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

Q 21 :

The largest value of $n$ , for which $40^{n}$ divides $60!,$ is [2026]

13
14
12
11

1

2

3

4

(2)

$40^{n} = 2^{3 n} \times 5^{n}$

$E_{2} (60!) = [\frac{60}{2}] + [\frac{60}{2^{2}}] + [\frac{60}{2^{3}}] + [\frac{60}{2^{4}}] + [\frac{60}{2^{5}}]$

$= 30 + 15 + 7 + 3 + 1 = 56$

$E_{5} (60!) = [\frac{60}{5}] + [\frac{60}{5^{2}}]$

$= 12 + 2 = 14$

$40^{n} = {(2^{3})}^{n} \times 5^{n} = {(2^{3} \times 5)}^{n}$

$60! = 2^{56} \times 5^{14} . . . = 2^{14} \cdot {(2^{3} \cdot 5)}^{14}$

$∴$ $Maximum value of n is 14 .$

Q 22 :

The letters of the word "UDAYPUR" are written in all possible ways with or without meaning and these words are arranged as in a dictionary. The rank of the word "UDAYPUR" is [2026]

1579
1581
1580
1578

1

2

3

4

(3)

$ADIPRUU$

$A \to \frac{6!}{2!} = 360$

$D \to \frac{6!}{2!} = 360$

$P \to \frac{6!}{2!} = 360$

$R \to \frac{6!}{2!} = 360$

$U A \to 5! = 120$

$U D A P \to 3! = 6$

$U D A R \to 3! = 6$

$U D A U \to 3! = 6$

$U D A Y P R U \to 1$

$U D A Y P U R \to 1$

$Total = 1580$

Q 23 :

The largest $n \in N,$ for which $7^{n}$ divides $101!$ , is : [2026]

18
19
15
16

1

2

3

4

(4)

$Exponent of 7 in 101!$

$= [\frac{101}{7}] + [\frac{101}{7^{2}}] + [\frac{101}{7^{3}}] + \dots$

$= 14 + 2 = 16$

Q 24 :

Let S = {1,2,3,4,5,6,7,8,9}. Let $x$ be the number of 9-digit numbers formed using the digits of the set S such that only one digit is repeated and it is repeated exactly twice. Let y be the number of 9-digit numbers formed using the digits of the set S such that only two digits are repeated and each of these is repeated exactly twice. Then, [2026]

$45 x = 7 y$
$21 x = 4 y$
$29 x = 5 y$
$56 x = 9 y$

1

2

3

4

(2)

$S = {1, 2, 3, \dots, 9}$

$x = {}^{9}C_{1} \cdot {}^{8}C_{7} \times \frac{9!}{2} = \frac{9 \times 8 \times 9!}{2}$

$y = {}^{9}C_{2} \cdot {}^{7}C_{5} \times \frac{9!}{2! 2!} = \frac{9 \times 8}{2} \times \frac{7 \times 6}{2} \times \frac{9!}{2! 2!}$

$\Rightarrow \frac{x}{y} = \frac{4}{21}$

$\Rightarrow 21 x = 4 y$

Q 25 :

The number of ways of arrangement of 5 letters of the word “IITJEE” is

(180)

$Coefficient of x^{5} in 5! {(1 + x + \frac{x^{2}}{2})}^{2} {(1 + x)}^{2}$

Q 26 :

If ${}^{2 n + 1}P_{n - 1} : {}^{2 n - 1}P_{n} = 3 : 5$ , then the value of $n$ is equal to

4
3
2
1

1

2

3

4

(1)

We have

$\frac{{}^{2 n + 1}P_{n - 1}}{{}^{2 n - 1}P_{n}} = \frac{3}{5}$ $\Rightarrow \frac{(2 n + 1)!}{(n + 2)!} \times \frac{(n - 1)!}{(2 n - 1)!} = \frac{3}{5}$ $\Rightarrow \frac{2 (2 n + 1)}{(n + 2) (n + 1)} = \frac{3}{5}$

$\Rightarrow 10 (2 n + 1) = 3 (n + 2) (n + 1) \Rightarrow 3 n^{2} - 11 n - 4 = 0$

$\Rightarrow (3 n + 1) (n - 4) = 0$

$∴$ $n = 4$ $[∵ n \neq \frac{- 1}{3}]$

Q 27 :

The total number of 9-digit numbers which have all different digits is

10!
9!
9 × 9!
10 × 10!

