Class 11 RD Sharma Solutions- Chapter 16 Permutations - Exercise 16.3 | Set 1

Question 1. Evaluate each of the following:

(i) ⁸P₃

(ii) ¹⁰P₄

(iii) ⁶P₆

(iv) P (6, 4)

Solution:

(i) ⁸P₃

As we know that, ⁸P₃ can be written as P (8, 3)

After applying the formula,

P (n, r) = \frac{n!}{(n-r)!}

P (8, 3)  =\frac{8!}{(8-3)!}\\ =\frac{8!}{5!}\\ =\frac{(8\times7\times6\times5!)}{5!}

= 8 × 7 × 6

= 336

∴ ⁸P₃ = 336

(ii) ¹⁰P₄

As we know that, ¹⁰P₄ can be written as P (10, 4)

After applying the formula,

P (n, r) = \frac{n!}{(n-r)!}\\

P (10, 4) = \frac{10!}{(10-4)!}\\ =\frac{10!}{6!}\\ =\frac{(10\times9\times8\times7\times6!)}{6!}

= 10 × 9 × 8 × 7

= 5040

∴ ¹⁰P₄ = 5040

(iii) ⁶P₆

As we know that, ⁶P₆ can be written as P (6, 6)

After applying the formula,

P (n, r) = \frac{n!}{(n-r)!}

P (6, 6) = \frac{6!}{(6-6)!}\\ =\frac{6!}{0!}\\ =\frac{(6\times5\times4\times3\times2\times1)}{1}   {Since, 0! = 1}

= 6 × 5 × 4 × 3 × 2 × 1

= 720

∴ ⁶P₆ = 720

(iv) P (6, 4)

After applying the formula,

P (n, r) = \frac{n!}{(n-r)!}

P (6, 4) = \frac{6!}{(6-4)!}\\ =\frac{6!}{2!}\\ =\frac{6\times5\times4\times3\times2!}{2!}

= 6 × 5 × 4 × 3

= 360

∴ P (6, 4) = 360

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

Solution:

Given:

P (5, r) = P (6, r – 1)

After applying the formula,

P (n, r) = \frac{n!}{(n-r)!}

P (5, r) = \frac{5!}{(5-r)!}

P (6, r-1) = \frac{6!}{(6-(r^{-1}))!}\\ =\frac{6!}{(6-(r-1))!}\\ =\frac{6!}{(7-r)!}

So, from the question,

P (5, r) = P (6, r – 1)

So, after substituting the values in above expression we will get,

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

Upon evaluating,

\frac{(7-r)!}{(5-r)!}=\frac{6!}{5!}\\ \frac{[(7-r)(7-r-1)(7-r-2)!]}{(5-r)!}=\frac{(6\times5!)}{5!}\\ \frac{[(7-r)(6-r)(5-r)!]}{(5-r)!}=6

(7 – r) (6 – r) = 6

42 – 6r – 7r + r² = 6

42 – 6 – 13r + r² = 0

r² – 13r + 36 = 0

r² – 9r – 4r + 36 = 0

r(r – 9) – 4(r – 9) = 0

(r – 9) (r – 4) = 0

r = 9 or 4

For, P (n, r): r ≤ n

∴ r = 4 [for, P (5, r)]

Question 3. If 5 P(4, n) = 6 P(5, n – 1), find n.

Solution:

Given:

5 P(4, n) = 6 P(5, n – 1)

After applying the formula,

P (n, r) = \frac{n!}{(n-r)!}

P (4, n) = \frac{4!}{(4-n)!}

P (5, n-1) = \frac{5!}{(5-(n-1))!}\\ \frac{5!}{(5-n+1)!}\\ \frac{5!}{(6-n)!}

So, from the question,

5 P(4, n) = 6 P(5, n – 1)

So, after substituting the values in above expression we will get,

\frac{5 × 4!}{(4 – n)!} = \frac{6 × 5!}{(6 – n)!}

Upon evaluating,

\frac{(6 – n)!}{(4 – n)!}= \frac{6}{5} × \frac{5!}{4!}\\ \frac{[(6 – n) (6 – n – 1) (6 – n – 2)!]}{(4 – n)!} = \frac{(6 × 5 × 4!)}{ (5 × 4!)}\\ \frac{[(6 – n) (5 – n) (4 – n)!]}{(4 – n)!} = 6

