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

Last Updated : 09 Mar, 2022

### 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) =

P (8, 3)

= 8 Ã— 7 Ã— 6

= 336

âˆ´ 8P3 = 336

(ii) 10P4

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

After applying the formula,

P (n, r) =

P (10, 4) =

= 10 Ã— 9 Ã— 8 Ã— 7

= 5040

âˆ´ 10P4 = 5040

(iii) 6P6

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

After applying the formula,

P (n, r) =

P (6, 6) =  {Since, 0! = 1}

= 6 Ã— 5 Ã— 4 Ã— 3 Ã— 2 Ã— 1

= 720

âˆ´ 6P6 = 720

(iv) P (6, 4)

After applying the formula,

P (n, r) =

P (6, 4) =

= 6 Ã— 5 Ã— 4 Ã— 3

= 360

âˆ´ P (6, 4) = 360

### Question 2. If P (5, r) = P (6, r â€“ 1), find r.

Solution:

Given:

P (5, r) = P (6, r â€“ 1)

After applying the formula,

P (n, r) =

P (5, r) =

P (6, r-1) =

So, from the question,

P (5, r) = P (6, r â€“ 1)

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

Upon evaluating,

(7 â€“ r) (6 â€“ r) = 6

42 â€“ 6r â€“ 7r + r2 = 6

42 â€“ 6 â€“ 13r + r2 = 0

r2 â€“ 13r + 36 = 0

r2 â€“ 9r â€“ 4r + 36 = 0

r(r â€“ 9) â€“ 4(r â€“ 9) = 0

(r â€“ 9) (r â€“ 4) = 0

r = 9 or 4

For, P (n, r): r â‰¤ n

âˆ´ r = 4 [for, P (5, r)]

### Question 3. If 5 P(4, n) = 6 P(5, n â€“ 1), find n.

Solution:

Given:

5 P(4, n) = 6 P(5, n â€“ 1)

After applying the formula,

P (n, r) =

P (4, n) =

P (5, n-1) =

So, from the question,

5 P(4, n) = 6 P(5, n â€“ 1)

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

Upon evaluating,

(6 â€“ n) (5 â€“ n) = 6

30 â€“ 6n â€“ 5n + n2 = 6

30 â€“ 6 â€“ 11n + n2 = 0

n2 â€“ 11n + 24 = 0

n2 â€“ 8n â€“ 3n + 24 = 0

n(n â€“ 8) â€“ 3(n â€“ 8) = 0

(n â€“ 8) (n â€“ 3) = 0

n = 8 or 3

For, P (n, r): r â‰¤ n

âˆ´ 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) =

P (n, 5) =

P (n, 3) =

So, from the question,

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

After substituting the values in above expression we will get,

Upon evaluating,

(n â€“ 3) (n â€“ 4) = 20

n2 â€“ 3n â€“ 4n + 12 = 20

n2 â€“ 7n + 12 â€“ 20 = 0

n2 â€“ 7n â€“ 8 = 0

n2 â€“ 8n + n â€“ 8 = 0

n(n â€“ 8) â€“ 1(n â€“ 8) = 0

(n â€“ 8) (n â€“ 1) = 0

n = 8 or 1

For, P(n, r): n â‰¥ r

âˆ´ 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) =

P (n, 4) =

So, from the question,

nP4 = P (n, 4) = 360

After substituting the values in above expression we will get,

= 360

= 360

n (n â€“ 1) (n â€“ 2) (n â€“ 3) = 360

n (n â€“ 1) (n â€“ 2) (n â€“ 3) = 6Ã—5Ã—4Ã—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) =

P (9, r) =

So, from the question,

P (9, r) = 3024

Substituting the obtained values in above expression we get,

= 3024

(9 â€“ r)! = 5!

9 â€“ r = 5

-r = 5 â€“ 9

-r = -4

âˆ´ The value of r is 4.

### Question 7. If P (11, r) = P (12, r â€“ 1), find r.

Solution:

Given:

P (11, r) = P (12, r â€“ 1)

After applying the formula,

P (n, r) =

P (11, r) =

P (12, r-1) =

So, from the question,

P (11, r) = P (12, r â€“ 1)

After substituting the values in above expression we will get,

Upon evaluating,

= 12

(13 â€“ r) (12 â€“ r) = 12

156 â€“ 12r â€“ 13r + r2 = 12

156 â€“ 12 â€“ 25r + r2 = 0

r2 â€“ 25r + 144 = 0

r2 â€“ 16r â€“ 9r + 144 = 0

r(r â€“ 16) â€“ 9(r â€“ 16) = 0

(r â€“ 9) (r â€“ 16) = 0

r = 9 or 16

For, P (n, r): r â‰¤ n

âˆ´ 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) =

P (n, 4) =

P (n, 2) =

So, from the question,

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

After substituting the values in above expression we will get,

Upon evaluating,

= 12

= 12

= 12

(n â€“ 2) (n â€“ 3) = 12

n2 â€“ 3n â€“ 2n + 6 = 12

n2 â€“ 5n + 6 â€“ 12 = 0

n2 â€“ 5n â€“ 6 = 0

n2 â€“ 6n + n â€“ 6 = 0

n (n â€“ 6) â€“ 1(n â€“ 6) = 0

(n â€“ 6) (n â€“ 1) = 0

n = 6 or 1

For, P (n, r): n â‰¥ r

âˆ´ n = 6 [for, P (n, 4)]

### Question 9. If P(n â€“ 1, 3) : P(n, 4) = 1 : 9, find n.

Solution:

Given:

P (n â€“ 1, 3): P (n, 4) = 1 : 9

After applying the formula,

P (n, r) =

P (n â€“ 1, 3) =

P (n, 4) =

So, from the question,

After substituting the values in above expression we will get,

n = 9

âˆ´ The value of n is 9.

### Question 10. If P(2n â€“ 1, n) : P(2n + 1, n â€“ 1) = 22 : 7 find n.

Solution:

Given:

P(2n â€“ 1, n) : P(2n + 1, n â€“ 1) = 22 : 7

After applying the formula,

P (n, r) =

P (2n â€“ 1, n) =

P (2n + 1, n â€“ 1) =

So, from the question,

After substituting the values in above expression we will get,

7(n + 2) (n + 1) = 22Ã—2 (2n + 1)

7(n2 + n + 2n + 2) = 88n + 44

7(n2 + 3n + 2) = 88n + 44

7n2 + 21n + 14 = 88n + 44

7n2 + 21n â€“ 88n + 14 â€“ 44 = 0

7n2 â€“ 67n â€“ 30 = 0

7n2 â€“ 70n + 3n â€“ 30 = 0

7n(n â€“ 10) + 3(n â€“ 10) = 0

(n â€“ 10) (7n + 3) = 0

n = 10,

As we know that, n â‰

âˆ´ The value of n is 10.

Previous
Next