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

Part $1$: Finding number of ways in which $5$ girls can sit in $5$ places.

___G___G___G___G____G____

Number of ways girls can sit = ${}^5{P}_5$

= $\frac{5!}{(5-5)!}$

= $\frac{5!}{0!}$

$=5!$

$=5\times 4\times 3\times 2\times 1$

$=120$

Step $2$: Finding numbers of ways in which $3$ boys can sit in $6$ places.

Required number of ways ${}^6{P}_3$

= $\frac{6!}{(6-3)!}$

= $\frac{6!}{3!}$

= $\frac{(6\times 5\times 4\times 3\times 2\times 1)}{3\times 2\times 1}$

$=120$

Step $3$ : Finding total number of ways.

$\therefore$ Total number of required ways

$=120\times 120$

$=14400$