Question

$10$ different toys are to be distributed among $10$ children. If the total number of ways of distributing all these toys so that exactly two children do not get any toy, is $k(10!),$ then the value of $k$ is

• A. $45$
• B. $360$
• C. $375$
• D. None of these

Answer 1

The correct option is C $375$

$\text{Case I :}0011111113$
Number of ways $=\frac{10!}{7!3!2!}$

$\text{Case II :}0011111122$
Number of ways $=\frac{10!}{6!2!2!2!2!}$

Hence, required number of ways
$=10!(\frac{10!}{7!3!2!}+\frac{10!}{6!2!2!2!2!})$
$=10!(\frac{7!\times 8\times 9\times 10}{7!\times 3\times 2\times 1\times 2\times 1}+\frac{10\times 9\times 8\times 7\times 6!}{6!\times 2\times 2\times 2\times 2})$
$=10!(60+315)=375\cdot 10!$
$\Rightarrow k=375$