Buying DoughnutsDate: 03/22/2002 at 22:24:18 From: Wendy Subject: Probability I understand the question but I'm stuck on the equations. I know we can guess and test to find out, but I was just wondering if there is a formula/equation we can use for this question. Janine wants to buy three doughnuts, and there are five varieties to choose from. She wants each doughnut to be a different variety. How many combinations are there? Date: 03/23/2002 at 11:23:34 From: Doctor Ian Subject: Re: Probability Hi Wendy, Let's call the different kinds of doughnuts A, B, C, D, and E. There are 5 ways that Janine can make her first choice: A _ _ B _ _ C _ _ D _ _ E _ _ In each case, there are 4 ways to make the next choice: A _ _ A B _ A C _ A D _ A E _ B _ _ B A _ B C _ B D _ B E _ C _ _ C A _ C B _ C D _ C E _ D _ _ D A _ D B _ D C _ D E _ E _ _ E A _ E B _ E C _ E D _ 5 5*4 In each case, there are three ways to make the final choice. I'm not going to list all of them, but you can see the the total number of choices will be 5*4*3, right? In general, if you have N things, and you want to choose K of them, there are N * (N-1) * ... * (N-K+1) ways to make those choices. If you're going to choose all the items eventually, this becomes N * (N-1) * ... * 2 * 1 which is abbreviated N! (pronounced N factorial). Note that N * (N-1) * ... * (N-K+1) * (N-K) * ... * 1 N * (N-1) * ... * (N-K+1) = ------------------------------------------- (N-K) * ... * 1 because the stuff at the end cancels out. So we can write N! N * (N-1) * ... * (N-K+1) = ------ (N-K)! You may have to look at this for a while, and work out a few examples before it really sinks in, but believe me, if you do that it will be time well spent! I strongly recommend that you do it while you have time to experiment, rather than when you won't, for example, during a test. Anyway, this tells us the number of 3-doughnut PATTERNS that we can make. But is there really a difference between ABC and CBA? In some cases, it makes a difference (for example, if we're generating locker combinations); but in our case, it does not. So we need to find some way to ignore the duplicates, a way that is easier than looking through the list to find them. Here's one way to do that. Let's think about doughnuts A, B, and C. Given the way we produced the patterns, we're going to have these three doughnuts show up in every possible order. Do you see why? Well, how many orders would that be? Do you see that it's really the same problem we just solved? The number of patterns that we can make from three doughnuts is 3! . So for each combination of three doughnuts, we're going to have 3! patterns of those doughnuts. Which means that we have 3! times more patterns than we need. So to get the number of combinations, we divide the number of patterns by this 'excess factor', to get a general formula for the number of sets of K elements you can choose from N candidates: N! ------ (N-K)! ------------ K! which can be written more simply as N! -------- K!(N-K)! So if there are 8 types of donuts, and she wants to choose 5, Janine could do that in 8! 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 -------- = -------------------------------------------- 5!(8-5)! (5 * 4 * 3 * 2 * 1) * (3 * 2 * 1) 8 * 7 * 6 = --------- 3 * 2 * 1 = 56 different ways. Does this help? - Doctor Ian, The Math Forum http://mathforum.org/dr.math/ |
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