Question

10 different toys are to be distributed among 10 children. Total number of ways of distributing these toys so that exactly 2 children do not get any toy, is equal to –

• A. (10!)² $\left( \frac{1}{3!2!7!} + \frac{1}{(2!)^{5}6!} \right)$
• B. (10!)²$\left( \frac{1}{3!2!7!} + \frac{1}{(2!)^{4}6!} \right)$
• C. (10!)²$\left( \frac{1}{3!7!} + \frac{1}{(2!)^{5}6!} \right)$
• D. (10!)²$\left( \frac{1}{3!7!} + \frac{1}{(2!)^{4}6!} \right)$

Answer 1

Answer: (B)

(10!)²$\left( \frac{1}{3!2!7!} + \frac{1}{(2!)^{4}6!} \right)$

Answer 2

It is possible in two mutually exclusive cases

Case 1. 2 children get none, one child gets three and all remaining children get one each.

Case 2. Children get none, 2 children get 2 each and all remaining children get one each.

In case one number of ways

= $\frac{\text{10}!}{3!2!7!}$· 10!

In second case number of ways

= $\frac{\text{10}!}{(2!)^{4}.6!}$· 10!

Thus total ways

= (10!)² $\left( \frac{1}{3!2!7!} + \frac{1}{(2!)^{4}6!} \right)$

Hence (2) is correct answer.