# Class 11 RD Sharma Solutions – Chapter 16 Permutations – Exercise 16.3 | Set 2

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

Solution:

Given:

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

After applying the formula,

P (n, r) =

P (n, 5) =

P (n, 3) =

So, from the question,

After substituting the values in above expression we will get,

(n â€“ 3)(n â€“ 4) = 2

n2 â€“ 3n â€“ 4n + 12 = 2

n2 â€“ 7n + 12 â€“ 2 = 0

n2 â€“ 7n + 10 = 0

n2 â€“ 5n â€“ 2n + 10 = 0

n (n â€“ 5) â€“ 2(n â€“ 5) = 0

(n â€“ 5) (n â€“ 2) = 0

n = 5 or 2

For, P (n, r): n â‰¥ r

âˆ´ n = 5 [for, P (n, 5)]

### 1. P (1, 1) + 2. P (2, 2) + 3 . P (3, 3) + â€¦ + n . P(n, n) = P(n + 1, n + 1) â€“ 1.

Solution:

By using the formula,

P (n, r) =

P (n, n) =

=

= n! [Since, 0! = 1]

Consider LHS:

= 1. P(1, 1) + 2. P(2, 2) + 3. P(3, 3) + â€¦ + n . P(n, n)

= 1.1! + 2.2! + 3.3! +â€¦â€¦â€¦+ n.n! [Since, P(n, n) = n!]

= (2! â€“ 1!) + (3! â€“ 2!) + (4! â€“ 3!) + â€¦â€¦â€¦ + (n! â€“ (n â€“ 1)!) + ((n+1)! â€“ n!)

= 2! â€“ 1! + 3! â€“ 2! + 4! â€“ 3! + â€¦â€¦â€¦ + n! â€“ (n â€“ 1)! + (n+1)! â€“ n!

= (n + 1)! â€“ 1!

= (n + 1)! â€“ 1 [Since, P (n, n) = n!]

= P(n+1, n+1) â€“ 1

= RHS

Hence Proved.

### Question 13. If P(15, r â€“ 1) : P(16, r â€“ 2) = 3 : 4, find r.

Solution:

Given:

P(15, r â€“ 1) : P(16, r â€“ 2) = 3 : 4

After applying the formula,

P (n, r) =

P (15, r â€“ 1) =

P (16, r â€“ 2) =

So, from the question,

After substituting the values in above expression we will get,

(18 â€“ r) (17 â€“ r) = 12

306 â€“ 18r â€“ 17r + r2 = 12

306 â€“ 12 â€“ 35r + r2 = 0

r2 â€“ 35r + 294 = 0

r2 â€“ 21r â€“ 14r + 294 = 0

r(r â€“ 21) â€“ 14(r â€“ 21) = 0

(r â€“ 14) (r â€“ 21) = 0

r = 14 or 21

For, P(n, r): r â‰¤ n

âˆ´ r = 14 [for, P(15, r â€“ 1)]

### Question 14. n+5Pn+1 = 11(n â€“ 1)/2 n+3Pn, find n.

Solution:

Given:

n+5Pn+1 = 11(n â€“ 1)/2 n+3Pn

P (n +5, n + 1) = 11(n â€“ 1)/2 P(n + 3, n)

By using the formula,

P (n, r) =

P(n + 5, n=1) =

P(n + 3, n) =

So, from the question

P(n + 5, n + 1) = 11(n -1)/2P(n + 3, n)

After substituting the values in above expression we get,

(n + 5) (n + 4) = 22 (n â€“ 1)

n2 + 4n + 5n + 20 = 22n â€“ 22

n2 + 9n + 20 â€“ 22n + 22 = 0

n2 â€“ 13n + 42 = 0

n2 â€“ 6n â€“ 7n + 42 = 0

n(n â€“ 6) â€“ 7(n â€“ 6) = 0

(n â€“ 7) (n â€“ 6) = 0

n = 7 or 6

âˆ´ The value of n can either be 6 or 7.

### Question 15. In how many ways can five children stand in a queue?

Solution:

Number of arrangements of â€˜nâ€™ things taken all at a time = P (n, n)

Hence,

After applying the formula,

P (n, r) =

The total number of ways in which five children can stand in a queue = the number of arrangements of 5 things taken all at a time = P (5, 5)

Thus,

P (5, 5) =

= 5! [Since, 0! = 1]

= 5 Ã— 4 Ã— 3 Ã— 2 Ã— 1

= 120

Therefore, Number of ways in which five children can stand in a queue are 120.

### 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?

Solution:

Given:

The total number of teachers in a school = 36

As we know that, number of arrangements of n things taken r at a time = P(n, r)

After applying the formula,

P (n, r) =

âˆ´ The total number of ways in which this can be done = the number of arrangements of 36 things taken 2 at a time = P(36, 2)

P (36, 2) =

= 36 Ã— 35

= 1260

Hence, Number of ways in which one principal and one vice-principal are to be appointed out of total 36 teachers in school are 1260.

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