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

Q 31 :
The number of seven digits odd numbers, that can be formed using all the seven digits 1, 2, 2, 2, 3, 3, 5 is _______ . [2023]

0%

(240)

Given digits, 1, 2, 2, 2, 3, 3, 5

Total digits = 7

Total number of seven digits $= \frac{7!}{3! 2!} = 420$

Total number of seven digit even numbers $= \frac{3 \times 6!}{2! 3!} = 180$

$∴$ Total number of seven digit odd numbers = 420 - 180 = 240

Q 32 :
Let 5 digit numbers be constructed using the digits 0, 2, 3, 4, 7, 9 with repetition allowed, and are arranged in ascending order with serial numbers. Then the serial number of the number 42923 is _______ . [2023]

0%

(2997)

Number starting with 2 and 3 = $2 \times 6^{4}$

Number starting with 40 = $6^{3}$

Number starting with 420, 422, 423, 424, 427 = $5 \times 6^{2}$

Similarly, number starting with 4290 = 6

$Total = 2 \times 6^{4} + 6^{3} + 5 \times 6^{2} + 6 + 3 = 2997$

Q 33 :
If ${}^{2 n + 1}{P_{n - 1} : {}^{2 n - 1}{P_{n} = 11 : 21,}}$ then $n^{2} + n + 15$ is equal to ______ . [2023]

0%

(45)

Given, ${}^{2 n + 1}{P_{n - 1}} :^{2 n - 1} P_{n} = 11 : 21$

$\Rightarrow \frac{(2 n + 1)!}{(2 n + 1 - n + 1)!} \times \frac{(n - 1)!}{(2 n - 1)!} = \frac{11}{21}$

$\Rightarrow \frac{(2 n + 1)!}{(n + 2)!} \times \frac{(n - 1)!}{(2 n - 1)!} = \frac{11}{21}$

$\Rightarrow \frac{(2 n + 1) (2 n) (2 n - 1)!}{(n + 2) (n + 1) (n) (n - 1)!} \times \frac{(n - 1)!}{(2 n - 1)!} = \frac{11}{21}$

$\Rightarrow \frac{(2 n + 1) (2 n)}{(n + 2) (n + 1) (n)} = \frac{11}{21} \Rightarrow \frac{4 n + 2}{n^{2} + 3 n + 2} = \frac{11}{21}$

$\Rightarrow 84 n + 42 = 11 n^{2} + 33 n + 22 \Rightarrow 11 n^{2} - 51 n - 20 = 0$

$\Rightarrow (n - 5) (11 n + 4) = 0 \Rightarrow n = 5, \frac{- 4}{11}$

The value of $n$ can never be negative.

So, $n = 5$

$∴$ $n^{2} + n + 15 = {(5)}^{2} + 5 + 15 = 45$

Q 34 :
If all the words with or without meaning made using all the letters of the word "NAGPUR" are arranged as in a dictionary, then the word at $315^{th}$ position in this arrangement is [2024]

NRAPGU
NRAGPU
NRAPUG
NRAGUP

1

2

3

4

(1)

Letter in the word NAGPUR = 6

If we fix A _ _ _ _ _

Number of words start with A = 5! = 120

If we fix G _ _ _ _ _

Number of words start with G = 5! = 120

If we fix N A _ _ _ _

Number of words start with NA = 4! = 24

If we fix N G _ _ _ _

Number of words start with NG = 4! = 24

If we fix N P _ _ _ _

Number of words start with NP = 4! = 24

Next word i.e., $313^{th}$ word in dictionary will be NRAGPU, $314^{th}$ word will be NRAGUP, $315^{th}$ word will be NRAPGU.

Q 35 :
60 words can be made using all the letters of the word BHBJO, with or without meaning. If these words are written as in a dictionary, then the 50th word is [2024]

OBBJH
HBBJO
OBBHJ
JBBOH

1

2

3

4

(1)

Let us fix B first (Dictionary Order)

B _ _ _ _ (We have 4 distinct letters left, to be arranged in 4 distinct ways)
Number of ways = 4! = 24

H _ _ _ _ (We have B, B, J, O to be arranged in 4 ways when H is fixed)

Number of ways = $\frac{4!}{2!} = 12$

J _ _ _ _ (B, B, H, O to be arranged in 4 ways)

Number of ways = $\frac{4!}{2!} = 12$

49th word will be OBBHJ
50th word will be OBBJH

Q 36 :
If $n$ is the number of ways in which five different employees can sit into four indistinguishable offices where any office may have any number of persons including zero, then $n$ is equal to [2024]

43
47
53
51

1

2

3

4

(4)

