Question

$10$ different toys are to be distributed among $10$ children. The total number of ways of distributing these toys, so that exactly two children do not get any toy, is equal to:

• A. $\frac{10!}{2!3!7!}$
• B. $\frac{10!}{{\left(2!\right)}^4\times 6!}$
• C. ${\left(10!\right)}^2\left[\frac{1}{{\left(2!\right)}^4\times 6!}+\frac{1}{2!\times 3!}\right]$
• D. $\frac{10!\times 10!}{{\left(2!\right)}^2\times 6!}\left[\frac{25}{84}\right]$

Answer 1

The correct option is D

$\frac{10!\times 10!}{{\left(2!\right)}^2\times 6!}\left[\frac{25}{84}\right]$

Explanation for the correct option:

Step 1. We need to distribute $10$ toys among ten children in which $2$ children do not get any toys.
So, in this condition two cases arises such that in the first case, two children do not get any toys, one child gets three toys and all the other children get one toy each.

In the second case, two children do not get any toy, two children get two toys each and the remaining children get one toy each.

Step 2. Evaluate both the cases separately:
Case I. $2$ children get nothing and one gets three and other gets one each.
Then, all number of ways $N_1=\frac{10!}{2!\times 3!\times 7!}\times 10!$ ……….(1)
Case II. $2$ children get nothing and two get $2$ each and the other get one each.
Then, all number of ways $N_2=\frac{10!}{{(2!)}^4\times 6!}\times 10!$ ……….(2)

Step 3: The total number of ways is : $N=N_1+N_2$

Putting equation (1) and equation (2) together, we have:
$\Rightarrow N=\frac{10!\times 10!}{2!3!7!}+\frac{10!\times 10!}{{(2!)}^4\times 6!}$

Taking $\frac{10!\times 10!}{{(2!)}^2\times 6!}$ as a common term, we get:

$\Rightarrow N=\frac{10!\times 10!}{{(2!)}^2\times 6!}\left[\frac{1}{21}+\frac{1}{4}\right]$

$\Rightarrow N=\frac{10!\times 10!}{{(2!)}^2\times 6!}\left[\frac{4+21}{84}\right]$

$\Rightarrow N=\frac{\mathbf{10}\mathbf{!}\mathbf{\times }\mathbf{10}\mathbf{!}}{{\mathbf{(}\mathbf{2}\mathbf{!}\mathbf{)}}^{\mathbf{2}}\mathbf{\times }\mathbf{6}\mathbf{!}}\left[\frac{25}{84}\right]$

Hence, Option ‘D’ is correct.