Question

4 Boys and 3 girls are standing in a straight line. Find the probability that girls and boys stand alternatively. Please answer as soon as possible.

Answer 1

$P\left(BGBGBGB\right)=\frac{4}{7}\times \frac{3}{6}\times \frac{3}{5}\times \frac{2}{4}\times \frac{2}{3}\times \frac{1}{2}\times \frac{1}{1}=\frac{4!3!}{7!}=\frac{(3\times 2)}{(7\times 6\times 5)}=\frac{1}{35}$
That's because at first there are 4 boys out of 7 children.
Then there are 6 children left and 3 are girls so the next probability is $\frac{3}{6}$, and so on.