, \dots, 20}$

$Number of polynomials in S will be 10 .$

Q 24 :

Which of the following must be the true value of the statements below in that order.

Statement I: There are $'' n'',$ $(n \geq 3)$ straight lines, drawn in a plane, no two of which are parallel and no three pass through the same point, and the points of intersection of these lines are joined. Then the number of fresh lines, hence made is $\frac{n (n - 1) (n - 2) (n - 3)}{8}$

Statement II: Vertices of $n$ -sided regular polygon, joined to form a triangle then number of obtuse angled triangle, is $n .^{\frac{n}{2} - 1} C_{2}$ , if $n$ is even $(n > 4)$

TT
TF
FT
FF

1

2

3

4

(1)

Refer P & C synapses Integrated and objective

Q 25 :

Given that $x + y + z + t = 30$ . Match the number of solutions to this equation in List-II with constraints placed on $x, y, z, t$ in List-I.

List-I
List-II

P) $x, y, z, t \in I; x \geq - 1, y \geq 3, z \geq 5, t \geq 0$ I) 585

Q) $x, y, z, t \in I; x \geq 0, 2 \leq y \leq 7, 4 \leq z \leq 8, t \geq 1$ II) 680

R) $x, y, z, t \in W and x, y are odd while z, t are even$ III) 816

S) $x, y, z, t \in N and t is a multiple of 5$ IV) 580

V) 2600

The correct option is:

P → (V), Q → (I), R → (II), S → (IV)
P → (I), Q → (V), R → (III), S → (II)
P → (II), Q → (V), R → (III), S → (I)
P → (III), Q → (V), R → (IV), S → (IV)

1

2

3

4

(1)

$(P) {}^{28}C_{3} = 2600$

$(Q) {}^{26}C_{3} - {}^{20}C_{3} - {}^{21}C_{3} + {}^{15}C_{3} = 585$

$(R) {}^{17}C_{3} = 680$

$(S) {}^{24}C_{2} + {}^{19}C_{2} + {}^{14}C_{2} + {}^{9}C_{2} + {}^{4}C_{2} = 580$

Q 26 :

Match the following

Column 1 Column - 2

A) Number of triangle that can be made using the vertices of a polygon of 10 sides as their vertices and having exactly one side common with the polygon is p) 75

B) Number of triangle that can be made using the vertices of a polygon of 10 sides as their vertices and having exactly 2 sides common with the polygon is q) 110

C) Number of quadrilaterals that can be made using the vertices of a polygon of 10 sides as their vertices and having exactly 2 sides common with the polygon is r) 60

D) Number of quadrilaterals that can be made using the vertices of a polygon of 10 sides as their vertices and having 3 sides common with the polygon is s) 10

A-r, B-s, C-p, D-q
A-s, B-r, C-p, D-s
A-r, B-s, C-p, D-s
None of these

1

2

3

4

(3)

(A) No. of such triangles = $10 \times {}^{6}C_{1} = 60$

(B) No. of such triangles = 10

(C) No. of such quadrilaterals $= 10 \times 5 + \frac{10 \times 5}{2}$ $= 50 + 25$ $= 75$

(D) No. of such quadrilaterals = 10 (when four consecutive points are taken)

Q 27 :

If $p, q, r$ are prime numbers and $α, β, γ$ are positive integers such that L.C.M. of $α, β, γ$ is $p^{3} q^{2} r$ and greatest common divisor of $α, β, γ$ is $p q r$ , then the number of possible triplets $(α, β, γ)$ will be:

36
72
144
60

1

2

3

4

(2)

$LCM of α, β, γ = p^{3} q^{2} r & H C F = p q r$ $∴$ $α = p^{m_{1}} q^{n_{1}} r$

$β = p^{m_{2}} q^{n_{2}} r, γ = p^{m_{3}} q^{n_{3}} r$

$Minimum of (m_{1}, m_{2}, m_{3}) = 1 & maximum of (m_{1}, m_{2}, m_{3}) = 3$

$∴$ $Number of possibilities for m_{1}, m_{2}, m_{3} = 12$

$And minimum of n_{1}, n_{2}, n_{3} = 1 and maximum (n_{1}, n_{2}, n_{3}) = 2$

$∴$ $Number of possibilities = 6$

$∴$ $Total Number of ordered triplets = 12 \times 6 = 72$

Q 28 :

The number of seven-digit numbers, that can be formed by using the digits 1, 2, 3, 4, 5 and 7 such that each digit is used at least once, is: [2026]

15400
17800
16800
29400

1

2

3

4

(3)

Case 1: One digit appears 3 times and other four digits appear once each.

