Question 1: Evaluate each of the following:

i) ^8P_3          ii) ^{10}P_4            iii) ^6P_6           iv) P(6, 4)

Answer:

We know, ^nP_r = \frac{n!}{(n-r)!}

i) ^8P_3 = \frac{8!}{(8-3)!} = \frac{8!}{5!} = \frac{8 \times 7 \times 6 \times 5!}{5!} = 8 \times 7 \times 6 = 336

ii) ^{10}P_4 = \frac{10!}{(10-4)!} = \frac{10!}{6!} = \frac{10 \times 9 \times 8 \times 7 \times 6!}{6!} = 10 \times 9 \times 8 \times 7 = 5040

iii) ^6P_6 = \frac{6!}{(6-6)!} = \frac{6!}{0!} =  6! = 720

iv) P(6, 4) = ^6P_4 = \frac{6!}{(6-4)!} = \frac{6!}{2!} = \frac{6 \times 5 \times 4 \times 3 \times 2!}{2!} = 6 \times 5 \times 4 \times 3 = 360

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Question 2: If P (5, r) = P (6, r - 1) , find r .

Answer:

P (5, r) = P (6, r - 1)

\Rightarrow \frac{5!}{(5-r)!} = \frac{6!}{(6-r+1)!}

\Rightarrow \frac{5!}{(5-r)!} = \frac{6 \times 5!}{(7-r)!}

\Rightarrow \frac{5!}{(5-r)!} = \frac{6 \times 5!}{(7-r)(6-r)(5-r)!}

\Rightarrow \frac{1}{1} = \frac{6 }{(7-r)(6-r)}

\Rightarrow (7-r)(6-r) = 6

\Rightarrow 42 - 6r - 7 r + r^2 = 6

\Rightarrow r^2 - 13 r + 36 = 0

\Rightarrow ( r-9)(r-4) = 0

\Rightarrow r = 9 \text{ or } r = 4

Since \Rightarrow r \leq n,  \therefore r \neq 9

Hence \Rightarrow r = 4

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Question 3: If 5 P(4, n) = 6 \cdot P (5, n - 1) , find n .

Answer:

5 P(4, n) = 6 \cdot P (5, n - 1)

\Rightarrow 5 \cdot ^4P_n = 6 \cdot ^5P_{n-1}

\Rightarrow 5 \cdot \frac{4!}{(4-n)!} = 6 \cdot \frac{5!}{(5-n+1)!}

 \Rightarrow \frac{1}{(4-n)!} = \frac{6}{(6-n)!}

\Rightarrow \frac{1}{(4-n)!} = \frac{6}{(6-n)(5-n)(4-n)!}

\Rightarrow (6-n)(5-n) = 6

\Rightarrow (6-n)(5-n) = 3 \times 2

Comparing LHS and RHS we get

6 - n = 3 \Rightarrow n = 3

5-n = 2 \Rightarrow n = 3

Hence n = 3

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Question 4: If P (n,5) =20 \cdot P(n, 3) , find n .

Answer:

P (n,5) =20 \cdot P(n, 3)

\Rightarrow  ^nP_5 = 20 \cdot ^nP_3

\Rightarrow \frac{n!}{(n-5)!} = 20 \cdot \frac{n!}{(n-3)!}

 \Rightarrow \frac{1}{(n-5)!} = \frac{20}{(n-3)(n-4)(n-5)!}

\Rightarrow (n-3)(n-4) = 20

\Rightarrow (n-3)(n-4) = 5 \times 4

Comparing LHS and RHS we get

n-3 = 5 \Rightarrow n = 8

n-4 = 4 \Rightarrow n = 8

Hence n = 8

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Question 5: If ^nP_4 = 360 , find the value of n .

