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 $6\cdot5\cdot2\cdot4!\cdot4! = 34560$
 A: 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.
A: To preserve my mental sanity I wrote a Ruby program to brute force this.
It reports
n = 86400

combinations. Of the $10!$ permutations
ne = 2419200

were invalid due to guys sticking to a table end and
n2 = 1123200

due to guys sitting next to each other and creating trouble.
This result is half of my original attempt to join $5$ women with $4$ guys ($5! \cdot 4!)$ and then letting the $6$-th woman choose one of ten positions (inflating it by $10$).
Looking at a simpler example ($3$ women and $1$ man), I noted that half the permutations plus choice result in a repetition, like $1m2, 3 \to 31m2$ ($3$ choosing first position) and $3m2, 1 \to 31m2$ ($1$ choosing second position). 
It were just the permutations which had a different sign, which hints to a difference of one transposition. So there might be some algebraic argument possible. 
Note: Beware of programming errors :-)
A: There are $\binom{6}{2}$ to choose two women who have to be kept at the end of the row and $2!$ ways of arranging them. The rest of the 8 places can be filled by women in 8 places as $4!$ and similarly, the men can be arranged in $4!$ ways.
So, the final answer on simplification turns out to be:  $6.5.4!.4! = 17280$ ways.