Number of such arrangements

$= {}^{5}C_{1} \times \frac{7!}{3!}$

$= \frac{5 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{3 \times 2 \times 1} = 4200$

Case 2: Two digits appear 2 times each and other three digits appear once each.

Number of such arrangements

$= {}^{5}C_{2} \times \frac{7!}{2! 2!}$

$= \frac{5 \times 4}{2} \times \frac{7 \times 6 \times 5 \times 4 \times 3 \times 2}{2 \times 2} = 12600$

Total number of arrangements

= 4200 + 12600 = 12800

Q 29 :

The number of elements in the set $S = {(r, k) : k \in Z and {}^{36}C_{r + 1} = \frac{6 ({}^{35}C_{r})}{(k^{2} - 3)}}$ , is: [2026]

2
4
8
16

1

2

3

4

(2)

${}^{36}C_{r + 1} = \frac{6 {}^{35}C_{r}}{(k^{2} - 3)}$

$\Rightarrow \frac{{}^{36}C_{r + 1}}{{}^{35}C_{r}} = \frac{6}{k^{2} - 3}$

$\Rightarrow \frac{\frac{36!}{(r + 1)! \times (35 - r)!}}{\frac{35!}{r! (35 - r)!}} = \frac{6}{k^{2} - 3}$

$\Rightarrow \frac{36}{r + 1} = \frac{6}{k^{2} - 3} \Rightarrow 6 (k^{2} - 3) = r + 1$

$\Rightarrow 6 k^{2} - r = 19 \Rightarrow r = 6 k^{2} - 19$

Now, ${}^{35}C_{r}$ is defined, so $0 \leq 6 k^{2} \leq 35$

$\Rightarrow 0 \leq k^{2} - 19 \leq 35 \Rightarrow 19 \leq 6 k^{2} \leq 54$

$\Rightarrow \frac{19}{6} \leq 6 k^{2} \leq 9$

$\Rightarrow 3.1 \leq k^{2} \leq 9 \Rightarrow k \in {- 3, - 2, 2, 3}$

$\Rightarrow r \in {35, 5}$

So, (r, k) = {(35, –3), (35,3), (5, –2), (5, 2)}

Number of elements = 4.

Q 30 :

If for $3 \leq r \leq 30, ({}^{30}C_{30 - r}) + 3 ({}^{30}C_{31 - r}) + 3 ({}^{30}C_{32 - r}) + ({}^{30}C_{33 - r}) = {}^{m}C_{r + 1}$ , then m equals: [2026]

31
32
33
34

1

2

3

4

(3)

Since, ${}^{30}C_{30 - r} = {}^{30}C_{r}$ , ${}^{30}C_{31 - r} = {}^{30}C_{r - 1}$ , ${}^{30}C_{32 - r} = {}^{30}C_{r - 2}$ , and ${}^{30}C_{33 - r} = {}^{30}C_{r - 3}$

$∴ {}^{30}C_{30 - r} + 3 ({}^{30}C_{31 - r}) + 3 ({}^{30}C_{32 - r}) + {}^{30}C_{33 - r} = {}^{m}C_{r}$

$\Rightarrow {}^{30}C_{r} + 3 [{}^{30}C_{r - 1} + {}^{30}C_{r - 2}] + {}^{30}C_{r - 3} = {}^{m}C_{r}$

$\Rightarrow ({}^{30}C_{r} + {}^{30}C_{r - 1}) + 2 ({}^{30}C_{r - 1} + {}^{30}C_{r - 2}) + ({}^{30}C_{r - 2} + {}^{30}C_{r - 3}) = {}^{m}C_{r}$

$\Rightarrow {}^{31}C_{r} + 2 \cdot ({}^{31}C_{r - 1}) + {}^{31}C_{r - 2} = {}^{m}C_{r}$

