Two People, Three ObjectsDate: 07/14/2001 at 03:02:14 From: Nupur Srivastava Subject: Combination Find the number of ways in which 2n objects of type A, 2n objects of type B, and 2n objects of type C can be divided between two persons, giving 3n objects to each. I started solving this problem thinking that we can select some or all objects in 2n+1 ways, but I got stuck at the point that each selection from type again yields a variable case for selection from type B - so how to calculate the cases where selection for the second case depends on the selection in the first case? Thanking you, Nupur Srivastava Date: 07/14/2001 at 07:49:38 From: Doctor Anthony Subject: Re: Combination If we select 3n objects for one person we automatically select the other 3n objects for the second person. We can find the number of ways of selecting 3n objects from 6n objects of the given colours by finding the coefficient of x^(3n) in the expansion of the the generating function [1 + x + x^2 + x^3 + .... + x^(2n)]^3 [1 - x^(2n+1)]^3 1 - 3.x^(2n+1) + 3.x^(4n+2) + x^(6n+3) ---------------- = --------------------------------------- (1-x)^3 (1-x)^3 Since we are only interested in powers of x up to 3n we can ignore the last two terms of the numerator. (1 - 3.x^(2n+1)).[1 + C(3,1)x + C(4,2)x^2 + ...+ C(n+1,n-1)x^(n-1) + C(3n+2,3n)x^(3n) + ...] and picking out the terms in x^(3n) we have C(3n+2,2) - 3.C(n+1,2) (3n+2)(3n+1) - 3(n+1)(n) 9n^2 + 9n + 2 - 3n^2 - 3n = ------------------------ = ------------------------- 2 2 6n^2 + 6n + 2 = -------------- = 3n^2 + 3n + 1 2 So the number of ways of dividing the 6n objects into two groups of 3n is 3n^2 + 3n + 1. We can do a simple check with n = 1. Suppose we have 2 red, 2 blue, 2 green objects. The formula gives 7 ways of dividing the objects into two groups of 3. We could choose 3 objects from 6 in the following ways RRB, RRG, BBR, BBG, GGR, GGB, RBG = 7 ways So the answer is correct for n = 1 and is a check that there are no arithmetic mistakes. We conclude that the number of arrangements is 3n^2 + 3n + 1. - Doctor Anthony, The Math Forum http://mathforum.org/dr.math/ |
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