In how many ways can 5 boys and 5 girls sit in a circle so that no two boys sit together?

Question

In how many ways can 5 boys and 5 girls sit in a circle so that no two boys sit together? (A) 5! × 5! (B) 4! × 5! (C) (5! × 5!/2) (D) ((4!)(4!)/2!)

Tardigrade Admin · Accepted Answer

First we fix the alternate position of girls and they can be arranged in 4! ways and in the five places five boys can be arranged in 5P5 ways. ∴ Total number of ways = 4! × 5P5 = 4! ×5!

Q. In how many ways can $5$ boys and $5$ girls sit in a circle so that no two boys sit together?

$5! \times 5!$

$4! \times 5!$

$\frac{5 ! \times 5 !}{2}$

$\frac{( 4 ! ) ( 4 ! )}{2 !}$

First we fix the alternate position of girls and they can be arranged in $4!$ ways and in the five places five boys can be arranged in $^{5} P_{5}$ ways.
$∴$ Total number of ways $= 4! \times^{5} P_{5} = 4! \times 5!$

Q. In how many ways can 5 boys and 5 girls sit in a circle so that no two boys sit together?

5!×5!

4!×5!

25!×5!​

2!(4!)(4!)​