In how many ways can $4$ men and $3$ women be arranged at a round tale if:

i) the women always sit together?
ii) the women never sit together?

I attempted both the questions but the answers I got don't match the ones given in the source. This is what I did:

For (i) I considered the women to be one unit, who can be arranged in $3!$ ways, and so the total no. of units would be $5$, which we could arrange in $4!$ ways. So the total is $4! \cdot 3!$

For (ii), it would translate to women sitting in the $3$ spaces between men, so I considered the table as two separate ones where the men sit in $3!$ ways, and the women sit in $2!$ ways, so the total would be $3! \cdot 2!$

I'd like to know where my reasoning fell apart.

This is being repurposed in an effort to cut down on duplicates, see here:

Coping with abstract duplicate questions

and here: List of abstract duplicates

See the related links on this Question, where variants involving seating in a row may be found.

  • $\begingroup$ HINT: The key is that the seven people are seated at a round table, not in a row. $\endgroup$ – Ant May 30 '15 at 21:42
  • 1
    $\begingroup$ Does the absolute positioning count, or since it's a "round table" does it not matter where someone is? That is, is one of the seats at the "round" table nearest the bathroom? $\endgroup$ – user1717828 May 30 '15 at 21:42
  • $\begingroup$ No, absolute positioning does not count. $\endgroup$ – GinaJay May 30 '15 at 21:44
  • $\begingroup$ And about sitting in a round table - yes, it should be (n-1)! ways, but I did consider that while attempting it. $\endgroup$ – GinaJay May 30 '15 at 21:45

Your answer to the first question is correct. Here is another way to see it: In a circle, if the three women sit together, so do the four men. Thus, there are two blocks, which can be arranged in $(2 - 1)! = 1! = 1$ way around the table. Within the block of women, the women can be arranged in $3!$ ways. Within the block of men, the men can be arranged in $4!$ ways. Hence, there are $3!4!$ arrangements of three women and four men around a circular table in which all the women sit together.

For the second problem, seat the men first. Since the four men are sitting in a circle, there are $(4 - 1)! = 3!$ distintinguishable arrangements of the men. This leaves four spaces in which to place the three women, one to the right of each man. We can select three of these four spaces in $C(4, 3)$ ways, and arrange the women within the selected spaces in $3!$ ways. Hence, the number of possible seating arrangements of four men and three women around a circular table in which no two of the three women sit together is $3! \cdot C(4, 3) \cdot 3! = 6 \cdot 4 \cdot 6 = 144$.

Your error in the second problem was to apply the rule $(n - 1)!$ twice. Once the men are seated, rotating the women produces a new arrangement. To see this, suppose Andrew, Bruce, Charles, and David are seated around the table in counterclockwise order. Suppose Elizabeth sits to the right of Andrew, Fiona sits to the right of Bruce, and Gretchen sits to the right of Charles. If each women moves to her right (past a man), Elizabeth is now to the right of Bruce, Fiona is to the right of Charles, and Gretchen is now to the right of David. Clearly, this is a different arrangement.


Part I is correct and there is another way you can solve part II. See if this is something you can understand easily.

Choose any two men in ${4\choose2}$ and fix them as one M1. They could be permuted in 2 ways. Now you have three men and three women can be placed M1-W1-M2-W2-M3-W3.This is the only way all women don't sit together. Now fix M1 and this particular arrangement has remainder of the men permuted by 2! and three women permuted by 3!. Thus the total number of ways that women don't sit next to each other is ${4\choose2}\times 2\times2!\times3! =144$


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