Circular Arrangement with numbers The number of ways of arranging 2 women and 7 men around a circular table containing nine numbered chairs such that the women are not together.
I am getting answer as 7!*7c2(arranging 2 women in the 7 blank spaces formed by 7 men)*2(arrangement of 2 women among themselves)= 42*7!
But the answer given is 54*7!. Can someone please help me in where I am getting wrong?
 A: The women are not interchangeable.  You seat the first one in one of $9$ seats.  The second has $6$ choices as three seats are not available.  $9 \cdot 6=54$.  Then seat the men in $7!$ ways.
A: Two answers matching with the correct one have already been given.
I am only explaining why your method needs slight tweaking to give the correct answer.
The tweaking needed is simply to first treat the chairs as unnumbered, using your method,
and then number the chairs.
The men can be seated in $6!$ ways, and the women arranged in the gaps in $\binom72 \cdot2$ ways
Now numbering the chairs, ans $= 6!\cdot\binom72\cdot 2\cdot 9$
A: Ross Millikan has provided you with a simple solution in which the women are placed first.
You attempted to solve the problem by placing the men first, then inserting the women.  Below you will find a modification of that approach.
First, we solve the corresponding problem for a linear arrangement.   
There are $7!$ ways of arranging the men in a row.  This creates eight spaces, six between successive men and two at the ends of the row.  Since the women cannot be adjacent, we choose two of these eight spaces for the women, then arrange them in these spaces in one of two ways.  Therefore, there are 
$$7! \cdot \binom{8}{2} \cdot 2! = 7! \cdot \frac{8!}{6!2!} \cdot 2! = 7! \cdot \frac{8 \cdot 7 \cdot 6!}{6!2!} \cdot 2! = 7! \cdot 8 \cdot 7 = 56 \cdot 7!$$
linear arrangements in which the two women are not adjacent.  
However, if we form a circle, those linear arrangements in which the two women are at both ends of the row result in them sitting in adjacent seats.  There are $2!$ ways to sit the women at the ends of the row and $7!$ ways to arrange the men between them.  Hence, there are 
$$2! \cdot 7! = 2 \cdot 7!$$
linear arrangements in which the two women occupy both ends of the row.
Since each permissible circular arrangement corresponds to a permissible linear arrangement in which the women do not occupy the seats at both ends of the row, the number of permissible circular arrangements is 
$$56 \cdot 7! - 2 \cdot 7! = 54 \cdot 7!$$
A: Well first of all, you need to place the $2$ women: you have $9$ chairs to place the first on some chair and to place the other, you will have to choose from $6$ chairs since she can't sit on any of the two chairs next to the first and obviously can't sit on the chair of the first. Then to place the women you have $9*6$.
Once yo did that you can place the 7 men on the 7 left chairs and there are $7!$ ways to do it.
Therefore your final answer is $9*6*7!=54*7!$.
