There are 5 boys and 4 girls in my class. All of them are distinguishable.

In how many ways can they be seated in a row of 9 chairs such that at least 2 boys are next to each other?

Guest Mar 5, 2015

#1**+5 **

First consider the number of ways they can be seated so that *no* boys are next to each other. The only way this can happen is if the boys are on seats 1, 3, 5, 7 and 9. There are 5 boys who could go on seat 1. For each of these there are 4 boys who could go on seat 3 ...etc., so there are 5! ways of arranging the 5 boys on these seats. For each of these there are 4! ways of arranging the girls on seats 2, 4, 6 and 8. So in total there are 5!*4! ways of arranging everyone on 9 chairs such that no boys sit together.

Now there are 9! ways of arranging everyone on the 9 chairs, regardless of who sits next to whom.

This means there must be 9! - 5!*4! ways of arranging everyone on the 9 chairs such that at least two boys are next to each other.

$${\mathtt{9}}{!}{\mathtt{\,-\,}}{\mathtt{5}}{!}{\mathtt{\,\times\,}}{\mathtt{4}}{!} = {\mathtt{360\,000}}$$

.

Alan
Mar 5, 2015

#1**+5 **

Best Answer

First consider the number of ways they can be seated so that *no* boys are next to each other. The only way this can happen is if the boys are on seats 1, 3, 5, 7 and 9. There are 5 boys who could go on seat 1. For each of these there are 4 boys who could go on seat 3 ...etc., so there are 5! ways of arranging the 5 boys on these seats. For each of these there are 4! ways of arranging the girls on seats 2, 4, 6 and 8. So in total there are 5!*4! ways of arranging everyone on 9 chairs such that no boys sit together.

Now there are 9! ways of arranging everyone on the 9 chairs, regardless of who sits next to whom.

This means there must be 9! - 5!*4! ways of arranging everyone on the 9 chairs such that at least two boys are next to each other.

$${\mathtt{9}}{!}{\mathtt{\,-\,}}{\mathtt{5}}{!}{\mathtt{\,\times\,}}{\mathtt{4}}{!} = {\mathtt{360\,000}}$$

.

Alan
Mar 5, 2015