ywilfred wrote:

Well, I solved it this way.

Two possible ways: 4W2M OR 3W3M

I'll call the two men who refuse to serve together John and Dick

Case 1: 4W2M. Can be broken into three subgroups.

Subgroup A: 2 men (excluding John and Dick) and 4 women

So we have 6C2*5C4 = 75 ways

Subgroup B: 2 men (John and one other) and 4 women

So we have 6C1*5C4 = 30

Subgroup C: 2 men (Dick and one other) and 4 women

SO we have another 30 ways

Case 2: 3M3W. This case can be further broken down in 3 subgroups.

Subgroup 1: 3 men (excluding John and Dick) and 3 women

So we have 6C3*5C3 = 200

SUbgroup 2: 3 men (John and 2 others) and 3 women

SO we have 6C2*5C3 = 150

Subgroup 3: 3 men (Dick and 2 others) and 3 women

So we have 6C2*5C3 = 150

Total = 635 ways

Of course, this method is rather long, but it's a logical way of thinking. But I'm game for any shortcuts. I'll probably only use this method if I don't have a short-cut, or I can't remember the short-cut during the test.

Good explanation. thanks.

I guess the number of steps can be reduced if we find the opposite first:

That is, find all combinations when two persons are ALWAYS together:

to chose 2M 4W, number of such arrangements will be:

6C0 * 5C4 = 5

{Number of combinations of n different things taken r at a time when p particular things always occur is n -pCr -p. }

similarly, to chose 3M 3W, number of arrangement when two men always together:

6C1 * 5C3 = 60

Total such combinations : 60 + 5 = 65

We need to subtract this from all arrangements (700 : see fresinha2 post below) to get required number:

700 - 65 = 635