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Can someone please explain how to solve circular permutation sums. I just cannot seem to understand them.

eg.

$\text{(4)}$ The number of ways in which $6$ men and $5$ women can dine at a round table if no two women are to sit together is given by

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$\text{(a)}$ $6!*5!$

$\text{(b)}$ $50$

$\text{(c)}$ $5!*4!$

$\text{(d)}$ $7!*5!$

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First seat the $6$ men. If they were seated in a row, there would be $6!$ possible permutations of them. However, they’re actually seated around a circular table, which changes things. The six permutations $ABCDEF,BCDEFA,CDEFAB,DEFABC,EFABCD$, and $FABCDE$ are the same when the men are sitting in a circle: the only differences is where in the circle you started the list. Thus, each arrangement of the men around the table corresponds to $6$ permutations in a straight line, and there are therefore only $\frac{6!}6=5!$ possible arrangements of the men around the table. (This is a useful general fact: there are $(n-1)!$ circular permutations of $n$ things.)

Now we seat the $5$ women. There are $6$ positions available between adjacent men, so we have $6$ possible ways to seat the first woman. If we put another woman between the same two men, she’d be sitting next to the first woman; we don’t want that, so we have to seat her in one of the other $5$ gaps between adjacent men. That leaves $4$ open gaps for the third women, $3$ for the fourth woman, and $2$ for the last woman, so altogether we can seat the women in $6\cdot5\cdot4\cdot3\cdot2=6!$ different ways.

Combining the two results, we see that the $11$ people can be seated in $5!\cdot6!$ different orders around the table if no two women sit next to each other.

Firstly let the 6 men sit at the table, this can happen in $(6 – 1)!$ ways. Then there are 6 spots between the men for the women to sit. This can be done in $6.5.4.3.2 = 6!$ ways, hence you have the answer $6!*5!$.

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