Question

In how many ways can 5 girls and 3 boys be seated in a row so that no two boys are together ?

Answer 1

First, we seat the $5$ girls in $5!$ ways. For each such a arrangement, the three boys can be seated only at the $6$ cross marks places X G X G X G X G X G X Three boys can be selected in ${}^6{P}_3$ ways. Hence total number of ways

$=5!{\times }^6{P}_3=(5\times 4\times 3\times 2\times 1)\times (6\times 5\times 4)$

$=14400$