Discrete math: In how many ways can $5$ of the girls and $3$ of the boys be lined up in a row so that no two of the boys are side-by-side?

Question

There are $12$ girls and $9$ boys at a party. For a certain game, one must line up $5$ of the girls and $3$ of the boys in a row, left-to-right, in such a way that no two boys are side-by-side. In how many ways can this be done?

Can someone please give me some hints?

Answer 1

Strategy:

1. Choose which five of the girls will be in the line.
2. Arrange those five girls in a line. This creates six spaces in which a boy can be placed, four between successive girls and two at the ends of the row. $$\square g \square g \square g \square g \square g \square$$
3. To ensure that no two of the boys are adjacent, choose three of those six spaces for the boys.
4. Choose which three boys will be placed in the selected spaces.
5. Arrange the selected boys in those spaces.

Answer 2

I am thinking the lineup arrangement would be shown as following(G stands girl and B stands boy)

BGBGBGBGBGB

5 girls can stand in any 5 spots, so 5!=120

3 boys can stand in any 6 spots, so p(6,3) = 120

picking random 5 girls from total 12 girls , so 12 choose 5 = 792

picking random 3 boys from total 9 boys, so 9 choose 3 = 84

result is 84 * 792 *120 * 120 = 958003200