(6 – n) (5 – n) = 6

30 – 6n – 5n + n² = 6

30 – 6 – 11n + n² = 0

n² – 11n + 24 = 0

n² – 8n – 3n + 24 = 0

n(n – 8) – 3(n – 8) = 0

(n – 8) (n – 3) = 0

n = 8 or 3

For, P (n, r): r ≤ n

∴ n = 3 [for, P (4, n)]

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

Solution:

Given:

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

After applying the formula,

P (n, r) = \frac{n!}{(n – r)!}

P (n, 5) = \frac{n!}{(n – 5)!}

P (n, 3) = \frac{n!}{(n – 3)!}

So, from the question,

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

After substituting the values in above expression we will get,

\frac{n!}{(n – 5)!} = 20 × \frac{n!}{(n – 3)!}

Upon evaluating,

\frac{n! (n – 3)!}{n! (n – 5)!} = 20\\ \frac{[(n – 3) (n – 3 – 1) (n – 3 – 2)!]} {(n – 5)!} = 20\\ \frac{[(n – 3) (n – 4) (n – 5)!]}{(n – 5)!} = 20

(n – 3) (n – 4) = 20

n² – 3n – 4n + 12 = 20

n² – 7n + 12 – 20 = 0

n² – 7n – 8 = 0

n² – 8n + n – 8 = 0

n(n – 8) – 1(n – 8) = 0

(n – 8) (n – 1) = 0

n = 8 or 1

For, P(n, r): n ≥ r

∴ n = 8 [for, P(n, 5)]

Question 5. If ⁿP₄ = 360, find the value of n.

Solution:

Given:

ⁿP₄ = 360

ⁿP₄ can be written as P (n , 4)

After applying the formula,

P (n, r) = \frac{n!}{(n – r)!}

P (n, 4) = \frac{n!}{(n – 4)!}

So, from the question,

ⁿP₄ = P (n, 4) = 360

After substituting the values in above expression we will get,

\frac{n!}{(n – 4)! }   = 360

\frac{[n(n – 1) (n – 2) (n – 3) (n – 4)!]} {(n – 4)!}   = 360

n (n – 1) (n – 2) (n – 3) = 360

n (n – 1) (n – 2) (n – 3) = 6×5×4×3

Upon comparing,

The value of n is 6.

Question 6. If P(9, r) = 3024, find r.

Solution:

Given:

P (9, r) = 3024

After applying the formula,

P (n, r) = \frac{n!}{(n – r)!}

P (9, r) = \frac{9!}{(9 – r)!}

So, from the question,

P (9, r) = 3024

Substituting the obtained values in above expression we get,

\frac{9!}{(9 – r)! }  = 3024

\frac{1}{(9 – r)!} = \frac{3024}{9!}

= \frac{3024}{(9×8×7×6×5×4×3×2×1)}

= \frac{3024}{(3024×5×4×3×2×1)}

= \frac{1}{5!}

(9 – r)! = 5!

9 – r = 5

-r = 5 – 9

-r = -4

∴ The value of r is 4.

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

Solution:

Given:

P (11, r) = P (12, r – 1)

After applying the formula,

P (n, r) = \frac{n!}{(n – r)!}

P (11, r) = \frac{11!}{(11 – r)!}

P (12, r-1) = \frac{12!}{(12 – (r-1))!}

= \frac{ 12!}{(12 – r + 1)!}

= \frac{12!}{(13 – r)!}

So, from the question,

P (11, r) = P (12, r – 1)

After substituting the values in above expression we will get,

\frac{11!}{(11 – r)!} =\frac{12!}{(13 – r)!}

Upon evaluating,

\frac{(13 – r)! }{(11 – r)!} = \frac{12!}{11!}

\frac{[(13 – r) (13 – r – 1) (13 – r – 2)!] }{(11 – r)!} = \frac{(12×11!)}{11!}

\frac{[(13 – r) (12 – r) (11 -r)!] }{(11 – r)! }  = 12

(13 – r) (12 – r) = 12

156 – 12r – 13r + r² = 12

156 – 12 – 25r + r² = 0

r² – 25r + 144 = 0

r² – 16r – 9r + 144 = 0

r(r – 16) – 9(r – 16) = 0

(r – 9) (r – 16) = 0

r = 9 or 16

For, P (n, r): r ≤ n

∴ r = 9 [for, P (11, r)]

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

Solution:

Given:

P (n, 4) = 12. P (n, 2)

