PS: permutation

amitdgr wrote:There are 5 married couples and a group of three is to be formed out of them; how many arrangements are there if a husband and wife may not be in the same group?

Stuart Kovinsky wrote:

amitdgr wrote:There are 5 married couples and a group of three is to be formed out of them; how many arrangements are there if a husband and wife may not be in the same group?

Let's try this one without using a formula.

For our first person, we have 10 choices.

Once we've chosen a person, we also eliminate his or her spouse. So, for our second person, we have 8 choices.

Now we've eliminated both of those people and their spouses, so for our third person, we have 6 choices.

Therefore, we have 10*8*6 = 480 possible trios.

Ian Stewart wrote:The phrasing of the question is not good- in mathematical counting, the word 'arrangements' implies that order matters, but there is no reason to think order should matter in this question. A real GMAT question will never be ambiguous about whether order matters.

Ian Stewart wrote: Alternatively you could look at the problem as follows (assuming order does not matter):

-there are 10C3 = 120 ways to choose three people;
-if we choose a group of three which *does* include a married couple, we have five choices for the married couple, and 8 choices for the person to join them, or 5*8 = 40 choices in total;
-we thus must have 120-40 = 80 ways of choosing three people and *not* including a married couple.