Q65 / 200 · Manipal (MET) 2014 · Mathematics

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
    $(10!)^2\left[\frac{1}{(2!)^4 \times 6!}+\frac{1}{2!\times 3!}\right]$
  • Correct answer
    D
    $\frac{10!\times 10!}{(2!)^2 \times 6!}\left[\frac{25}{84}\right]$

Solution

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 others get one each. Then, total number of ways $=\frac{10!}{(2!)^4 \times 6!} \times 10!$ Hence, total number of ways $=\frac{(10!)^2}{2!\times 3!\times 7!}+\frac{(10!)^2}{(2!)^4 \times 6!}=\frac{(10!)^2 \times 25}{(2!)^2 \times 6!\times 84}$

Hence, the correct answer is option: 'D'.

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