Combinatorics question: Boys and Girls around table

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.

• The statement is ambiguous. Two ways that differ on a rotation are considered different or not? – 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:

ABGBGBGB

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 $6 \cdot 24 = 144$

• Thank you, this ended up being correct. – 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.