Question

In how many ways can $10$ balls be distributed to two boys, one receiving two and the other eight balls?

• A. $45$
• B. $75$
• C. $90$
• D. $Noneofthese$

Answer 1

The correct option is C $90$

$10$ balls between two boys

Let two boys A & B

A receive $2$

B receive $8$

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

$=\frac{10\times 9\times 8!}{2\times 1\times 8!}$

$=45$

No. of distribution if A get $8$ & B get $2$

$=\frac{10!}{8!2!}$

$=\frac{10\times 9\times 8!}{8!\times 2\times 1}$

$=45$

Total number of ways$=45+45$

$=90$.