Question

10 different toys are to be distributed among 10 children such that exactly two children do not get any toy. Total number of ways is

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

Hint:

“Exactly none” → choose people first, then apply onto distribution.

Answer 1

The Correct Option is D