$\Rightarrow {}^{31}C_{r} + {}^{31}C_{r - 1} + {}^{31}C_{r - 1} + {}^{32}C_{r - 2} = {}^{m}C_{r}$

$\Rightarrow {}^{32}C_{r} + {}^{32}C_{r - 1} = {}^{m}C_{r} \Rightarrow + {}^{33}C_{r} = {}^{m}C_{r}$

$\Rightarrow m = 33$

Q 31 :

Let $p_{n}$ denote the total number of triangles formed by joining the vertices of an n-side regular polygon. If $p_{n + 1} - p_{n} = 66$ , then the sum of all distinct prime divisors of n is: [2026]

7
8
5
6

1

2

3

4

(3)

$p_{n} = {}^{n}C_{3}$

We have, $p_{n + 1} - p_{n} = 66$

${}^{n + 1}C_{3} - {}^{n}C_{3} = 66$

$\Rightarrow \frac{(n + 1) (n) (n - 1)}{6} - \frac{n (n - 1) (n - 2)}{6} = 66$

$\Rightarrow n (n - 1) [(n + 1) - (n - 2)] = 396$

$\Rightarrow 3 n (n - 1) = 396 \Rightarrow n^{2} - n - 132 = 0$

$\Rightarrow n (n - 12) (n + 11) = 0 \Rightarrow n = 12, - 11$

We know, $12 = 2^{2} 3^{1}$

$∴$ Divisor are 1, 2, 3, 4, 6, 12

Prime divisor are 2 and 3

$∴$ Sum = 5

Q 32 :

Let A = {(a, b, c) : a, b, c are non-negative integers and a + b + 2c = 22}. Then n(A) is equal to : [2026]

121
124
144
169

1

2

3

4

(3)

Let A = {(a, b, c) : a, b, c are non-negative integers and a + b + 2c = 22}

Put c = 0

a + b = 22 i.e., ${}^{23}C_{1}$

Put c = 1

a + b = 20 i.e., ${}^{21}C_{1}$

Put c = 2

a + b = 18 i.e., ${}^{19}C_{1}$

Put c = 11

a + b = 0 i.e., ${}^{1}C_{1}$

n(A) = ${}^{1}C_{1} + {}^{3}C_{1} + {}^{5}C_{1} + . . . + {}^{23}C_{1}$

$= 1 + 3 + 5 + . . . + 23 = {(12)}^{2} = 144$

Q 33 :

A box contains 5 blue, 6 yellow and 4 red balls. The number of ways, of drawing 8 balls containing at least two balls of each colour, is: [2026]

4100
4140
4230
4290

1

2

3

4

(1)

We first select 6 balls (2 Blue, 2 Yellow, 2 Red). Now we only need to select 2 more balls from the remaining balls.

For each colour, the coefficient of 'x' represents the number of ways to select and additional balls:

Case I: Blue (5 total): We already took 2, so 3 remaining.

Number of ways to select (0,1, or 2):

$(\begin{matrix} 4 \\ 2 \end{matrix}) + (\begin{matrix} 4 \\ 3 \end{matrix}) x + (\begin{matrix} 4 \\ 4 \end{matrix}) x^{2} = 6 + 4 x + x^{2}$

Case II: Yellow (6 total): We already took 2, or 4 remaining.

Number of ways to select (0, 1, or 2):

$(\begin{matrix} 6 \\ 2 \end{matrix}) + (\begin{matrix} 6 \\ 3 \end{matrix}) x + (\begin{matrix} 6 \\ 4 \end{matrix}) x^{2} = 15 + 20 x + 15 x^{2}$

Case III: Red (4 balls): We already took2, so 2 remaining.

Number of ways to pick (0, 1, or 2):

$(\begin{matrix} 4 \\ 2 \end{matrix}) + (\begin{matrix} 4 \\ 3 \end{matrix}) x + (\begin{matrix} 4 \\ 4 \end{matrix}) x^{2} = 6 + 4 x + x^{2}$

To get 2 more balls, we look for combinations of terms from each bracket where the powers of x add up to 2:

Combination (Blue, Yellow, Red)	Number of ways
2 Blue (0 Yellow, 0 Red)	5x15x6=450
2 Yellow (0 Blue, 0 Red)	15x10x6=900
2 Red (0 Blue, 0 Yellow)	1x10x15=150
1 Blue, 1 Yellow (0 Red)	10x20x6=1200
1 Blue, 1 Red (0 Yellow)	10x4015=600
1 Yellow, 1 Red (0 Red)	20x4x10=800

