Stuck with combinations!

[Uva@90]

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

OA E

I proceeded the above problem as below,
Committee of 6 members can be chosen by
(8C2*5C4) + (8C3*5C3) = 700
I am struck here,
I unable to handle this condition
two of the men refuse to serve together
But I know something to be subtracted from 700.
So I went for E, only option less than 700.
But I would like to learn how to calculate the above red colored one.

Thanks in advance,

Regards,
Uva.

[GMATGuruNY]

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

OA E

I proceeded the above problem as below,
Committee of 6 members can be chosen by
(8C2*5C4) + (8C3*5C3) = 700
I am struck here,
I unable to handle this condition
two of the men refuse to serve together
But I know something to be subtracted from 700.
So I went for E, only option less than 700.
But I would like to learn how to calculate the above red colored one.

Thanks in advance,

Regards,
Uva.

Let A and B = the two men who cannot serve together.
From the 700 committees you have counted, we must subtract the BAD committees: those that include both A and B.

Bad case 1: 3 women and 1 man are chosen to serve with A and B
Number of ways to choose 3 women from 5 options = 5C3 = (5*4*3)/(3*2*1) = 10.
Number of ways to choose 1 man from the 6 remaining men = 6C1 = 6.
To combine these options, we multiply:
10*6 = 60.

Bad case 2: 4 women are chosen to serve with A and B
Number of ways to choose 4 women from 5 options = (5*4*3*2)/(4*3*2*1) = 5.

Thus:
Good committees = 700-60-5 = 635.

[Uva@90]

Let A and B = the two men who cannot serve together.
From the 700 committees you have counted, we must subtract the BAD committees: those that include both A and B.

Bad case 1: 3 women and 1 man are chosen to serve with A and B
Number of ways to choose 3 women from 5 options = 5C3 = (5*4*3)/(3*2*1) = 10.
Number of ways to choose 1 man from the 6 remaining men = 6C1 = 6.
To combine these options, we multiply:
10*6 = 60.

Bad case 2: 4 women are chosen to serve with A and B
Number of ways to choose 4 women from 5 options = (5*4*3*2)/(4*3*2*1) = 5.

Thus:
Good committees = 700-60-5 = 635.

Got it.

Thanks Mitch you made it clear.