Number of ways $10$ men and $4$ women can sit at a round table so there is no block of $3$ consecutive women 
In how many ways can $10$ men and $4$ women sit at a round table so that there is no block of $3$ consecutive women?

Since there is no restriction on men, I thought they should sit first. Since the table is round, I fixed one seat for $1$ in 10 men and placed the rest of them $9!$ ways and then we have $10$ gaps in which we can place $4$ women, at most $2$ in each. I think these are the possibilities:

*

*One woman in each gap: we can choose those gaps in $\binom{10}4$ ways and then place women in $4!$ different ways.

*$2$ women in  $1$ gap and the remaining $2$ in each of their own. We need $3$ gaps altogether, which can be chosen in $\binom{10}3$ ways. Since women are different entities, I believe we should be able to place them in, again $4!$ ways.

*$2$ women in $2$ gaps: $2$ in each. We can choose $2$ gaps in $\binom{10}2$ ways and again, if I'm not wrong, place them in $4!$ ways.

Therefore, my answer to the number of possible ways is: $$9!\cdot 4!\left(\binom{10}4+\binom{10}3+\binom{10}2\right).$$
Can somebody check my answer as I'm not quite sure im my deduction?
 A: Your answer is missing factor of $3$ in the second term.
$ \displaystyle 9!\cdot 4!\left(\binom{10}4+ \color {blue} {3} \cdot \binom{10}3+\binom{10}2\right)$
Why $3$? That's because when you make groups of $2, 1$ and $1$ women and choose $3$ spaces between men to seat them which is ${10 \choose 3}$, you also have $3$ ways to choose which of those $3$ spaces will have two women.
Alternatively, subtract arrangements which have $3$ or $4$ consecutive women from total arrangements of $13!$. So the answer can also be written as,
$ \displaystyle 13! - (10! \cdot 9 + 10!) \cdot 4!$
Explanation:
First term being subtracted - three consecutive women. Let's first seat three women together and then men can be arranged in $10!$ ways. The lone woman can sit in $9$ spaces between men. Finally women can be arranged within in $4!$ ways.
Second term being subtracted - all four women sit together. Let's first seat all women together and then men can be arranged in $10!$ ways. Finally women can be arranged within in $4!$ ways.
A: Let's say the seats are numbered $1$ to $14$. Fix an arbitrary man at seat $1$. Now we have $9$ men left and $4$ women left to seat.
Now consider the remaining $13$ seats. Since the man in seat $1$ is blocking any possibility of a consecutive block of women wrapping around, we just need to find the number of ways to put the remaining $13$ people in a line so that we don't have a consecutive block of $3$ women.
We will use complimentary counting. There are of course $13!$ ways to seat the $13$ people with no attention to restrictions. The only way to have a consecutive block of $3$ women is if there is a block of exactly $3$ women or a block of exactly $4$ women.
For the first case, there are $\binom{4}{3}\cdot 3!$ ways to choose the $3$ women in the block and then order them. There are $11!$ ways to arrange the $9+2=11$ entities. However, $10\cdot 2\cdot 9!$ of these have a block of $4$ women. So there are a total of $216\cdot 10!$ ways to arrange this case.
For the second case, there are $4!$ ways to arrange the $4$ women in the block. There are $10!$ ways to arrange the $10$ entities. Hence the total is $24\cdot 10!$
So the total is $13!-240\cdot 10!=\boxed{1476\cdot 10!}$.
