Problem Solving (PS) | A committee of three people is to be chosen from four married couples.

How would the solution change if the question did not say A (1) team of 3 is to be selected?
Instead just how many committes..... would that be 3C8 - (1C4*6) + (1C3 *4)? or 3C8 - [(2C2)^4 * 1C6]?
2C2^4: there are 4 couple and we choose both members of the couple
1C6 ways to choose 1 member out of 6 persons

We could make 2 committes with 2 siblings + another person. Then there would be one spare couple of siblings unchosen.
Right? I am confused....
Any clarification would help a lot! Thanks!­

Statistics : Posted by ruis • on 11 May 2010, 12:35 • Replies 36 • Views 225539

---