Question

The number of ways in which $10$ boys can be divided into $2$ groups of $5,$ such that two tallest boys are in two different groups, is equal to

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

Answer 1

Answer: (A)

Excluding the $2$ tallest boys, we can divide the $8$ boys into $2$ groups of $4$ each by
$\frac{8!}{4!4!2!}$ ways.
The tallest boys can be assigned to the groups in $2$ ways.
The desired number of ways is
$\frac{8!}{4!4!2!}\times 2!=\frac{8\times 7\times 6\times 5}{4\times 3\times 2}=70$