Answer:

^nP_4 = 360

\Rightarrow \frac{n!}{(n-4)!} = 360

\Rightarrow n(n-1)(n-2)(n-3) = 360

\Rightarrow n(n-1)(n-2)(n-3) = 6 \times 5 \times 4 \times 3

Comparing LHS and RHS we get

n = 6 

n-1 = 5 \Rightarrow n = 6

n-2 = 4 \Rightarrow n = 6

n-3 = 3 \Rightarrow n = 6

Hence n = 6

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Question 6:  If P (9, r) = 360 , find r

Answer:

P (9, r) = 360

\Rightarrow ^9P_r = 360

\Rightarrow \frac{9!}{(9-r)!} = 360

\Rightarrow \frac{9!}{(9-r)!} = 9 \times 8 \times 7 \times 6

\Rightarrow \frac{9!}{(9-r)!} = \frac{9!}{5!}

\Rightarrow 9-r = 5

\Rightarrow r = 4

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Question 7: If P(11, r) = P (12, r - 1) find r .

Answer:

P(11, r) = P (12, r - 1)

\Rightarrow \frac{11!}{(11-r)!} = \frac{12!}{(12-r+1)!}

\Rightarrow \frac{1}{(11-r)!} = \frac{12}{(13-r)!}

\Rightarrow \frac{1}{(11-r)!} = \frac{12}{(13-r)(12-r)(11-r)!}

\Rightarrow (13-r)(12-r) = 12

\Rightarrow (13-r)(12-r) = 4 \times 3

Comparing LHS and RHS we get

13-r = 4 \Rightarrow r = 9

12 - r = 3 \Rightarrow r = 9

\therefore r = 9

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Question 8: If P (n, 4) =12 \cdot P (n,2) , find n .

Answer:

P (n, 4) =12 \cdot P (n,2)

\Rightarrow \frac{n!}{(n-4)!} = 12 \cdot \frac{n!}{(n-2)!}

\Rightarrow \frac{1}{(n-4)!} = \frac{12}{(n-2)(n-3)(n-4))!}

\Rightarrow (n-2)(n-3) = 12

\Rightarrow (n-2)(n-3) = 4 \times 3

Comparing LHS and RHS we get

n-2 = 4 \Rightarrow n = 6

n-3 = 3 \Rightarrow n=6

\therefore n = 6

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Question 9: If P (n -1, 3) : P (n, 4) =1 :9 , find n .

Answer:

P (n -1, 3) : P (n, 4) =1 :9

\Rightarrow \frac{(n-1)!}{(n-1-3)!} : \frac{(n)!}{(n-4)!} = 1 : 9

\Rightarrow \frac{(n-1)!}{(n-1-3)!} \times \frac{(n-4)!}{(n)!} = \frac{1}{9}

\Rightarrow \frac{1}{n-1} = \frac{1}{9}

\Rightarrow 9 = n - 1

\Rightarrow n = 10

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Question 10: If P (2n -1, n): P (2n +1, n -1) = 22:7 find n .

Answer:

P (2n -1, n): P (2n +1, n -1) = 22:7

\Rightarrow \frac{(2n-1)!}{(2n-1-n)!} : \frac{(2n+1)!}{(2n+1-n+1)! } = 22: 7

\Rightarrow \frac{(2n-1)!}{(n-1)!} : \frac{(2n+1)!}{(n+2)! } = 22: 7

\Rightarrow \frac{(2n-1)!}{(n-1)!} \times \frac{(n+2)! }{(2n+1)!} = \frac{22}{7}

\Rightarrow \frac{(2n-1)!}{(n-1)!} \times \frac{(n+2)(n+1)(n)(n-1)! }{(2n+1)(2n)(2n-1)!} = \frac{22}{7}

\Rightarrow \frac{(n+2)(n+1)(n)}{(2n+1)(2n)} = \frac{22}{7}

\Rightarrow 7 ( n+2)(n+1) = 44 ( 2n+1)

\Rightarrow 7n^2 + 14n + 7n + 14 = 88n +44

\Rightarrow 7n^2 + 21 n + 14 = 88 n + 44

\Rightarrow 7n^2 - 67n - 30 =0

\Rightarrow (n-10)(7n+3) = 0

\Rightarrow n = 10 or n = \frac{-3}{7} (this is not possible)

\Rightarrow \therefore n = 10

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Question 11: If P (n, 5):P (n,3)= 2:1 , find n .

