Question

The number of ways $10$ boys can be divided into $2$ groups of $5$, such that two particular boys are in the different groups, is

• A. $70$
• B. $35$
• C. $252$
• D. $126$

Answer 1

The correct option is A $70$

After excluding $2$ particular boys we can divide $8$ boys in to two equal groups in $\frac{8!}{2!\cdot 4!\cdot 4!}$ ways
and $2$ boys can be be distributed in $2!$ ways
Hence total possible ways $=\frac{8!}{2!\cdot 4!\cdot 4!}\times 2!=70$