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

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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.


Last Updated : 09 Mar, 2022
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