men refuse to serve together

IMO E

I will divide calculations into 2 parts and then I will add numbers

commitee of 6 contains at least 2 men and 3 women means that it contains 3 men and 3 women or 2 men and 4 women

let us assign symbols M1 and M2 respectively to two men who refuse to serve together

The number of combinations of r objects chosen from a set of n is
nCr = n! / ((n-r)! r!)

1) How many different committees of 6 can be chosen from 8 men and 5 women so as to contain 3 men and 3 women if two of the men refuse to serve together? [spoiler](6C2 + 6C2 + 6C3)5C3 = 500[/spoiler]

if M1 was chosen we have 2 other places and 6 free men; it is 6C2 possible choices
if M2 was chosen we have 2 other places and 6 free men; it is 6C2 possible choices
otherwise we have 3 places and 6 free men; it is 6C3 possible choices
the number of combinations is 6C2 + 6C2 + 6C3 = 50

We have 3 places and 5 women; it is 5C3 possible choices
the number of combinations is 5C3 = 10

we can choose 50*10 = 500 different commitees
--------
2) How many different committees of 6 can be chosen from 8 men and 5 women so as to contain 2 men and 4 women if two of the men refuse to serve together? [spoiler](6C1 + 6C1 + 6C2)5C4 = 135[/spoiler]

if M1 was chosen we have one place and 6 free men; it is 6C1 possible choices
if M2 was chosen we have one place and 6 free men; it is 6C1 possible choices
otherwise we have 2 places and 6 free men; it is 6C2 possible choices
the number of combinations is 6C1 + 6C1 + 6C2 = 27

We have 4 places and 5 women; it is 5C4 possible choices
the number of combinations is 5C4 = 5

we can choose 27*5 = 135 different commitees
--------
now we should add all possible outcomes
solution: [spoiler]500 + 135 = 635
IMO E[/spoiler]