1

2

3

4

(3)

The total number of 9-digit numbers, having all different digits $= 9 \times {}^{9}P_{8} = \frac{9 \times 9!}{1!} = 9 \times 9!$

Q 28 :

Find the number of different 8-letter arrangements that can be made from the letters of the word DAUGHTER so that

(i) all vowels occur together

(ii) all vowels do not occur together

4320, 36000
4300, 36000
4200, 36000
4300, 3600

1

2

3

4

(1)

(i) There are 8 different letters in the word DAUGHTER, in which there are 3 vowels, namely, A, U and E. Since the vowels have to occur together, we assume them as a single object (AUE). This single object together with 5 remaining letters (objects) will be counted as 6 objects. Then we count permutations of these 6 objects taken all at a time. This number would be ${}^{6}P_{6}$ = 6!. Corresponding to each of these permutations, we shall have 3! permutations of the three vowels A, U and E taken all at a time. Hence, by the multiplication principle, the required number of permutations is

= 6! × 3! = 4320.

(ii) If we have to count those permutations in which all vowels are never together, we first have to find all possible arrangements of 8 letters taken all at a time, which can be done in 8! ways. Then, we have to subtract from this number the number of permutations in which the vowels are always together.

Therefore, the required number

= 8! − 6! × 3!

= 6!(7 × 8 − 6) = 2 × 6!(28 − 3)

= 50 × 6! = 50 × 720 = 36000.

Q 29 :

Find the number of ways of choosing two red cards and two black cards from a pack of 52 playing cards.

105620
105624
105625
105600

1

2

3

4

(3)

There are 26 red cards and 26 black cards. Therefore, the required number of ways $= {}^{26}C_{2} \times {}^{26}C_{2}$

$= {(\frac{26!}{2! 24!})}^{2} = {(325)}^{2} = 105625$

Q 30 :

The number of arrangements that can be made taking 4 letters, at a time, out of the letters of the word PASSPORT is

606
626
666
686

1

2

3

4

(1)

Q 31 :

Total number of four digit odd numbers that can be formed using 0, 1, 2, 3, 5, 7 are

216
375
400
720

1

2

3

4

(4)

$Odd numbers are 1, 3, 5, 7$

$We have to fill up four places like T H H T U$

$(Case: If repetition is allowed)$

$Required number of 4-digit numbers$

$= {}^{5}C_{1} \times 6^{2} \times {}^{4}C_{1} = 5 \times 6^{2} \times 4 = 5 \times 36 \times 4 = 720$

Q 32 :

If $P (n, r) = 1680$ and $C (n, r) = 70$ , then $69 n + r! =$

128
576
256
625

1

2

3

4

(2)

$P (n, r) = 1680 and C (n, r) = 70$

$C (n, r) = \frac{n!}{r! (n - r)!} = 70$

$\Rightarrow \frac{P (n, r)}{r!} = 70 \Rightarrow \frac{1680}{r!} = 70 \Rightarrow r! = 24 = 4!$

$∴$ $r = 4$

$Now, P (n, r) = \frac{n!}{(n - r)!} = 1680$

$\frac{n!}{(n - 4)!} = 1680 \Rightarrow n (n - 1) (n - 2) (n - 3) = 1680$

$\Rightarrow$ $n (n - 1) (n - 2) (n - 3) = 8 \times 7 \times 6 \times 5$

$After comparing both sides, n = 8$

$Now, 69 n + r! = (69 \times 8) + 4! = 552 + 24 = 576$

Q 33 :

If ${}^{43}C_{r - 6} = {}^{43}C_{3 r + 1}$ , then the value of $r$ is

12
8
6
10

1

2

3

4

(1)

${}^{43}C_{r - 6} = {}^{43}C_{3 r + 1}$

$\Rightarrow r - 6 + 3 r + 1 = 43$ $(∵ {}^{n}C_{p} = {}^{n}C_{r} \Rightarrow p + r = n)$

$\Rightarrow 4 r - 5 = 43$

$\Rightarrow 4 r = 48$

$\Rightarrow r = \frac{48}{4} = 12$

Q 34 :

The number of ways in which 6 men and 5 women can dine at a round table if no two women are to sit together is given by

30
5! × 4!
7! × 5!
6! × 5!

