Question

Find the number of ways in which 5 boys and 5 girls may be seated in a row so that no two girls and no two boys are together.

Answer 1

Hint: There are only two types. of such representation where no two girls and no two boys are together. Permutation of girls and boys in seating arrangements will lead to the result.

Complete step-by-step answer:
We have to make an arrangement where no two boys or girls will sit together. Notice that if the row starts with a boy then the next one must be a girl for that case. Again, the next one a boy and after that a girl and so on. So, boy and girl will be seated alternatively. Similarly, if the row starts with a girl then the next one must be a boy for that case. Again, the next one a girl and after that a boy and so on.
Now there are a total of 5 boys and 5 girls.
Let we have placed 5 boys in a particular order where row starts with a boy with a gap between them. Therefore, the girls will fill the gaps. Now as there are 5 girls, the number of ways by which the gaps of a particular arrangement can be filled by 5 girls is simply 5!.
Now for each such arrangement, as there are 5 boys, they too can rearrange between themselves by 5! Ways. So, the total number of ways is \[5!\times 5!={{\left( 5! \right)}^{2}}\].
Similarly if the row starts with a girl it will also result in \[5!\times 5!={{\left( 5! \right)}^{2}}\]
So the total number of ways so that no two girls and no two boys are together is \[2\cdot {{\left( 5! \right)}^{2}}\].

Note: As every boy or girl is a distinct element, we use this process. If all boys or girls are the same then there would only be two types of ways to represent.