Required number of ways = 450 + 900 + 150 + 1200 + 600 + 800 = 4100

Q 34 :

The number of 4-letter words, with or without meaning, each consisting of two vowels and two consonants that can be formed from the letters of the word INCONSEQUENTIAL, without repeating any letter, is: [2026]

2670
2840
2920
3600

1

2

3

4

(4)

In word INCONSEQUENTIAL, distinct vowels are (I, E, O, A, U) and distinct consonants are (N, C, S, Q, T, L)

Total number of 4 letter words using two vowels and two consonants without repetition = ${}^{5}C_{2} \times {}^{6}C_{2} \times 4!$ = 3600

Q 35 :

A building has ground floor and 10 more floors. Nine persons enter in a lift at the ground floor. The lift goes up to the 10th floor. The number of ways, in which any 4 persons exit at a floor and the remaining 5 persons exit at a different floor, if the lift does not stop at the first and the second floors, is equal to: [2026]

2670
2840
7056
11340

1

2

3

4

(3)

The lift does not stop at 1st and 2nd floors.

Available floors are: 3, 4, 5, 6, 7, 8, 9, 10

Let group 1 have 4 persons and group 2 have 5 persons.

Number of ways in which two groups are formed = $\frac{9!}{4! 5!} = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1}$ = 126 ways

Number of ways of selecting 2 floors out of 8 floors

Total number of ways in which both groups can leave at different floors = 126 x 28 x 2 = 7056.

Q 36 :

A person has three different bags and four different books. The number of ways, in which he can put these books in the bags so that no bag is empty, is: [2026]

18
36
39
72

1

2

3

4

(2)

Bag 1: 2

Bag 2: 1

Bag 3: 1

Number of ways = $(\frac{(4!)}{2! {(1!)}^{2} . 2!}) \times 3!$

$= \frac{24 \times 6}{2 \times 2} = 36$

Q 37 :

Two players A and B play a series of games of badminton. The player, who wins 5 games first, wins the series. Assuming that no game ends in a draw, the number of ways, in which player A wins the series is ________. [2026]

(126)

Let B's wins = k

Total games played = 5 + k

Last game is won by A so among the first 4+ k games,

A wins 4 times

B wins k times

$∴$ Number of ways in which A wins = 4 + k

Sum of all possible values of k

$= \sum_{k = 0}^{4} {}^{4 + k}C_{4} + {}^{4}C_{4} + {}^{5}C_{4} + {}^{6}C_{4} + {}^{7}C_{4} + {}^{8}C_{4}$

= 1 + 5 + 15 + 35 + 70 = 126

Topic Question Set

Q 11 :

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 12 :

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 13 :

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 14 :

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]

Q 15 :

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

Q 16 :

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]

Q 17 :

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]

Q 18 :

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]

Q 19 :

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]

Q 21 :

The number of ways, in which 16 oranges can be distributed to four children such that each child gets at least one orange, is [2026]

Q 22 :

Let S denote the set of 4-digit numbers abcd such that $a > b > c > d$ and P denote the set of 5-digit numbers having product of its digits equal to 20. Then $n (S) + n (P)$ is equal to ______ [2026]

Q 23 :

Let $S = {x^{3} + a x^{2} + b x + c : a, b, c \in ℕ and a, b, c \leq 20}$ be a set of polynomials. Then the number of polynomials in S, which are divisible by $x^{2} + 2$ , is [2026]

Q 27 :

If $p, q, r$ are prime numbers and $α, β, γ$ are positive integers such that L.C.M. of $α, β, γ$ is $p^{3} q^{2} r$ and greatest common divisor of $α, β, γ$ is $p q r$ , then the number of possible triplets $(α, β, γ)$ will be:

Q 28 :

The number of seven-digit numbers, that can be formed by using the digits 1, 2, 3, 4, 5 and 7 such that each digit is used at least once, is: [2026]

Q 29 :

The number of elements in the set $S = {(r, k) : k \in Z and {}^{36}C_{r + 1} = \frac{6 ({}^{35}C_{r})}{(k^{2} - 3)}}$ , is: [2026]

