Question

In how many ways $10$ boys and $5$ girls can sit around a circular table, so that no two girls sit together?

• A. $10!{}^{10}{P}_5$
• B. $9!{}^9{P}_5$
• C. $10!{}^9{P}_5$
• D. $9!{}^{10}{P}_5$

Answer 1

Answer: (D)

$9!{}^{10}{P}_5$

First average the $10$ seats and let boys be placed

They may be arranged in $(n-1)!$ ways $=?$ ways

$\Rightarrow$now let there are $10$ places where girls can be placed but there are only $5$ girls so arrangements = ${}^{10}{P}_5$

$\Rightarrow$overall arrangements = $9!{\times }^{10}{P}_5$

Hence the correct answer is $9!{\times }^{10}{P}_5$