Question

There are three girls in a class of $10$ students. The number of different ways in which they can be seated in a row such that no two of the three girls are together is?

• A. $7!\times {}^6\mathrm{P}_3$
• B. $7!\times {}^8\mathrm{P}_3$
• C. $7!\times 3!$
• D. $\frac{10!}{3!\times 7!}$

Answer 1

The correct option is B

$7!\times {}^8\mathrm{P}_3$

Evaluate the number of ways using permutations

Given, out of $10$ there are three girls and $7$ boys

$7$ boys can be arranged in $7!$Ways

Since, no two girls should sit together.

Thus, there are $8$ positions that can be occupied by $3$ girls in ${}^8\mathrm{P}_3$ ways.

Therefore, the required number of ways $=7!\times {}^8\mathrm{P}_3$

Hence, option $\left(\mathrm{B}\right)$, $7!\times {}^8\mathrm{P}_3$ is the correct answer.