Answer:

P (n, 5):P (n,3) = 2:1

\Rightarrow \frac{n!}{(n-5)!} : \frac{n!}{(n-3)!} = 2:1

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

\Rightarrow \frac{(n-3)(n-4)(n-5)!}{(n-5)!} = 2

\Rightarrow (n-3)(n-4) = 2

\Rightarrow n^2 - 7n + 12 = 2

\Rightarrow n^2 - 7n + 10 = 0

\Rightarrow (n-5)(n-2) = 0

\Rightarrow n = 5 or n = 2 ( not possible as n cannot be letter than 3)

\therefore n = 5

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Question 12: Prove that: 1 \cdot P (1, 1) + 2 \cdot P (2, 2) + 3 \cdot P ( 3, 3) +\ldots + n \cdot P (n, n) = P (n + 1, n + 1) - 1

Answer:

LHS = 1 \cdot P (1, 1) + 2 \cdot P (2, 2) + 3 \cdot P ( 3, 3) +\ldots + n \cdot P (n, n)

= 1 \cdot ^1P_1 + 2 \cdot ^2P_2 + 3 \cdot ^3P_3 +\ldots + n \cdot ^nP_n

= 1 \cdot  \frac{1!}{(1-1)!} + 2 \cdot \frac{2!}{(2-2)!} + 3 \cdot \frac{3!}{(3-3)!} +\ldots + n \cdot \frac{n!}{(n-n)!}

= 1 \cdot 1! + 2 \cdot 2! + 3 \cdot 3! +\ldots + n \cdot n!

= \sum \limits_{r=1}^{n}  r \cdot r!

= \sum \limits_{r=1}^{n}  [(r+1)-1] \cdot r! 

= \sum \limits_{r=1}^{n}  [(r+1)! - r! ]

= [ ( 2! - 1!) + ( 3! - 2!) + \ldots + [(n+1)! - n!]

= ( n + 1 )! - 1!

= ( n + 1 )! - 1 = RHS. Hence proved.

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Question 13: If P (15, r -1) : P (16, r - 2) = 3 : 4 ,  find r

Answer:

P (15, r -1) : P (16, r - 2) = 3 : 4

\Rightarrow ^{15}P_{r-1} : ^{16}P_{r-2} = 3:4

\Rightarrow \frac{15!}{(15-r+1)!} : \frac{16!}{(16-r+2)!} = 3:4

\Rightarrow \frac{15!}{(15-r+1)!} \times \frac{(18-r)!}{16!} = \frac{3}{4}

\Rightarrow \frac{1}{(16-r)!} \times \frac{(18-r)(17-r)(16-r)!}{16!} = \frac{3}{4}

\Rightarrow (18-r)(17-r) = 12

\Rightarrow (18-r)(17-r) = 4 \times 3

Comparing LHS with RHS, we get

18-r = 4 \Rightarrow r = 14

17-r = 3 \Rightarrow r = 14

\therefore r = 14

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Question 14: If ^{n+5}P_{n+1} = \Big[ \frac{11(n-1)}{2} \Big] ^{n+3}P_n , find n

Answer:

^{n+5}P_{n+1} = \Big[ \frac{11(n-1)}{2} \Big] ^{n+3}P_n

\Rightarrow \frac{(n+5)!}{(n+5-n-1)!} = \Big[ \frac{11(n-1)}{2} \Big]  \frac{(n+3)!}{(n+3-n)!}

\Rightarrow \frac{(n+5)(n+4)(n+3)!}{4!} = \Big[ \frac{11(n-1)}{2} \Big] \frac{(n+3)!}{3!}

\Rightarrow \frac{(n+5)(n+4)}{4} = \frac{11(n-1)}{2}

\Rightarrow (n+5)(n+4) = 22 ( n-1)

\Rightarrow n^2 - 13n + 42 = 0

\Rightarrow (n-7)(n-6) = 0

\Rightarrow n = 7 \text{ or } n = 6

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Question 15: In how, many ways can five children stand in a queue?

Answer:

We need to permute 5 children out of 5 .

