Let us take an example:

Ex 1: Find the number of ways in which 5 boys and 5 girls be seated alternatively to form a line.

Solution: 5 boys can be arranged in a line in ⁵P₅ = 5! ways

Since the boys and girls are alternating, So corresponding each of 5! ways of arrangements of 5 boys, 

we obtain 5 places marked by the cross as given below

1. B₁ * B₂ * B₃ * B₄ * B₅ * 

2. * B₁ * B₂ * B₃ * B₄ * B₅ 

It is clear that 5 girls can be arranged in 5 places marked by the cross in (5! + 5!) ways 

Now, total number of ways of making a line

= 5! * (5! + 5!) = 5 ! + 2 * (5!)² 

Ex 2: If there are 6 girls and 5 boys who sit in a row, then find the probability that no two boys sit together.

Solution: 6 girls and 5 boys can sit in a row in 11! ways.

So, the exhaustive number of cases = 11!

Now, six girls can sit in a row in 6! and in each such arrangement, there are 7 places between them in which 5 boys can be

seated in ⁷P₅ ways. 

Therefore, total number of ways in which no two boys sit together  =  6! * ⁷P₅

Now, the required probability = (6! * ⁷P₅ )/11! = (6! * 7!)/(2! * 11!)

So, here we should keep in mind that how to do arrangements for given conditions as shown in the above examples.