Question

In how many ways can $10$ different balls be divided between two boys, one receiving two and the other eight balls

• A. $45$
• B. $75$
• C. $90$
• D. None of these

Answer 1

Answer: (C)

$90$

When $A$ gets $2$ and $B$ gets $8$ balls,

No. of distribution$=\frac{10!}{2!\times 8!}=45$

When $B$ gets $2$ and $A$ gets $8$ balls,

No. of distribution$=\frac{10!}{2!\times 8!}=45$

$\therefore \text{Total no. of ways}=45+45=90$.