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$5$ 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

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  • $\begingroup$ 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. $\endgroup$ – N. F. Taussig Oct 21 '18 at 19:00
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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 $$5!\binom{6}{4}4! \quad \text{ways}$$

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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
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