Q 30 :

If for $3 \leq r \leq 30, ({}^{30}C_{30 - r}) + 3 ({}^{30}C_{31 - r}) + 3 ({}^{30}C_{32 - r}) + ({}^{30}C_{33 - r}) = {}^{m}C_{r + 1}$ , then m equals: [2026]

Q 31 :

Let $p_{n}$ denote the total number of triangles formed by joining the vertices of an n-side regular polygon. If $p_{n + 1} - p_{n} = 66$ , then the sum of all distinct prime divisors of n is: [2026]

Q 32 :

Let A = {(a, b, c) : a, b, c are non-negative integers and a + b + 2c = 22}. Then n(A) is equal to : [2026]

Q 33 :

A box contains 5 blue, 6 yellow and 4 red balls. The number of ways, of drawing 8 balls containing at least two balls of each colour, is: [2026]

Q 34 :

The number of 4-letter words, with or without meaning, each consisting of two vowels and two consonants that can be formed from the letters of the word INCONSEQUENTIAL, without repeating any letter, is: [2026]

Q 36 :

A person has three different bags and four different books. The number of ways, in which he can put these books in the bags so that no bag is empty, is: [2026]

Q 37 :

Two players A and B play a series of games of badminton. The player, who wins 5 games first, wins the series. Assuming that no game ends in a draw, the number of ways, in which player A wins the series is ________. [2026]

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	List-I		List-II
P)	$x, y, z, t \in I; x \geq - 1, y \geq 3, z \geq 5, t \geq 0$	I)	585
Q)	$x, y, z, t \in I; x \geq 0, 2 \leq y \leq 7, 4 \leq z \leq 8, t \geq 1$	II)	680
R)	$x, y, z, t \in W and x, y are odd while z, t are even$	III)	816
S)	$x, y, z, t \in N and t is a multiple of 5$	IV)	580
		V)	2600

	Column 1		Column - 2
A)	Number of triangle that can be made using the vertices of a polygon of 10 sides as their vertices and having exactly one side common with the polygon is	p)	75
B)	Number of triangle that can be made using the vertices of a polygon of 10 sides as their vertices and having exactly 2 sides common with the polygon is	q)	110
C)	Number of quadrilaterals that can be made using the vertices of a polygon of 10 sides as their vertices and having exactly 2 sides common with the polygon is	r)	60
D)	Number of quadrilaterals that can be made using the vertices of a polygon of 10 sides as their vertices and having 3 sides common with the polygon is	s)	10

Topic Question Set

Q 11 : 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 12 : 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 13 : 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 14 : For n≥2, let Sn denote the set of all subsets of {1, 2, ..., n} with no two consecutive numbers. For example {1,3,5}∈S6 but {1,2,4}∉S6. Then n(S5) is equal to __________. [2025]

Q 15 : The number of singular matrices of order 2, whose elements are from the set {2, 3, 6, 9}, is __________. [2025]

Q 16 : If ∑r=05C2r+1112r+2=mn, gcd (m, n) = 1, then m – n is equal to __________. [2025]

Q 17 : If ∑r=130r2(Cr30)2Cr–130=α×229, then α is equal to __________. [2025]

Q 18 : Let S={p1,p2,......,p10} be the set of first ten prime numbers, Let A=S∪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∈S, y∈A, such that x divides y, is __________. [2025]

Q 19 : 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]

Q 21 : The number of ways, in which 16 oranges can be distributed to four children such that each child gets at least one orange, is [2026]

Q 22 : Let S denote the set of 4-digit numbers abcd such that a>b>c>d and P denote the set of 5-digit numbers having product of its digits equal to 20. Then n(S)+n(P) is equal to ______ [2026]

Q 23 : Let S={x3+ax2+bx+c: a,b,c∈ℕ and a,b,c≤20} be a set of polynomials. Then the number of polynomials in S, which are divisible by x2+2, is [2026]

Q 27 : If p,q,r are prime numbers and α,β,γ are positive integers such that L.C.M. of α,β,γ is p3q2r and greatest common divisor of α,β,γ is pqr, then the number of possible triplets (α,β,γ) will be:

Q 28 : The number of seven-digit numbers, that can be formed by using the digits 1, 2, 3, 4, 5 and 7 such that each digit is used at least once, is: [2026]

