$5$ 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.


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


  • $\begingroup$ 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

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.