Find the number of ways in which 6 persons out of 5 men and 5 women can be seated at a round table such that 2 men are never together. Find the number of ways in which 6 persons out of 5 men and 5 women can be seated at a round table such that 2 men are never together.

My attempt:
6 people may be 3 men and 3 women, 2 men and 4 women or 1 man and 5 women.
Then by the gap method, 3 men and 3 women can be seated in $\binom{5}{3}2!3!=120$ ways.
2 men and 4 women can be seated in $\binom{5}{4}3!\frac{4!}{2!}=360$ ways
1 men and 5 women can be seated in $\binom{5}{5}4!\frac{5!}{4!}=120$ ways
So the total number of ways is 600, but the answer given in my book is 5400. Where did I go wrong?

Label the chairs from $1$ to $6$ in increasing order, label the men and the women from $1$ to $5$.
Consider three cases:
1) There is 1 man: choose a man in $5$ ways, place the man at chair $1$, and permute the $5$ women in the remaining chairs in $5!$ ways. Therefore the number of ways is $5\cdot 5!=600$.
2) There are 2 men: choose two men in $\binom{5}{2}$ ways, place the man with the smallest label at chair 1 and the other at chair $3$, $4$ or $5$ ($3$ ways), choose four women in $\binom{5}{4}$ ways and arrange them in the remaining chairs in $4!$ ways. Therefore the number of ways is $\binom{5}{2}\cdot 3\cdot \binom{5}{4}\cdot 4!=3600$.
3) There are 3 men: choose three men in $\binom{5}{3}$ ways, place the man with the smallest label at chair 1, and the other at chair $3$ and $5$ ($2$ ways), choose three women in $\binom{5}{3}$ ways and arrange them in the remaining chairs in $3!$ ways.
Therefore the number of ways is $\binom{5}{3}\cdot2 \cdot \binom{5}{3}\cdot 3!=1200$.
Hence the total number of ways is $600+3600+1200=5400$.
