A committee of 6 is chosen from 8 men and 5 women so as to contain at least 2 men and 3 women. How many different committees could be formed if two of the men refuse to serve together?
(A) 3510
(B) 2620
(C) 1404
(D) 700
(E) 635

This was a question that I got from another forum, however the way i solved it and the solutions that were posted in the forum did not match.

My solution:-
Since we must include at least 2 men and 3 women, we have just one variable, i.e, the 6th member. So, the sixth one can be both a man or a woman. hence the number of ways are-

1)6th man-(3 men and 3 women)
8C3 * 5C3 = 560

2) 6Th woman-(2 men and 4 women)
8C2 * 5C4 = 140
Combining these two, the answer comes to be 700.

However,the answer that I got in the forum is the following-

Committee can have either: 2 men and 4 women OR 3 men and 3 women (to meet the condition of at least 2 men and 3 women).

Ways to chose 6 members committee without restriction (two men refuse to server together):

Ways to chose 6 members committee with two particular men serve together:

700-65 = 635

Answer: E.

I would love some clarification as I did not understand why my method yields the wrong answer. Also, I am not quite fond of the way the problem was solved in the forum.

Any thoughts would be helpful.