After applying the formula,

P (n, r) = \frac{n!}{(n – r)!}

P (n, 4) = \frac{n!}{(n – 4)!}

P (n, 2) = \frac{n!}{(n – 2)!}

So, from the question,

P (n, 4) = 12. P (n, 2)

After substituting the values in above expression we will get,

\frac{n!}{(n – 4)!} = 12 × \frac{n!}{(n – 2)!}

Upon evaluating,

\frac{n! (n – 2)! }{ n! (n – 4)! }   = 12

\frac{[(n – 2) (n – 2 -1) (n – 2 – 2)!] }{(n – 4)! }  = 12

\frac{[(n – 2) (n – 3) (n – 4)!] }{ (n – 4)!}   = 12

(n – 2) (n – 3) = 12

n² – 3n – 2n + 6 = 12

n² – 5n + 6 – 12 = 0

n² – 5n – 6 = 0

n² – 6n + n – 6 = 0

n (n – 6) – 1(n – 6) = 0

(n – 6) (n – 1) = 0

n = 6 or 1

For, P (n, r): n ≥ r

∴ n = 6 [for, P (n, 4)]

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

Solution:

Given:

P (n – 1, 3): P (n, 4) = 1 : 9

\frac{P (n – 1, 3)}{ P (n, 4)} = \frac{1 }{ 9}

After applying the formula,

P (n, r) = \frac{n!}{(n – r)!}

P (n – 1, 3) = \frac{(n – 1)! }{ (n – 1 – 3)!}

= \frac{ (n – 1)! }{ (n – 4)!}

P (n, 4) = \frac{n!}{(n – 4)!}

So, from the question,

\frac{P (n – 1, 3)}{ P (n, 4)} = \frac{1 }{9}

After substituting the values in above expression we will get,

\frac{\frac{[(n – 1)!}{  (n – 4)!]}} { \frac{[n!}{(n – 4)!]}} = \frac{1}{9}\\ \frac{[(n – 1)! }{ (n – 4)!]} × \frac{[(n – 4)! }{ n!]} = \frac{1}{9}\\ \frac{(n – 1)!}{n!} = \frac{1}{9}\\ \frac{(n – 1)!}{n (n – 1)!} = \frac{1}{9}\\ \frac{1}{n} = \frac{1}{9}

n = 9

∴ The value of n is 9.

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

Solution:

Given:

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

\frac{P(2n – 1, n) }{ P(2n + 1, n – 1)} = \frac{22 }{ 7}

After applying the formula,

P (n, r) =\frac{ n!}{(n – r)!}

P (2n – 1, n) =\frac{ (2n – 1)! }{ (2n – 1 – n)!}

= \frac{(2n – 1)! }{ (n – 1)!}

P (2n + 1, n – 1) = \frac{(2n + 1)! }{ (2n + 1 – n + 1)!}

= \frac{(2n + 1)! }{ (n + 2)!}

So, from the question,

\frac{P(2n – 1, n) }{ P(2n + 1, n – 1)} = \frac{22 }{ 7}

After substituting the values in above expression we will get,

\frac{\frac{(2n – 1)! }{ (n – 1)!} }{\frac{ (2n + 1)! }{ (n + 2)!}} = \frac{22}{7}\\ \frac{(2n – 1)! }{ (n – 1)!} × \frac{(n + 2)! }{ (2n + 1)!} = \frac{22}{7}\\ \frac{(2n – 1)! }{ (n – 1)!} × \frac{[(n + 2) (n + 2 – 1) (n + 2 – 2) (n + 2 – 3)!] }{ [(2n + 1) (2n + 1 – 1) (2n + 1 – 2)]} = \frac{22}{7}\\ \frac{[(2n – 1)! }{ (n – 1)!]} × \frac{[(n + 2) (n + 1) n(n – 1)!] }{ [(2n + 1) 2n (2n – 1)!]} = \frac{22}{7}\\ \frac{[(n + 2) (n + 1)] }{ (2n + 1)2} = \frac{22}{7}\\

7(n + 2) (n + 1) = 22×2 (2n + 1)

7(n² + n + 2n + 2) = 88n + 44

7(n² + 3n + 2) = 88n + 44

7n² + 21n + 14 = 88n + 44

7n² + 21n – 88n + 14 – 44 = 0

7n² – 67n – 30 = 0

7n² – 70n + 3n – 30 = 0

7n(n – 10) + 3(n – 10) = 0

(n – 10) (7n + 3) = 0

n = 10, \frac{-3}{7}

As we know that, n ≠ \frac{-3}{7}

∴ The value of n is 10.