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• RD Sharma Class 11 Solutions for Maths

# 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 – 3)(n – 4) = 2

n2 – 3n – 4n + 12 = 2

n2 – 7n + 12 – 2 = 0

n2 – 7n + 10 = 0

n2 – 5n – 2n + 10 = 0

n (n – 5) – 2(n – 5) = 0

(n – 5) (n – 2) = 0

n = 5 or 2

For, P (n, r): n ≥ r

∴ n = 5 [for, P (n, 5)]

### 1. P (1, 1) + 2. P (2, 2) + 3 . P (3, 3) + … + n . P(n, n) = P(n + 1, n + 1) – 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) + … + n . P(n, n)

= 1.1! + 2.2! + 3.3! +………+ n.n! [Since, P(n, n) = n!] = (2! – 1!) + (3! – 2!) + (4! – 3!) + ……… + (n! – (n – 1)!) + ((n+1)! – n!)

= 2! – 1! + 3! – 2! + 4! – 3! + ……… + n! – (n – 1)! + (n+1)! – n!

= (n + 1)! – 1!

= (n + 1)! – 1 [Since, P (n, n) = n!]

= P(n+1, n+1) – 1

= RHS

Hence Proved.

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

Solution:

Given:

P(15, r – 1) : P(16, r – 2) = 3 : 4 After applying the formula,

P (n, r) = P (15, r – 1) =  P (16, r – 2) =  So, from the question, After substituting the values in above expression we will get, (18 – r) (17 – r) = 12

306 – 18r – 17r + r2 = 12

306 – 12 – 35r + r2 = 0

r2 – 35r + 294 = 0

r2 – 21r – 14r + 294 = 0

r(r – 21) – 14(r – 21) = 0

(r – 14) (r – 21) = 0

r = 14 or 21

For, P(n, r): r ≤ n

∴ r = 14 [for, P(15, r – 1)]

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

Solution:

Given:

n+5Pn+1 = 11(n – 1)/2 n+3Pn

P (n +5, n + 1) = 11(n – 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 – 1)

n2 + 4n + 5n + 20 = 22n – 22

n2 + 9n + 20 – 22n + 22 = 0

n2 – 13n + 42 = 0

n2 – 6n – 7n + 42 = 0

n(n – 6) – 7(n – 6) = 0

(n – 7) (n – 6) = 0

n = 7 or 6

∴ 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 ‘n’ 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 × 4 × 3 × 2 × 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) = ∴ 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 × 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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