1

2

3

4

(4)

Number of women = 5

Number of men = 6

Number of ways of arranging 6 men at a round table is $(n - 1)!$

= $(6 - 1)!$

= $5!$

Now we are left with six places between the men, and there are 5 women. These 5 women can be arranged in these places in ${}^{6}P_{5}$ ways.

$∴$ The required number of ways

= $5! \times {}^{6}P_{5}$

= $5! \times 6!$

Q 35 :

The number of ways in which 21 objects can be grouped into three groups of 8, 7 and 6 objects is

$\frac{20!}{8! + 7! + 6!}$
$\frac{21!}{8! 7!}$
$\frac{21!}{8! 7! 6!}$
$\frac{21!}{8! + 7! + 6!}$

1

2

3

4

(3)

Out of 21 objects, if we take 8 objects as 8 similar objects, 7 objects as 7 similar objects and 6 objects as 6 similar objects.

$∴$ Number of ways $= \frac{21!}{8! 7! 6!}$

Q 36 :

Eleven books consisting of 5 mathematics, 4 physics and 2 chemistry are placed on a shelf. The number of possible ways of arranging them on the assumption that the books on the same subject are all together is

$4! 2!$
$11!$
$5! 4! 3! 2!$
none of these

1

2

3

4

(3)

As books of the same subject are to be together, we have 1 + 1 + 1 = 3 groups of books.

So they can be arranged in $3!$ ways. But inside the group, say in Mathematics, 5 individual books can be arranged in $5!$ ways.

Similarly, 4 Physics and 2 Chemistry books can be arranged unit in themselves in $4!$ and $2!$ ways respectively.

$∴$ Total number of ways $= 5! \times 4! \times 3! \times 2!$

Q 37 :

The number of ways in which 9 persons can be divided into three equal groups is

1680
840
560
280

1

2

3

4

(4)

9 persons can be divided into three equal groups in $\frac{9!}{{(3!)}^{4}} = 280$ ways.

Q 38 :

The rank of the word MOTHER when the letters of the word are arranged alphabetically as in a dictionary, is:

261
343
309
273

1

2

3

4

(3)

Rank of the word MOTHER

$= 5! + 5! + 4! + 4! + 3! + 3! + 3! + 2! + 1$

$= 306 + 3$

$= 309$

Q 39 :

If the letters of the word SACHIN are arranged in all possible ways and these words are written out as in a dictionary, then the word SACHIN appears at serial number

602
603
600
601

1

2

3

4

(4)

$↓ Fixed$

$S A C H I N$

No. of words start with A = $5!$

No. of words start with C = $5!$

No. of words start with H = $5!$

No. of words start with I = $5!$

No. of words start with N = $5!$

Total words = $5! + 5! + 5! + 5! + 5!$ $= 5 (5!)$ $= 600$

Now, add the rank of SACHIN.

So, the required rank of SACHIN $= 600 + 1$ $= 601$

Q 40 :

The number of permutations of the letters of the word 'CONSEQUENCE' in which all the three E's are together is

$9! 3!$
$\frac{9!}{2!}$
$\frac{9!}{2! 2! 3!}$
$\frac{9!}{2! 2!}$

1

2

3

4

(4)

Considering all E's as one letter, 2 N's and 2 C's are also there.

So, number of permutations $= \frac{9!}{2! 2!}$

Q 41 :

If one person handshakes with the other only once and the number of handshakes is 66, then the number of persons will be

10
33
24
12

1

2

3

4

(4)

Let the number of persons be $n$

One person out of $n$ persons will handshake with $(n - 1)$ persons.

Second person out of $(n - 1)$ persons will handshake with $(n - 2)$ persons.

Third person out of $(n - 2)$ persons will handshake with $(n - 3)$ persons.

$\begin{matrix} . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . \\ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . \\ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . \end{matrix}$

${(n - 1)}^{th}$ person out of 2 persons will handshake with 1 person.

