Question

The number of ways in which $5$ boys and $5$ girls can be arranged in a row so that no two girls and no two boys are together is

• A. $2{(5!)}^2$
• B. ${(5!)}^2$
• C. $5!6!$
• D. $10!$

Answer 1

Answer: (A)

$2{(5!)}^2$

There are two ways of doing this

Case $I:1^{st}$ person is a boy.

$\underset{–}{B_1{G}_1{B}_2{G}_2{B}_3{G}_3{B}_4{G}_4{B}_5{G}_5}$

The no. of ways in which the boys can be rearranged among themselves is $5!.$ The same is for the girls.
So, the total number of ways
$=5!\times 5!$

Case $II:1^{st}$ person in the row is a girl.

$\underset{–}{G_1{B}_1{G}_2{B}_2{G}_3{B}_3{G}_4{B}_4{G}_5{B}_5}$

Same as above, $5!\times 5!$
So, the total no. of ways is $=2\times (5!)\times (5!)$

So, total number of ways $=2\times {(5!)}^2$

Hence, option A is correct.