# $5$ girls and $4$ boys sit in chairs. In how many ways can they sit such that any two boys won't be adjacent?

$$5$$ girls and $$4$$ boys sit in chairs. In how many ways can they sit such that any two boys won't be adjacent?

I'll be drawing a diagram for the purpose of understanding the question better.

$$G_1B_1G_2B_2G_3B_3G_4B_4G_5$$

Girls and boys can be permutated in $$5!$$ and $$4!$$ ways respectively. However, I believe that I've gone wrong somewhere. Could you assist?

Regards

• What you have counted is the number of ways the boys and girls can be seated so that no two girls sit in adjacent seats. – N. F. Taussig Oct 21 '18 at 19:00

First make the girls sit in any of the $$5!$$ possible ways. Once they are seated, there are $$6$$ spaces in between and on the sides of the girls: $$\times G_1 \times G_2 \times G_3\times G_4\times G_5 \times$$ Now each boy can be seated in any of these slots marked as $$\times$$. Thereby guaranteeing that no two boys are adjacent.

Now pick the $$4$$ slots for the boys, that can be done in $$\binom{6}{4}$$ ways. Once they have the slots picked, then they can be permuted among themselves in those slots in $$4!$$ ways. So in all $$5!\binom{6}{4}4! \quad \text{ways}$$

First the girls can sit in any order, giving us 5!=120 ways for the girls to sit.

Once the girls all sit there are 6 positions for the boys to choose from.

            x G x G x G x G x G x


Much like the other answer the "x" is a position that a boy can sit in.

(Since no boy can sit together) the first boy has 6 seats to choose, the second has 5 to choose from, the third has 4 seats to choose from, and the final boy has 3 seats to choose from. Giving us 6*5*4*3=360

Finally putting it all together we have

                        (5!)(6*5*4*3) = (120)(360)
= 43,200