Question

In a class of $10$ students there are $3$ girls $A,B,C$. In how many different ways can they be arranged in a row such that no two of the three girls are consecutive.

Answer 1

There are $7$ boys and $3$ girls. Seven boys can be arranged in a row in ${}^7{P}_7=7!$ ways. Now, we have $8$ places in which we can arrange $3$ girls in ${}^8{P}_3$ ways.

Hence, by fundamental principle of counting, the number of arrangements $=7!{\times }^8{P}_3=7!\times \frac{8!}{5!}=7!\times 336$