So total number of handshakes

$= (n - 1) + (n - 2) + \dots + 2 + 1 = \frac{(n - 1) n}{2}$

$∴$ $\frac{n (n - 1)}{2} = 66$ $\Rightarrow n^{2} - n - 132 = 0$

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

So, Total number of persons = 12

Q 42 :

The sum $1 \times 1! + 2 \times 2! + . . . + 50 \times 50!$ equals

$51!$
$51! - 1$
$51! + 1$
$2 \times 51!$

1

2

3

4

(2)

$1 (1!) + 2 (2!) + \dots + 50 (50!)$

$= (2 - 1) (1!) + (3 - 1) (2!) + (4 - 1) (3!) + \dots + (51 - 1) (50!)$

$= (2! - 1!) + (3! - 2!) + (4! - 3!) + \dots + (51! - 50!)$

$= 51! - 1! = 51! - 1$

Q 43 :

The number of words that can be written using all the letters of the word 'IRRATIONAL' is

$\frac{10!}{{(2!)}^{3}}$
$\frac{10!}{{(2!)}^{2}}$
$\frac{10!}{2!}$
$10!$

1

2

3

4

(1)

There are 10 letters in the word IRRATIONAL in which there are 2'I', 2'R' and 2'A'

$∴$ No. of words $= \frac{10!}{2! 2! 2!} = \frac{10!}{{(2!)}^{3}}$

Q 44 :

If $n$ is any positive integer, then $\frac{1}{2^{n}} ({}^{2 n}P_{n}) =$

$2 \cdot 4 \cdot 6$ ...... $(2 n)$
$1 \cdot 2 \cdot 3$ ...... $n$
$1 \cdot 3 \cdot 5$ ...... $(2 n - 1)$
$1 \cdot 2 \cdot 3$ ...... $(3 n)$

1

2

3

4

(3)

$\frac{1}{2^{n}} ({}^{2 n}P_{n}) = \frac{1}{2^{n}} (\frac{(2 n)!}{n!})$

$= \frac{(2 n) (2 n - 1) (2 n - 2) \dots 3 \cdot 2 \cdot 1}{2^{n} \times n!}$

$= \frac{[(2 n) (2 n - 2) \dots 6 \cdot 4 \cdot 2] [(2 n - 1) (2 n - 3) \dots 5 \cdot 3 \cdot 1]}{2^{n} \times n!}$

$= \frac{2^{n} (n (n - 1) \dots 3 \cdot 2 \cdot 1) (1 \cdot 3 \cdot 5 \dots (2 n - 1))}{2^{n} \cdot (n!)}$

$= \frac{2^{n} (n!) (1 \cdot 3 \cdot 5 \dots (2 n - 1))}{2^{n} \cdot (n!)}$

$= 1 \cdot 3 \cdot 5$ ...... $(2 n - 1)$

Q 45 :

If ${}^{12}P_{r} = {}^{11}P_{6} + 6 \cdot {}^{11}P_{5}$ , then $r =$

7
5
6
4

1

2

3

4

(3)

${}^{12}P_{r} = {}^{11}P_{6} + 6 \cdot {}^{11}P_{5} \Rightarrow r = 6$

Q 46 :

The value of ${}^{2}P_{1} + {}^{3}P_{1} + \dots + {}^{n}P_{1}$ is equal to

$\frac{n^{2} - n + 2}{2}$
$\frac{n^{2} + n + 2}{2}$
$\frac{n^{2} + n - 1}{2}$
$\frac{n^{2} + n - 2}{2}$

1

2

3

4

(4)

We have, ${}^{2}P_{1} + {}^{3}P_{1} + \dots + {}^{n}P_{1}$

$= 2 + 3 + \dots + n = \frac{n (n + 1)}{2} - 1 = \frac{n^{2} + n - 2}{2}$

Q 47 :

A polygon has 35 diagonals, then the number of its sides is

8
9
10
11

1

2

3

4

(3)

$Number of diagonals = {}^{n}C_{2} - n = 35 given,$ $n is the number of sides.$

$\Rightarrow \frac{n (n - 1)}{2} - n = 35$ $\Rightarrow \frac{n (n - 3)}{2} = 35$ $\Rightarrow n (n - 3) = 70$

$\Rightarrow n (n - 3) = 10 \times 7$ $\Rightarrow n^{2} - 3 n - 10 \times 7 = 0$

$\Rightarrow (n - 10) (n + 7) = 0$

$∴$ $n = 10$ $(As n + 7 \neq 0)$

Q 48 :

