Question

Six boys and six girls sit in a row randomly.The probability that the six girls sit together or the boys and girls sit alternately, is

• A. $\frac{3}{308}$
• B. $\frac{5!7!}{2}$
• C. $\frac{2}{205}$
• D. $\frac{4}{407}$

Answer 1

The correct option is B $\frac{5!7!}{2}$

Given 6 boys and 6 girls

There are 12 persons.

All of them can be seated in (12)! ways.

n(s)=(12)! ways where S is a sample space.

E= Event of the girls and boys to be seated alternately.

6 girls can be seated in 6! ways.

$G_1,G_2,G_3,G_4,G_5,G_6$

There are 7 alternate places for 6 boys

6 boys can be seated in 7 places in ${}^7{P}_6$ ways.

$\therefore n\left(E\right)=6!{.}^7{P}_6=\frac{6!7!}{(7-6)!}=6!.7!$

$\therefore$ The required probability = $\frac{n\left(E\right)}{n\left(S\right)}$

$=\frac{6!7!}{12}$