In how many ways can 7 women, 10 men sit at table such that no woman sits besides another? We have $7$ women and $10$ men; they sit at a table. I've been trying to solve how many ways can they sit excluding the case of women sitting next to each other. 
My reasoning was the following: I will have to alternate woman and men to certain extent so that no woman will sit next to another woman. The possible alternations are the following cases:
$a)$ One of $7$ woman takes the first sit; one of $10$ men take the second; one of $6$ remaining women take the third; one of $9$ remaining men take the fourth, etc. So I have $10*7*9*6*8*5*7*4*6*3*5*2*4*1=10!7!$ possibilties.
$b)$One of $10$ men takes the first place; one of $7$ woman the second; etc. Again, $10!7!$ possibilities, but the order has changed.
In either case, I have $3$ men remaining. Let's call $A$ and $A'$ the selections of alternated men and woman that I described on points $a)$ and $b)$,  and $S$ the selection of the three remaining men. The first man of $S$ is any of the three remaining; the second man of $S$ is any of the two remaining; the last one is the one that's neither of the previous two. So I have for $S$ $3*2*1=3!$ cases. 
So my posibilities are that $A$ is the case or $A'$ is the case, and $P$. Which leaves me with $(10!7!+10!7!)*3!$ posibilities.
Is my reasoning okay? With this type of problems it troubles me how hard it is to actually test or check if one's answer is right. (Excuse any error on my english, it's not my native language.)
 A: Your result is correct. More generally, assume that there are $m$ men and $w$ women with $m\geq w\geq 1$. Number the $m+n$ chairs around the circular table from $1$ to $m+n$ and let the "oldest" woman sit down at the $1$-st chair. In our final arrangement, let $x_i$ be the number of men between the $i$-th woman and the next one. Then
$$x_1+x_2+\dots +x_w=m$$
where each $x_i$ is a positive integer. The number of solutions of this equation is $\binom{m-1}{w-1}$ (see stars-and-bars). Each solution tells us which chair goes to a man or to a women. Now we have $(w-1)!$ ways to place the remaining $w-1$ women and $m!$ ways to place the $m$ men. Therefore the number of arrangements is
$$\binom{m-1}{w-1}(w-1)!m!=\frac{(m-1)!m!}{(m-w)!}.$$
For $m=10$ and $w=7$ the above formula gives
$$\frac{9!\cdot 10!}{3!}=219469824000$$
which coincides with your result $(10!7!+10!7!)3!$.
A: Robert Z has provided a nice solution.  Here is another approach.
Suppose Edward is one of the men.  We will use him as a reference point.  Seat Edward first.  Once Edward has been seated, there are $9!$ ways to seat the remaining men as we proceed clockwise around the table.  This creates ten spaces in which we can place the women, one to the left of each of the ten men.  To ensure that the women are separated, we choose seven of these ten spaces in which to place a woman.  The women can be arranged in those spaces in $7!$ ways.  Hence, the number of admissible arrangements is 
$$9!\binom{10}{7}7!$$
For $m$ men and $w$ women, a similar argument yields
$$(m - 1)!\binom{m}{w}w!$$
which is equal to zero when $w > m$, as we would expect.