Number of ways in which five different employees can sit are

$I$ $I I$ $I I I$ $I V$

$□$ $□$ $□$ $□$

5 0 0 0 $\to$ 5!/5! = 1

4 1 0 0 $\to$ 5!/4! = 5

3 2 0 0 $\to$ 5!/3!2! = 10

3 1 1 0 $\to$ 5!/3! 1! 1! 2! = 10

2 2 1 0 $\to$ $\frac{5!}{2! 2! 1! 2!} = 15$

2 1 1 1 $\to$ $\frac{5!}{2! 1! 1! 1! 3!} = 10$

$∴$ Total number of ways =1 + 5 + 10 + 10 + 15 + 10 = 51

Q 37 :
Let $α = \frac{(4!)!}{{(4!)}^{3!}}$ and $β = \frac{(5!)!}{{(5!)}^{4!}} .$ Then

$α \notin N$ and $β \in N$
$α \in N$ and $β \notin N$
$α \in N$ and $β \in N$
$α \notin N$ and $β \notin N$

1

2

3

4

(3)

Q 38 :
Number of ways of arranging 8 identical books into 4 identical shelves where any number of shelves may remain empty is equal to [2024]

18
15
12
16

1

2

3

4

(2)

Ways of arranging 8 identical books in 4 identical shelves = {8, 0, 0, 0}, {7, 1, 0, 0}, {6, 2, 0, 0}, {5, 3, 0, 0}, {4, 4, 0, 0}, {6, 1, 1, 0},{5, 2, 1, 0}, {4, 3, 1, 0}, {4, 2, 2, 0}, {3, 3, 2, 0}, {5, 1, 1, 1}, {4, 2, 1, 1}, {3, 3, 1, 1}, {3, 2, 2, 1}, {2, 2, 2, 2}.

Number of ways = 15

[As shelves and books are identical, so the arrangements (5, 2, 1, 0), (2, 5, 1, 0), and (1, 2, 5, 0) are considered as same]

Q 39 :
The number of ways of getting a sum 16 on throwing a dice four times is ______. [2024]

0%

(125)

We need sum of 16 on throwing a dice four times so possible ways are:
{(4, 4, 4, 4), (5, 4, 4, 3), (5, 5, 1, 5), (5, 5, 3, 3), (5, 5, 4, 2), (6, 4, 3, 3), (6, 4, 4, 2), (6, 5, 3, 2), (6, 5, 4, 1), (6, 6, 3, 1), (6, 6, 2, 2)}

Obtained Result	Number of Ways
(6, 6, 2, 2)	$\frac{4!}{2! 2!} = 6$
(6, 6, 3, 1)	$\frac{4!}{2!} = 12$
(6, 5, 4, 1)	$4! = 24$
(6, 5, 3, 2)	$4! = 24$
(6, 4, 4, 2)	$\frac{4!}{2!} = 12$
(6, 4, 3, 3)	$\frac{4!}{2!} = 12$
(5, 5, 4, 2)	$\frac{4!}{2!} = 12$
(5, 5, 3, 3)	$\frac{4!}{2! 2!} = 6$
(5, 5, 1, 5)	$\frac{4!}{3!} = 4$
(5, 4, 4, 3)	$\frac{4!}{2!} = 12$
(4, 4, 4, 4)	$1! = 1$
Total ways	125

Q 40 :
The number of integers, between 100 and 1000 having the sum of their digits equals to 14, is ______. [2024]

0%

(70)

Number between 100 to 1000 are 3-digit number.

Let number $a b c$ such that $a + b + c = 14$

where $a, b, c \in {0, 1, 2, . . . 9}$ and $a \geq 1$

Case I : All three digits are same ie., $a = b = c$

$\Rightarrow 3 a = 14$ , which is not possible

Case II : Two digits are same i.e., $a = b \neq c$

$\Rightarrow 2 a + c = 14$

$∴$ (a, c) = {(3, 8), (4, 6), (5, 4), (6, 2), (7, 0)}

In each subcase, number of ways of forming a 3-digit number = $\frac{3!}{2!}$

There are 5 such cases as (a, c) have 5 elements in which 0,7,7 is included

So, total number of 3 digits numbers when 2 digits are same $5 \times \frac{3!}{2!} - 1$

$= 15 - 1 = 14$

Case III : All digits are different

$∴$ (a, b, c) = {(1, 4, 9), (2, 4, 8), (2, 3, 9), (1, 5, 8), (3, 4, 7), (2, 5, 7), (1, 6, 7), (3, 5, 6), (5, 9, 0), (6, 8, 0)}

Total number of 3 digits number formed by above triplet = $10 \times 3! - 2 \times 2! = 60 - 4 = 56$

Total number of required number = 56 + 14 = 70

Topic Question Set

Q 31 :
The number of seven digits odd numbers, that can be formed using all the seven digits 1, 2, 2, 2, 3, 3, 5 is _______ . [2023]

