If 5 boys and 4 girls sit in a row at random then the probability the boys and girls sit alternatively is

Given that 5 boys and 4  girls sit in a row at random.

Total number of ways in which 9 persons can be arranged in a row = 9 $$\times$$ 8 $$\times$$ 7 $$\times$$ 6 $$\times$$ 5 $$\times$$ 4 $$\times$$ 3 $$\times$$ 2 $$\times$$ 1

=  5 boys can be arranged themselves in  5 $$\times$$ 4 $$\times$$ 3 $$\times$$ 2 $$\times$$ 1 ways.

 B 1 B 2 B 3 B 4 B 

Now,

Then there are 4 positions left(1,2,3,4,).

The 4 girls can be seated in (1,2,3,4) =  4 $$\times$$ 3 $$\times$$ 2 $$\times$$ 1  ways.

The girls can be arranged themselves in   4 $$\times$$ 3 $$\times$$ 2 $$\times$$ 1 ways.

Therefore, the number of ways both girls and boys can sit alternatively = 5 $$\times$$ 4 $$\times$$ 3 $$\times$$ 2 $$\times$$ 1 $$\times$$   4 $$\times$$ 3 $$\times$$ 2 $$\times$$ 1

Hence,

Required probability = 

(5 $$\times$$ 4 $$\times$$ 3 $$\times$$ 2 $$\times$$ 1 $$\\times 4 $$\times$$ 3 $$\times$$ 2 $$\times$$ 1)\( 9 $$\times$$ 8 $$\times$$ 7 $$\times$$ 6 $$\times$$ 5 $$\times$$ 4 $$\times$$ 3 $$\times$$ 2 $$\times$$ 1)

= (120 $$\times$$ 24)\(9 $$\times$$ 8 $$\times$$ 7 $$\times$$ 6 $$\times$$ 5 $$\times$$ 4 $$\times$$ 3 $$\times$$ 2 $$\times$$ 1)

=  4/504

=  1/126.    Answer