In how many ways can 4 boys and 4 girls sit around a circle table if each boy sits between two girls? (Rotations of the same arrangement are still considered the same. Each boy and girl is unique, not interchangeable.)

I said that it has to be of the form BGBGBGBG, so there would be (4!)(4!) ways to order the boys and the girls. However, you have to divide by 8 for overcounting, because there are 8 ways to rotate any order. This would be 576/8=72 ways. However, this is wrong. Could I get some help? Thank you.

  • 2
    $\begingroup$ The statement is ambiguous. Two ways that differ on a rotation are considered different or not? $\endgroup$ – leonbloy Jul 21 '14 at 14:21

I think of it this way. One of the girls is the Queen, and one of the chairs is a throne. The Queen of course sits on the throne. The rest of the girls can be permuted in $3!$ ways, and the boys in $4!$.

Remark: The problem with deciding that we will divide by something is that then we are doing manipulation and not visualization. I prefer to keep things concrete, to retain control over the counting process.


You have to decide what you want to define as a distinct way here.

For example if every one moves one seat clockwise is that a distinctly different way?

If we assume not and choose our set based on one particular girl, Alice sits going clockwise then the only way it can be done is:


Where A is alice, B is any boy, and G any girl.

We can put the three remaining girls in any order so that's $3!=6$

And the four boys in any order $4!=24$

Making the total number of ways $6 \cdot 24 = 144$

  • $\begingroup$ Thank you, this ended up being correct. $\endgroup$ – Bob Jul 21 '14 at 14:44

Since rotations are considered the same, we only have to find the permutations relative to one person. Counting henceforth should be easy.


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.