Can anyone check if this is correct? There are 6 woman and 4 men. how many ways can they be arranged so that there is a woman at the start and at the end of the row and no two men are together?
what i did is $6\cdot5\cdot2\cdot4!\cdot4! = 34560$

I think the answer is $86400$: 
Arrange the women: $6!$ ways.  There are now $5$ possible spots for men: W-W-W-W-W-W (we must pick four of the five spots between two women to place the men).  There are $5$ ways to choose $4$ of these $5$ spots.  There are then $4!$ ways to arrange the men.
Total: $6!\cdot 5\cdot 4!=86400$.
This agrees with mvw's answer, but arrived at differently.
