Veritas Prep GMAT Instructor
Joined: 28 Jul 2010
Affiliations: Veritas Prep
Posts: 11
Re: Confusing Combination Problem
[#permalink]
Updated on: 02 Aug 2010, 22:37
Hi there,
The way this question is phrased is confusing, too. I think that the original problem means to say that the constraint about not having the same two neighbors only applies to one specific person, for example, person B if you start with people ABCDE. This means that the answer actually should be 6.
Let's start with the first step. It's a permutation, because the order of the elements matters. Normally, to arrange 5 people in 5 seats, you simply calculate 5!. But in circular arrangements, ABCDE is the same arrangement whether A sits in seat 1, 2, 3, 4, or 5. So first off, you have to divide out those 5 redundant possibilities from the factorial calculation. That's why it's (5-1)!, or 4!. (As a rule, with circular arrangements, the formula for the number of arrangements is (n-1)!)
So that gives you 24 possibilities. But we also have the additional constraint. If person B regards ABC as the same as CBA, then the question actually becomes a combination, in which order doesn't matter. Now we have to figure out how many 2-person combinations can be created out of the remaining 4 people, once we've stuck B in place. That means 4!/2!2!, or 6. So of those 24 initial possibilities, only 6 are unique.
Another way to think about this last step is to understand that within the 24 unconstrained possibilities, you have a clockwise and counter-clockwise redundancy. (In other words, ABC = CBA.) So divide by 2 to get 12. But within each of the clockwise and counter-clockwise arrangements, there are an additional 2 possibilities, because the remaining two guests can be arranged DE or ED. That's why you have to divide by 2 again to get 6.
Think of it this way: A in seat 1, B in seat 2, C in seat 3, D in seat 4, and E in seat 5 is the same, according to the terms of this problem, as A-1, B-2, C-3, E-4, D-5, as C-1, B-2, A-3, D-4, E-5, and as C-1, B-2, A-3, E-4, D-5. Once you've placed B, there are three other redundant scenarios. Therefore you divide the original 24 possibilities by 4 to get 6.