Question

There are twelve seats in a row and six boys and six girls occupy the seats at random. Find the probability that the boys and girls sit alternatively.

• A. $\frac{1}{132}$
• B. $\frac{1}{462}$
• C. $\frac{1}{924}$
• D. $\frac{1}{231}$

Answer 1

The correct option is B $\frac{1}{462}$

The total number of arrangements of six boys and six girls = n(S)=12!
Boys can be seated alternately in 6! ways
Girls can be seated alternately in 6! ways
Also, the total arrangement can happen in 2 ways as the seats are even.
The number of favourable outcomes n(E) = 2 $\times$ 6! $\times$ 6!
$P\left(E\right)=\frac{2\times 6!\times 6!}{12!}=\frac{1}{462}$