A polygon has 54 diagonals. Number of sides of this polygon is

12
15
16
9

1

2

3

4

(1)

$Polygon has 54 diagonals.$

$Number of diagonals in a polygon of n sides = \frac{n (n - 3)}{2}$

$Here, \frac{n (n - 3)}{2} = 54$

$\Rightarrow n (n - 3) = 108$ $\Rightarrow n^{2} - 3 n - 108 = 0$

$\Rightarrow n^{2} - 12 n + 9 n - 108 = 0$

$\Rightarrow n (n - 12) + 9 (n - 12) = 0$ $\Rightarrow (n + 9) (n - 12) = 0$

$∴$ $n = 12$ $[∵ n = - 9 is not possible]$

Q 49 :

Let $n \in N$ with $n \geq 3$ . Then $(\binom{n}{3})$ can be expressed as a polynomial

$\frac{1}{6} n^{3} - \frac{1}{2} n^{2} + \frac{1}{3} n$
$n^{3} - n^{2} + n$
$n^{3} - 3 n^{2}$
$n^{3} - 3 n^{2} + n$

1

2

3

4

(1)

$(\binom{n}{3}) = {}^{n}C_{3} = \frac{n (n - 1) (n - 2)}{3!} = \frac{n (n - 1) (n - 2)}{6}$

$= \frac{n^{3} - 3 n^{2} + 2 n}{6} = \frac{1}{6} n^{3} - \frac{1}{2} n^{2} + \frac{n}{3}$

Q 50 :

The value of ${}^{50}C_{4} + \sum_{r = 1}^{6} {}^{56 - r}C_{3}$ is

${}^{56}C_{4}$
${}^{56}C_{3}$
${}^{55}C_{3}$
${}^{55}C_{4}$

1

2

3

4

(1)

${}^{50}C_{4} + \sum_{r = 1}^{6} {}^{56 - r}C_{3}$

$Putting r = 6, 5, 4, 3, 2, 1, we get$

${}^{50}C_{4} + {}^{50}C_{3} + {}^{51}C_{3} + {}^{52}C_{3} + {}^{53}C_{3} + {}^{54}C_{3} + {}^{55}C_{3}$ $(∵ {}^{n}C_{r} + {}^{n}C_{r + 1} = {}^{n + 1}C_{r + 1})$

$= {}^{51}C_{4} + {}^{51}C_{3} + {}^{52}C_{3} + {}^{53}C_{3} + {}^{54}C_{3} + {}^{55}C_{3}$

$= {}^{52}C_{4} + {}^{52}C_{3} + {}^{53}C_{3} + {}^{54}C_{3} + {}^{55}C_{3}$

$= {}^{53}C_{4} + {}^{53}C_{3} + {}^{54}C_{3} + {}^{55}C_{3}$

$= {}^{54}C_{4} + {}^{54}C_{3} + {}^{55}C_{3}$

$= {}^{55}C_{4} + {}^{55}C_{3} = {}^{56}C_{4}$

Topic Question Set

Q 21 :

The largest value of $n$ , for which $40^{n}$ divides $60!,$ is [2026]

Q 22 :

The letters of the word "UDAYPUR" are written in all possible ways with or without meaning and these words are arranged as in a dictionary. The rank of the word "UDAYPUR" is [2026]

Q 23 :

The largest $n \in N,$ for which $7^{n}$ divides $101!$ , is : [2026]

Q 25 :

The number of ways of arrangement of 5 letters of the word “IITJEE” is

Q 26 :

If ${}^{2 n + 1}P_{n - 1} : {}^{2 n - 1}P_{n} = 3 : 5$ , then the value of $n$ is equal to

Q 27 :

The total number of 9-digit numbers which have all different digits is

Q 28 :

Find the number of different 8-letter arrangements that can be made from the letters of the word DAUGHTER so that

(i) all vowels occur together

(ii) all vowels do not occur together

Q 29 :

Find the number of ways of choosing two red cards and two black cards from a pack of 52 playing cards.

