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

Concept: Distribution with restriction (exactly two get nothing)

Step 1: Choose 2 children who get nothing: \[ \binom{10}{2} \]

Step 2: Remaining 8 children must get at least one toy.

Step 3: Distribute 10 distinct toys among 8 children: onto function Stirling concept.

Step 4: Using formula: \[ \text{Ways} = \binom{10}{2} \times \textcolor{red}{(\text{onto distributions})} \]

Step 5: Simplifies to given option (D).
Conclusion:
Answer = option (D)