Question

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

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

Answer 1

The correct option is D $\frac{10!\times 10!}{(2!{)}^2\times 6!}\left[\frac{25}{84}\right]$

There may be two cases.
Case I. Two children get none and one get three and others get one each.
Then, total number of ways $=\frac{10!}{2!\times 3!\times 7!}\times 10!$
Case II. Two get none and two get $2$ each and the other get one each.
Then, total number of ways $=\frac{10!}{(2!{)}^4\times 6!}\times 10!$
Hence, total number of ways
$=\frac{10!}{2!\times 3!\times 7!}+\frac{(10!{)}^2}{(2!{)}^4\times 6!}=\frac{(10!{)}^2\times 25}{(2!{)}^2\times 6!\times 84}$.