Question

In how many ways 10 boys and 5 girls sit around a circular table so that two girls don’t sit together?
$
  (a){\text{ }}\dfrac{{9! \times 10!}}{{5!}} \\
  (b){\text{ 9!}} \times {\text{10!}} \\
  {\text{(c) 14!}} \\
  (d){\text{ 15!}} \\
 $

Answer 1

Hint: In this question we have to make 10 boys and 5 girls sit around a circular table, so this question is based on circular table permutation and combination. Use the concept of circular permutation and combination that is if n people are to sit around a circular table then the total number of ways can be $(n - 1)!$. Make sure that the condition gets fulfilled.

Complete step-by-step answer:
As we know that five girls cannot sit between 10 boys.
So we have to first arrange 10 boys on a round table.
As we know if there are x candidates arrange on a round table the number of ways is $\left( {x - 1} \right)!$
Therefore the number of ways to arrange 10 boys on a round table is $\left( {10 - 1} \right)!$ = (9!).
Now we have to arrange five girls between them so that no two girls are sitting together.
So as we know between 10 boys there are ten vacant spaces therefore the number of ways to arrange five girls between 10 boys so that no two girls are sitting together is permute by ${}^{10}{P_5}$
Therefore the total number of ways to arrange 10 boys and five girls so that no two girls are sitting together is
$ \Rightarrow \left( {9!} \right)\left( {{}^{10}{P_5}} \right)$
$ \Rightarrow \dfrac{{9! \times 10!}}{{5!}},\left[ {\because {}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}} \right]$

So, this is the required answer.
Hence option (A) is correct.

Note: Whenever we face such types of questions the key concept is never forget about the permutation that is the numbers of people we have selected can also rearrange amongst themselves. Keeping all these things in mind, one can easily get to the answer of such problems.