Question

5 girls and 5 boys are to be seated around a circular table such that no 2 girls will sit together. The number of ways in which they can be seated around the table is

• A. $5!\times 5!$
• B. $5!\times 4!$
• C. $4!\times 4!$
• D. $5!\times {}^5{P}_4$

Answer 1

The correct option is B

$5!\times 4!$

First arrange 5 boys around the table. This can be done in 4! ways.

Now, there are 5 gaps in the circular arrangement as shown above.

Let us represent 5 girls as $G_1,G_2,G_3,G_4,G_5$ and 5 boys as $B_1,B_2,B_3,B_4,B_5$

Consider figures 1 and 2. As $G_1,G_2,G_3,G_4,G_5$ arranged in a circle, both figures represent same circular permutation

But in figures 3 and 4, though $G_1,G_2,G_3,G_4,G_5$ are arranged same as in figures 1 and 2, but when boys are arranged first, one can easily see that both permutations are different.

So, seating 5 girls in 5 gaps in circular table can be considered as normal permutation and number of ways they can be seated = 5!

So, total number of ways in which seating can be done = 4! $\times$ 5!