Replying to a PM re: what the division by 3! is doing here.
We need to divide by 3! here because this problem is asking about combinations, not permutations. Let's say we start off just filling three slots from the original 10 people (5 couples).
Then I'd have 10 choices for who to put in my first slot. After I put someone there, I'd have 8 choices for who to put in my next slot (since whoever I initially chose AND that person's spouse would no longer be eligible). Then I'd have 6 remaining choices for who to put in my third, final slot. So, so far we're at 10*8*6 possibilities. Now, IF I were actually assigning specific ROLES -- e.g. if I were picking one person to be the leader of the committee, one person to be the secretary, and one person to be the alternate -- I'd stop there, and say there were 480 (that is, 10*8*6) possibilities for how I could assign these roles to three people, choosing from among my initial 10 with the no-more-than-one-from-each-couple restriction.
But that's not what I'm doing in this problem -- this problem just wants me to form a committee -- there are no specific roles. So now the thing is, say we look at one of our 10*8*6 possibilities, and it goes Alfred, Doris, Edward. Well, somewhere in that list of 10*8*6 possibilities, I'm also going to find a possible committee consisting of Alfred, Edward, Doris. And I'm also somewhere going to find a possible committee consisting of Doris, Alfred, Edward. And one consisting of Doris, Edward, Alfred. Then I'll find one consisting of Edward, Alfred, Doris... and finally one consisting of Edward, Doris, Alfred. And the catch is that these really shouldn't be counted as 6 different possible committees, because they're all actually the same. Since for any set of three people I chose, I would have found that set 6 times on my list, I divide through by 6 to get rid of all those extra copies.
Specifically, it wound up being 6 here that I needed to divide through by because 6 -- 3! -- is the number of possible arrangements of any given group of three people. If I were choosing 4-person committees instead, I'd instead divide whatever initial product I'd gotten by 4!, because my initial list would have had any given group of four people showing up in 4! different arrangements. In short, any given group of n particular people can show up in n! different arrangements, so when you're dealing with combinations (rather than permutations) problems, in which all of those n! arrangements amount to the same exact group, you divide through by n! so that you only count that group once instead of n! times.
Let me know if that clears it up!
Ashley Newman-Owens
GMAT Instructor
Veritas Prep
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