Therefore the required number of ways = ^5P_5 = \frac{5!}{(5-5)!} = 5! = 120

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Question 16: From among the 36 teachers in a school, one principal and one vice-principal are to be appointed. In how many ways can this be done?

Answer:

We need to permute 2 teachers out of 36 .

Therefore the required number of ways = ^{36}P_2 = \frac{36!}{(36-2)!} = 36 \times 35 = 1260

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Question 17: Four letters \text{E, K, S} and \text{ V } , one in each, were purchased from a plastic warehouse. How many ordered pairs of letters, to be used as initials, can be formed from them?

Answer:

We need to permute 2 letters out of 4 .

Therefore the required number of ways = ^4P_2 = \frac{4!}{(4-2)!} = 4 \times 3 = 12

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Question 18: Four books, one each in Chemistry, Physics, Biology and Mathematics, are to be arranged in a shelf. In how many ways can this be done?

Answer:

We need to permute 4 books out of 4 .

Therefore the required number of ways = ^4P_4 = \frac{4!}{(4-4)!} = 4! = 24

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Question 19: Find the number of different 4- letter words, with or without meanings, that can be formed from the letters of the word \text{ 'NUMBER'} .

Answer:

We need to permute 4 letters out of 6 .

Therefore the required number of ways = ^6P_4 = \frac{6!}{(6-4)!} = 6 \times 5 \times 4 \times 3 = 360

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Question 20: How many three-digit numbers are there, with distinct digits, with each digit odd?

Answer:

The odd digits are 1, 3, 5, 7, 9

We need to permute 3 digits out of 5 .

Therefore the required number of ways = ^5P_3 = \frac{5!}{(5-3)!} = 5 \times 4 \times 3  = 60

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Question 21: How many words, with or without meaning, can be formed by using all the letters of the word \text{'DELHI'} , using each letter exactly once?

Answer:

Number of letters in \text{DELHI} are 5

We need to permute 5 letters out of 5 .

Therefore the required number of ways = ^5P_5 = \frac{5!}{(5-5)!} = 5! = 120

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Question 22: How, many words, with or without meaning, can be formed by using the letters of the word \text{'TRIANGLE'} ?

Answer:

Number of letters in \text{TRIANGLE} are 8

We need to permute 8 letters out of 8 .

Therefore the required number of ways = ^8P_8 = \frac{8!}{(8-8)!} = 8!

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Question 23: There are two works each of 3 volumes and two works each of 2 volumes; In how many ways can the 10 books be placed on a shelf so that the volumes of the same work are not separated?

Answer:

There are 4 different types of works.

Number of ways in which these 4 works can be arranged = ^4P_4 = 4! 

Within the works the volumes can be arranged.

Therefore the total number of ways arrangement can be made = 4! \times ( 3! \times 3!) \times ( 2! \times 2!) = 3456 .

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Question 24: There are 6 items in column A and 6 items in column B . A student is asked to match each item in column A with an item in column B . How many possible, correct or incorrect, answers are there to this question?

Answer:

There are 6 items in column A and 6 items in column B .

The first item from column A can be matched with any of the 6 items from column B .

Similarly, once the first items from column A has been matched, the  2nd item from the column A can only be matched with 5 items of the column B . and so on….

Hence the number of ways the match can be done = ^6P_6 = 6! = 720

The other way to look at this is to keep column A fixed and arrange 6 items from column B which would be ^6P_6 = 6! = 720

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Question 25: How many three-digit numbers are there, with no digit repeated?

Answer:

Total number of 3 digit number = ^{10}P_3 = \frac{10!}{(10-3)!} = \frac{10!}{7!} = 10 \times 9 \times 8 = 720

Total number of 3 digit numbers starting with 0 = ^9P_2 =  \frac{9!}{(9-2)!} = \frac{9!}{7!} = 9 \times 8 = 72

Hence the three-digit numbers are there, with no digit repeated = 720 - 72 = 648 

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Question 26: How many 6- digit telephone numbers can be constructed with digits 0,1,2,3,4,5,6,7,8,9 if each number starts with 35 and no digit appears more than once?

