# Two round table combinatorics problems.

I am struggling to answer the following two problems:

How many ways can 4 men and 8 women be seated at a round table if there are to be two women and between each man?

How many ways can 15 people be seated at a round table if B refuses to sit next to A?

• This isn't homework I am revising for an exam. I think the first one is 7!3! but i'm not sure. – CaptainMathSparrow Apr 21 '14 at 15:16
• No clue how to do the second one – CaptainMathSparrow Apr 21 '14 at 15:16

## 1 Answer

Let the men be A, B, C, D. In problems about circular permutations, two permutations that differ by a rotation are usually considered to be the same. Equivalently, we can assume that there is a special chair, and that A is sitting on it.

Then the set of positions occupied by the men is determined, and the remaining men can be arranged in these in $3!$ ways. For each such way, the women can be arranged in $8!$ ways.

For the second problem, you will find the following approach useful. Call a placement bad if A and B are next to each other. Count the number of bad placements, and subtract this from the total number of possible placements.

Another way of solving the problem is to assume as before that A sits in the special chair. How many ways are there to seat B? And now how many ways are there to seat the rest?

Remark: It is sometimes a good idea to solve a problem in two different ways. That can provide a (partial) check of correctness.

• What if the conditions of both problem 1 and 2 have to hold at the same time?What if A and B are male and female respectively? – rah4927 Apr 21 '14 at 15:29
• Ok, I think that if A goes in the fixed seat there are 14! ways to arrange the table around them. Then if B sits on the left or right of A this is a bad position. So there are $2\times13!$ bad positions. This means there are $14! - 2\times13!$ ways for them all to sit together and be happy. – CaptainMathSparrow Apr 21 '14 at 15:31
• @rah4927: That's clearly not the intended interpretation, since $4+8\ne 15$. A problem like the one you refer to can be solved, but it would have to be an explicit problem. – André Nicolas Apr 21 '14 at 15:32
• @CaptainMathSparrow: Yes, for the second problem an answer is $14!-2\cdot 13!$. – André Nicolas Apr 21 '14 at 15:35
• @rah4927: Yes, as you suggest take the "unrestricted" answer of the current first question, and remove the bad ones, of which there are $(3!)(2)(7!)$. – André Nicolas Apr 21 '14 at 15:50