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 Subject: Permutation / Combination Question Category: Science > Math Asked by: michned-ga List Price: \$10.00 Posted: 20 Jul 2004 15:18 PDT Expires: 19 Aug 2004 15:18 PDT Question ID: 376855
 ```I have a group of ten people and I need to select six of them to pose for a photograph. Two of the ten people are named Kristen and Karen. How many ways can I arrange six people out of the group of ten for the photograph if Kristen and Karen must be in the photograph? Additionally, how can I arange six people out of the group of ten for the photograph if Kristen and Karen must be in the photograph and they must appear next to each other in the photo?``` Request for Question Clarification by smudgy-ga on 20 Jul 2004 15:28 PDT ```Hi michned, Do you want information on how to arrive at the numbers you're looking for, or do you just want the numbers themsleves? What level of detail are you looking for, basically? Thanks, smudgy.``` Clarification of Question by michned-ga on 20 Jul 2004 16:47 PDT ```Hi, I would like a brief look at how to arrive at the answers. For example, if I just needed to arrange 6 people out of a group of 10, I know 10!/6!4! would give me the answer, but I am getting confused when attemting to put particular people into the group of 6. Thanks for your quick reply.```
 ```Hi michned, I hope you find the following answer satisfactory. If anything is unclear or if you want more explanation for any part, please request a clarification before rating and I wll do my best to clear up any confusion. First a couple of definitions: nPk is the permutation of k elements from a set of n--that is, the number of ways we can pick k things from a group of n in such a way that order matters. In terms of factorials, nPk = n!/(n-k)! nCk is the combination of k elements from a group of n, that is, the number of ways we can choose k elements from a group of n in such a way that order doesn't matter. In factorials, nCk = n!/((n-k)! * k!) For more information about combination, permutation, and factorials, see < http://www.themathpage.com/aPreCalc/permutations-combinations.htm > We'll start with part A. We know two of the people in our group are Kristen and Karen, so just have them stand over to the side. They're already in our picture group. Now there are eight people left. We need to choose a further four people from those eight to go in the picture taking group. It doesn't matter which order we pick the people to photograph--they will be ordered later when we arrange the photo. This means we are in an nCk situation. There are 8C4 ways of choosing four people from a group of eight. So that means that there are 8C4 possible groups that we might be taking pictures of. 8C4 = 8!/((8-4)! * 4!) = 70 possible groups to photograph. Now each of those 70 groups has six people, counting Karen and Kristen. So we need to figure out how many different orders those six people can be put in, say, from left to right. The number of ways to arrange all n members of a group in such a way that order matters is n!. So that means if we are arranging all six people in some particular order, there are 6! = 720 ways to do it. Now if there are 70 possible groups, and 720 ways to order each group, that means there are 70*720=50,400 different possible photograph arrangements that must include Kristen and Karen. The next part of the question is pretty interesting, because it requires us to use a very clever trick. Note that there are still 70 possible groups for us to photograph. The question is now, what if Kristen and Karen have to be next to each other? Well, let's do this, then. Pretend that instead of Kristen and Karen being separate, we tie them together with a bit of rope, as though they were in a three-legged race. Now wherever Kristen goes, Karen has to go, and vice versa. So that means, for the purposes of arranging the people in the photograph, we can treat KristenKaren as a single person. That means that instead of arranging 6 people, we are now effectively arranging 5--since Kristen and Karen kind of count as a single person. So the number of ways we can arrange the "5" people in some order is 5!=120. So the number of possible arrangements of the group that has the KristenKaren monster is 70*120=8400 ways. Now here's the thing: We originally might have tied Kristen on the left and Karen on the right, or vice versa. Let's say our arrangement was, say, something like: C D B KK A with Kristen and Karen tied together. This actually counts as two different arrangements, since we could have bound the pair so that either Kristen or Karen is on the left. So that number 8400 needs to be doubled--each arrangement involving the KristenKaren monster actually counts for two possible arrangements: one with Kristen on left and Karen on right, and one with Karen on left and Kristen on right. So the total number of ways to arrange the photo with Kristen and Karen next to each other is 8400*2 = 16,800 possible arrangements. I hope this explanation makes sense to you. If you are dissatisfied, or if you need further explanation in any aspect of this problem, please request a clarificaiton and I will do my best to make things clearer. Thanks, smudgy.``` Clarification of Answer by smudgy-ga on 20 Jul 2004 19:56 PDT ```Hi again michned, I should add: Google search strategy: Thanks, smudgy.```
 michned-ga rated this answer: ```Thanks for your help! I can understand the concepts much better now through your explanation.```