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

Answer 1

Answer: (D)

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
$=\frac{(10 !)^{2}}{2 ! \times 3 ! \times 7 !}+\frac{(10 !)^{2}}{(2 !)^{4} \times 6 !}$
$=\frac{(10 !)^{2} \times 25}{(2 !)^{2} \times 6 ! \times 84}$