0%

Q 32 :
Let 5 digit numbers be constructed using the digits 0, 2, 3, 4, 7, 9 with repetition allowed, and are arranged in ascending order with serial numbers. Then the serial number of the number 42923 is _______ . [2023]

0%

Q 33 :
If ${}^{2 n + 1}{P_{n - 1} : {}^{2 n - 1}{P_{n} = 11 : 21,}}$ then $n^{2} + n + 15$ is equal to ______ . [2023]

0%

Q 34 :
If all the words with or without meaning made using all the letters of the word "NAGPUR" are arranged as in a dictionary, then the word at $315^{th}$ position in this arrangement is [2024]

Q 35 :
60 words can be made using all the letters of the word BHBJO, with or without meaning. If these words are written as in a dictionary, then the 50th word is [2024]

Q 36 :
If $n$ is the number of ways in which five different employees can sit into four indistinguishable offices where any office may have any number of persons including zero, then $n$ is equal to [2024]

Q 37 :
Let $α = \frac{(4!)!}{{(4!)}^{3!}}$ and $β = \frac{(5!)!}{{(5!)}^{4!}} .$ Then

(3)

Q 38 :
Number of ways of arranging 8 identical books into 4 identical shelves where any number of shelves may remain empty is equal to [2024]

Q 39 :
The number of ways of getting a sum 16 on throwing a dice four times is ______. [2024]

0%

Q 40 :
The number of integers, between 100 and 1000 having the sum of their digits equals to 14, is ______. [2024]

0%

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Topic Question Set

Q 31 : The number of seven digits odd numbers, that can be formed using all the seven digits 1, 2, 2, 2, 3, 3, 5 is _______ . [2023]

0%

Q 32 : Let 5 digit numbers be constructed using the digits 0, 2, 3, 4, 7, 9 with repetition allowed, and are arranged in ascending order with serial numbers. Then the serial number of the number 42923 is _______ . [2023]

0%

Q 33 : If Pn-1:Pn=11:21,2n-12n+1 then n2+n+15 is equal to ______ . [2023]

0%

Q 34 : If all the words with or without meaning made using all the letters of the word "NAGPUR" are arranged as in a dictionary, then the word at 315th position in this arrangement is [2024]

Q 35 : 60 words can be made using all the letters of the word BHBJO, with or without meaning. If these words are written as in a dictionary, then the 50th word is [2024]

Q 36 : If n is the number of ways in which five different employees can sit into four indistinguishable offices where any office may have any number of persons including zero, then n is equal to [2024]

Q 37 : Let α=(4!)!(4!)3! and β=(5!)!(5!)4!. Then

(3)

Q 38 : Number of ways of arranging 8 identical books into 4 identical shelves where any number of shelves may remain empty is equal to [2024]

Q 39 : The number of ways of getting a sum 16 on throwing a dice four times is ______. [2024]

0%

Q 40 : The number of integers, between 100 and 1000 having the sum of their digits equals to 14, is ______. [2024]

0%

Want Us to Email you About Special Offers & Updates?

Q 31 :
The number of seven digits odd numbers, that can be formed using all the seven digits 1, 2, 2, 2, 3, 3, 5 is _______ . [2023]

Q 32 :
Let 5 digit numbers be constructed using the digits 0, 2, 3, 4, 7, 9 with repetition allowed, and are arranged in ascending order with serial numbers. Then the serial number of the number 42923 is _______ . [2023]

Q 33 :
If ${}^{2 n + 1}{P_{n - 1} : {}^{2 n - 1}{P_{n} = 11 : 21,}}$ then $n^{2} + n + 15$ is equal to ______ . [2023]

Q 34 :
If all the words with or without meaning made using all the letters of the word "NAGPUR" are arranged as in a dictionary, then the word at $315^{th}$ position in this arrangement is [2024]

Q 35 :
60 words can be made using all the letters of the word BHBJO, with or without meaning. If these words are written as in a dictionary, then the 50th word is [2024]

Q 36 :
If $n$ is the number of ways in which five different employees can sit into four indistinguishable offices where any office may have any number of persons including zero, then $n$ is equal to [2024]

Q 37 :
Let $α = \frac{(4!)!}{{(4!)}^{3!}}$ and $β = \frac{(5!)!}{{(5!)}^{4!}} .$ Then

Q 38 :
Number of ways of arranging 8 identical books into 4 identical shelves where any number of shelves may remain empty is equal to [2024]

Q 39 :
The number of ways of getting a sum 16 on throwing a dice four times is ______. [2024]

Q 40 :
The number of integers, between 100 and 1000 having the sum of their digits equals to 14, is ______. [2024]