1. Given twenty couples...

Given twenty couples, how many different three-member committees can be formed that do not contain both members of any of these couples?

Edit: Sorry for the identical thread violation.

2. You've got 20 couples and you need to pick people so no two can be from a couple.

Easy, you have 40 choices for the first committee member... you can pick anyone you like.

Then, out of the reamaining 39 people, you cannot pick the other half of the couple which you just picked. So you have 38 choices.

Once you pick out 2 people that aren't from the same couple, you've immediately ruled out their other half as well, so you can only choose from the remaining 18 couples, therefore 36 choices.

$40*38*36 = 54720$

3. The answer above, 54720, is an over-count by a factor of (3!).
The correct answer is 9120 or ${20 \choose 3} 2^3$
Choose three couples and then one from each of the couples.

4. Yes, you're quite right, pardon me... gotta say... Probability and chance has always been my weakness... what I really wanna know is what was wrong with my logic... and how can I fix my logic to eliminate the factor of 6?