Q 30 :

The number of arrangements that can be made taking 4 letters, at a time, out of the letters of the word PASSPORT is

Q 31 :

Total number of four digit odd numbers that can be formed using 0, 1, 2, 3, 5, 7 are

Q 32 :

If $P (n, r) = 1680$ and $C (n, r) = 70$ , then $69 n + r! =$

Q 33 :

If ${}^{43}C_{r - 6} = {}^{43}C_{3 r + 1}$ , then the value of $r$ is

Q 34 :

The number of ways in which 6 men and 5 women can dine at a round table if no two women are to sit together is given by

Q 35 :

The number of ways in which 21 objects can be grouped into three groups of 8, 7 and 6 objects is

Q 36 :

Eleven books consisting of 5 mathematics, 4 physics and 2 chemistry are placed on a shelf. The number of possible ways of arranging them on the assumption that the books on the same subject are all together is

Q 37 :

The number of ways in which 9 persons can be divided into three equal groups is

Q 38 :

The rank of the word MOTHER when the letters of the word are arranged alphabetically as in a dictionary, is:

Q 39 :

If the letters of the word SACHIN are arranged in all possible ways and these words are written out as in a dictionary, then the word SACHIN appears at serial number

Q 40 :

The number of permutations of the letters of the word 'CONSEQUENCE' in which all the three E's are together is

Q 41 :

If one person handshakes with the other only once and the number of handshakes is 66, then the number of persons will be

Q 42 :

The sum $1 \times 1! + 2 \times 2! + . . . + 50 \times 50!$ equals

Q 43 :

The number of words that can be written using all the letters of the word 'IRRATIONAL' is

Q 44 :

If $n$ is any positive integer, then $\frac{1}{2^{n}} ({}^{2 n}P_{n}) =$

Q 45 :

If ${}^{12}P_{r} = {}^{11}P_{6} + 6 \cdot {}^{11}P_{5}$ , then $r =$

Q 46 :

The value of ${}^{2}P_{1} + {}^{3}P_{1} + \dots + {}^{n}P_{1}$ is equal to

Q 47 :

A polygon has 35 diagonals, then the number of its sides is

Q 48 :

A polygon has 54 diagonals. Number of sides of this polygon is

Q 49 :

Let $n \in N$ with $n \geq 3$ . Then $(\binom{n}{3})$ can be expressed as a polynomial

Q 50 :

The value of ${}^{50}C_{4} + \sum_{r = 1}^{6} {}^{56 - r}C_{3}$ is

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Topic Question Set

Q 21 : The largest value of n, for which 40n divides 60!, is [2026]

Q 22 : The letters of the word "UDAYPUR" are written in all possible ways with or without meaning and these words are arranged as in a dictionary. The rank of the word "UDAYPUR" is [2026]

Q 23 : The largest n∈N, for which 7n divides 101!, is : [2026]

Q 25 : The number of ways of arrangement of 5 letters of the word “IITJEE” is

Q 26 : If Pn-12n+1:Pn2n-1=3:5, then the value of n is equal to

Q 27 : The total number of 9-digit numbers which have all different digits is

Q 28 : Find the number of different 8-letter arrangements that can be made from the letters of the word DAUGHTER so that (i) all vowels occur together (ii) all vowels do not occur together

Q 29 : Find the number of ways of choosing two red cards and two black cards from a pack of 52 playing cards.

Q 30 : The number of arrangements that can be made taking 4 letters, at a time, out of the letters of the word PASSPORT is

Q 31 : Total number of four digit odd numbers that can be formed using 0, 1, 2, 3, 5, 7 are

Q 32 : If P(n,r)=1680 and C(n,r)=70, then 69n+r!=

Q 33 : If Cr-643=C3r+143, then the value of r is

Q 34 : The number of ways in which 6 men and 5 women can dine at a round table if no two women are to sit together is given by

Q 35 : The number of ways in which 21 objects can be grouped into three groups of 8, 7 and 6 objects is

Q 36 : Eleven books consisting of 5 mathematics, 4 physics and 2 chemistry are placed on a shelf. The number of possible ways of arranging them on the assumption that the books on the same subject are all together is

Q 37 : The number of ways in which 9 persons can be divided into three equal groups is

Q 38 : The rank of the word MOTHER when the letters of the word are arranged alphabetically as in a dictionary, is:

Q 39 : If the letters of the word SACHIN are arranged in all possible ways and these words are written out as in a dictionary, then the word SACHIN appears at serial number

Q 40 : The number of permutations of the letters of the word 'CONSEQUENCE' in which all the three E's are together is

