Perm-Comb: husband/wife in committee

[sam77]

The below problem posted on Page 3 of Tips

Please help...

A committee of three people is to be chosen from four married couples. What is the number of different committees that can be chosen if two people who are married to each other can't both serve on the committee?

a) 16
b) 24
c) 26
d) 30
e) 32

[spoiler]I like to get away from using formulas on permutation/combination problems. Most GMAT problems are easier if you just think about them logically. In this problem, we have three "slots" on the team, so draw a little picture:

_____ ______ ______

And ask yourself: how many choices do I have for each slot?

For the first slot, you can pick anyone, so 8. For the second slot, you can pick anyone except the person you already picked and his or her spouse, so 6. For the last slot, using the same reasoning, we have 4 choices.

So: 8x6x4

But, since order doesn't matter in this problem, we have to divide by 3x2x1 (an easy way to think of it is to divide by N!, where "N" is the number of items in the numerator of your fraction).

(8x6x4)/(3x2x1) = 32

[/spoiler]

Can someone explain me the latter portion of the solution? I understood uptill 8x6x4. But, where does the order comes into picture (division by3x2x1)? In every case, the first slot will always have 8 options, second 6 and third 4. So, I think the solution ends right there.
Obviously, since it is not one of the answer choices, so I am wrong

[DavoodBeater]

i used to be familiar with these kinds of question, but it seems i can help you in that way.
but here is mine:

couple 1:c1 couple 2:c2 couple 3:c3 couple 4:c4

c1:2 * c2:2 * c3:2
c1:2 * c3:2 * c4:2
c1:2 * 2 c3:* c4:2
c2:2 * c3:2 * c4:2

it means that you have select couple 1 you have 2 options, male or female, and then for couple 2 and for couple 3, so 8 kinds of committees.
do it with the rest, and you have 32 ways.