Q 29 : The number of elements in the set S={(r,k):k∈Z and Cr+136=6(Cr35)(k2–3)}, is: [2026]

Q 30 : If for 3≤r≤30, (C30–r30)+3(C31–r30)+3(C32–r30)+(C33–r30)=Cr+1m, then m equals: [2026]

Q 31 : Let pn denote the total number of triangles formed by joining the vertices of an n-side regular polygon. If pn+1–pn=66, then the sum of all distinct prime divisors of n is: [2026]

Q 32 : Let A = {(a, b, c) : a, b, c are non-negative integers and a + b + 2c = 22}. Then n(A) is equal to : [2026]

Q 33 : A box contains 5 blue, 6 yellow and 4 red balls. The number of ways, of drawing 8 balls containing at least two balls of each colour, is: [2026]

Q 34 : The number of 4-letter words, with or without meaning, each consisting of two vowels and two consonants that can be formed from the letters of the word INCONSEQUENTIAL, without repeating any letter, is: [2026]

Q 36 : A person has three different bags and four different books. The number of ways, in which he can put these books in the bags so that no bag is empty, is: [2026]

Q 37 : Two players A and B play a series of games of badminton. The player, who wins 5 games first, wins the series. Assuming that no game ends in a draw, the number of ways, in which player A wins the series is ________. [2026]

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Q 11 :

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 12 :

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 13 :

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 14 :

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]

Q 15 :

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

Q 16 :

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]

Q 17 :

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]

Q 18 :

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]

Q 19 :

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]

Q 21 :

The number of ways, in which 16 oranges can be distributed to four children such that each child gets at least one orange, is [2026]

Q 22 :

Let S denote the set of 4-digit numbers abcd such that $a > b > c > d$ and P denote the set of 5-digit numbers having product of its digits equal to 20. Then $n (S) + n (P)$ is equal to ______ [2026]

Q 23 :

Let $S = {x^{3} + a x^{2} + b x + c : a, b, c \in ℕ and a, b, c \leq 20}$ be a set of polynomials. Then the number of polynomials in S, which are divisible by $x^{2} + 2$ , is [2026]

Q 27 :

If $p, q, r$ are prime numbers and $α, β, γ$ are positive integers such that L.C.M. of $α, β, γ$ is $p^{3} q^{2} r$ and greatest common divisor of $α, β, γ$ is $p q r$ , then the number of possible triplets $(α, β, γ)$ will be:

Q 28 :

The number of seven-digit numbers, that can be formed by using the digits 1, 2, 3, 4, 5 and 7 such that each digit is used at least once, is: [2026]

Q 29 :

The number of elements in the set $S = {(r, k) : k \in Z and {}^{36}C_{r + 1} = \frac{6 ({}^{35}C_{r})}{(k^{2} - 3)}}$ , is: [2026]

Q 30 :

If for $3 \leq r \leq 30, ({}^{30}C_{30 - r}) + 3 ({}^{30}C_{31 - r}) + 3 ({}^{30}C_{32 - r}) + ({}^{30}C_{33 - r}) = {}^{m}C_{r + 1}$ , then m equals: [2026]

Q 31 :

Let $p_{n}$ denote the total number of triangles formed by joining the vertices of an n-side regular polygon. If $p_{n + 1} - p_{n} = 66$ , then the sum of all distinct prime divisors of n is: [2026]

Q 32 :

Let A = {(a, b, c) : a, b, c are non-negative integers and a + b + 2c = 22}. Then n(A) is equal to : [2026]

Q 33 :

A box contains 5 blue, 6 yellow and 4 red balls. The number of ways, of drawing 8 balls containing at least two balls of each colour, is: [2026]

Q 34 :

The number of 4-letter words, with or without meaning, each consisting of two vowels and two consonants that can be formed from the letters of the word INCONSEQUENTIAL, without repeating any letter, is: [2026]

Q 36 :

A person has three different bags and four different books. The number of ways, in which he can put these books in the bags so that no bag is empty, is: [2026]

Q 37 :

Two players A and B play a series of games of badminton. The player, who wins 5 games first, wins the series. Assuming that no game ends in a draw, the number of ways, in which player A wins the series is ________. [2026]