Q 41 : If one person handshakes with the other only once and the number of handshakes is 66, then the number of persons will be

Q 42 : The sum 1×1!+2×2!+...+50×50! equals

Q 43 : The number of words that can be written using all the letters of the word 'IRRATIONAL' is

Q 44 : If n is any positive integer, then 12n(Pn2n)=

Q 45 : If Pr12=P611+6·P511, then r=

Q 46 : The value of P12+P13+⋯+P1n is equal to

Q 47 : A polygon has 35 diagonals, then the number of its sides is

Q 48 : A polygon has 54 diagonals. Number of sides of this polygon is

Q 49 : Let n∈N with n≥3. Then (n3) can be expressed as a polynomial

Q 50 : The value of C450+∑r=16C356-r is

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Q 21 :

The largest value of $n$ , for which $40^{n}$ divides $60!,$ is [2026]

Q 22 :

The letters of the word "UDAYPUR" are written in all possible ways with or without meaning and these words are arranged as in a dictionary. The rank of the word "UDAYPUR" is [2026]

Q 23 :

The largest $n \in N,$ for which $7^{n}$ divides $101!$ , is : [2026]

Q 25 :

The number of ways of arrangement of 5 letters of the word “IITJEE” is

Q 26 :

If ${}^{2 n + 1}P_{n - 1} : {}^{2 n - 1}P_{n} = 3 : 5$ , then the value of $n$ is equal to

Q 27 :

The total number of 9-digit numbers which have all different digits is

Q 28 :

Find the number of different 8-letter arrangements that can be made from the letters of the word DAUGHTER so that

(i) all vowels occur together

(ii) all vowels do not occur together

Q 29 :

Find the number of ways of choosing two red cards and two black cards from a pack of 52 playing cards.

Q 30 :

The number of arrangements that can be made taking 4 letters, at a time, out of the letters of the word PASSPORT is

Q 31 :

Total number of four digit odd numbers that can be formed using 0, 1, 2, 3, 5, 7 are

Q 32 :

If $P (n, r) = 1680$ and $C (n, r) = 70$ , then $69 n + r! =$

Q 33 :

If ${}^{43}C_{r - 6} = {}^{43}C_{3 r + 1}$ , then the value of $r$ is

Q 34 :

The number of ways in which 6 men and 5 women can dine at a round table if no two women are to sit together is given by

Q 35 :

The number of ways in which 21 objects can be grouped into three groups of 8, 7 and 6 objects is

Q 36 :

Eleven books consisting of 5 mathematics, 4 physics and 2 chemistry are placed on a shelf. The number of possible ways of arranging them on the assumption that the books on the same subject are all together is

Q 37 :

The number of ways in which 9 persons can be divided into three equal groups is

Q 38 :

The rank of the word MOTHER when the letters of the word are arranged alphabetically as in a dictionary, is:

Q 39 :

If the letters of the word SACHIN are arranged in all possible ways and these words are written out as in a dictionary, then the word SACHIN appears at serial number

Q 40 :

The number of permutations of the letters of the word 'CONSEQUENCE' in which all the three E's are together is

Q 41 :

If one person handshakes with the other only once and the number of handshakes is 66, then the number of persons will be

Q 42 :

The sum $1 \times 1! + 2 \times 2! + . . . + 50 \times 50!$ equals

Q 43 :

The number of words that can be written using all the letters of the word 'IRRATIONAL' is

Q 44 :

If $n$ is any positive integer, then $\frac{1}{2^{n}} ({}^{2 n}P_{n}) =$

Q 45 :

If ${}^{12}P_{r} = {}^{11}P_{6} + 6 \cdot {}^{11}P_{5}$ , then $r =$

Q 46 :

The value of ${}^{2}P_{1} + {}^{3}P_{1} + \dots + {}^{n}P_{1}$ is equal to

Q 47 :

A polygon has 35 diagonals, then the number of its sides is

Q 48 :

A polygon has 54 diagonals. Number of sides of this polygon is

Q 49 :

Let $n \in N$ with $n \geq 3$ . Then $(\binom{n}{3})$ can be expressed as a polynomial

Q 50 :

The value of ${}^{50}C_{4} + \sum_{r = 1}^{6} {}^{56 - r}C_{3}$ is