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# 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 × 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) = P (10, 4) = = 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) = P (6, 6) = {Since, 0! = 1}

= 6 × 5 × 4 × 3 × 2 × 1

= 720

6P6 = 720

(iv) P (6, 4)

After applying the formula,

P (n, r) = P (6, 4) = = 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) = P (5, r) = P (6, r-1) = So, from the question,

P (5, r) = P (6, r – 1)

So, after substituting the values in above expression we will get, Upon evaluating, (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) = P (4, n) = P (5, n-1) = So, from the question,

5 P(4, n) = 6 P(5, n – 1)

So, after substituting the values in above expression we will get, Upon evaluating, (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) = 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 – 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) = 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 – 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) = P (9, r) = So, from the question,

P (9, r) = 3024

Substituting the obtained values in above expression we get, = 3024    (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) = P (11, r) = P (12, r-1) =   So, from the question,

P (11, r) = P (12, r – 1)

After substituting the values in above expression we will get, Upon evaluating,   = 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) = 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 – 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 After applying the formula,

P (n, r) = P (n – 1, 3) =  P (n, 4) = So, from the question, After substituting the values in above expression we will get, 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 After applying the formula,

P (n, r) = P (2n – 1, n) =  P (2n + 1, n – 1) =  So, from the question, After substituting the values in above expression we will get, 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, As we know that, n ≠ ∴ The value of n is 10.

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