Answer:

In total there are 10 digits to chose from.

However, the first two digits are already fixed to 35 .

Therefore we can now chose 4 numbers from 8 in

= ^8P_4 = \frac{8!}{4!} = 8 \times 7 \times 6 \times 5 = 1680 ways.

Therefore 1680 , 6- digit telephone numbers can be constructed with digits 0,1,2,3,4,5,6,7,8,9 if each number starts with 35 and no digit appears more than once.

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Question 27: In how many ways can 6 boys and 5 girls be arranged for a group photograph if the girls are to sit on chairs in a row, and the boys are to stand in a row behind them?

Answer:

Number of ways you can arrange the girls = 5!

Number of ways you can arrange the boys = 6!

Hence the total number of arrangements = 5! \times 6! = 86400

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Question 28: If a denotes the number of permutations of (x + 2) things taken all at a time, b the number of permutations of x things taken 11 at a time and c the number of permutations of x - 11 things taken all at a time such that a = 182 bc , find the value of x .

Answer:

a = ^{x+2}P_{x+2} = \frac{(x+2)!}{x+2 - x - 2} = (x+2)!

b = ^{x}P_{11} = \frac{x!}{(x-11)!}

c = ^{x-11}P_{x-11 - x + 11} = \frac{(x-11)!}{x-11-x+11} = (x-11)!

Given a = 182 bc

\Rightarrow (x+2)! = 182 \cdot \frac{x!}{x-11} \cdot (x-11)!

\Rightarrow (x+2)! = 182 \cdot x!

\Rightarrow (x+2)(x+1) x! = 182 \cdot x!

\Rightarrow (x+2)(x+1) = 182

\Rightarrow (x+2)(x+1) = 14 \times 13

Comparing LHS with RHS we get

x+2 = 14 \Rightarrow x = 12

x+ 1 = 13 \Rightarrow x = 12

\therefore x = 12

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Question 29: How many 3- digit numbers can be formed by using the digits 1 to 9 if no digit is repeated?

Answer:

The total number of 3 digit number that can be formed = ^9P_3 = \frac{9!}{6!} = 9 \times 8 \times 7 = 504

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Question 30: How many 3- $digit even numbers can be made using the digits 1,2,3,4, 5,6,7 if no digits is repeated?

Answer:

The even digits are 2, 4, 6

The number of 3-digit numbers when the last digit is 2 = ^6P_2 = \frac{6!}{4!} = 6 \times 5 = 30

The number of 3-digit numbers when the last digit is 4 = ^6P_2 = \frac{6!}{4!} = 6 \times 5 = 30

The number of 3-digit numbers when the last digit is 6 = ^6P_2 = \frac{6!}{4!} = 6 \times 5 = 30

Hence the total number of 3 digit even numbers that can be formed = 30 + 30 + 30 = 90

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Question 31: Find the numbers of 4- digit numbers that can be formed using the digits 1, 2,3,4,5 if no digit is repeated? How many of these will be even?

Answer:

The number of 4 digit numbers that can be formed = ^5P_4 = \frac{5!}{1!} = 5 \times 4 \times 3 \times 2 = 120

The even digits are 2, 4

The number of 4-digit numbers when the last digit is 2 = ^4P_3 = \frac{4!}{1!} = 24

The number of 4-digit numbers when the last digit is 4 = ^4P_3 = \frac{4!}{1!} = 24

Therefore the total number of even numbers = 24 + 24 = 48

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Question 32: All the letters of the word \text{'EAMCOT'} are arranged in different possible ways. Find the number of arrangements in which no two vowels are adjacent to each other.

Answer:

Number of consonants = 3  \text{( M, C, T)}

Number of vowels = 3  \text{( E, A, O)}

No two vowels can be together.

Number of ways the vowels can be arranged = ^4P_3 = \frac{4!}{1} = 24

Number of ways consonants can be arranged = 3P_3 = 3! = 6

Hence the total number of arrangements in which no two vowels are adjacent to each other =  24 \times 6 = 144