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Unformatted text preview: Stat 230 Assignment 1  Solutions The first three questions consider the process of arranging coloured marbles in a row from left to right. Two marbles of the same colour are to be considered indistinguishable when counting arrangements. 1. (a) Suppose that there are 2 white and 4 black marbles. List all 15 ways to arrange these marbles. (b) What combinatorial number describes the number of ways to arrange n white and 4 black marbles? Check your answer for the case n = 2 . To pick an arrangement, choose 4 out of n + 4 positions to hold black marbles. This can be done in ( n +4 4 ) ways. In the case of n = 2 , this corresponds to the ( 6 4 ) = 6 5 4 3 4! = 15 arrangements listed in part (a). (c) In which of the arrangements in 1a (list them) is every black marble adjacent to at least one other black marble? (d) List the ways to arrange 2 white and 2 red marbles. (e) Explain why the lists in 1c and 1d have the same size? If 4 blue marbles are arranged so that each is adjacent to at least one other, then they must appear either in two groups of 2 or in a single group of 4 . In both cases, the black marbles appear in even groups and each group may be replaced with half as many red marbles. The process may be reversed by replacing each red marble with two adjacent black marbles. (f) How many ways are there to arrange n white and 4 black marbles so that every black marble is adjacent to at least one other black marble? (Check that your answer agrees with part 1c.) Using the logic from part (e), the required arrangements may be counted by counting, instead, all arrangements of n white and 2 red marbles, of which there are ( n +2 2 ) (using the logic from (b)) . This agrees with part (c), since ( 2+2 2 ) = 4 3 2! = 6 . (g) If 20 white and 4 black marbles are arranged at random, what is the probability that every black marble is adjacent to at least one other black marble? By part (f), there are ( 20+2 2 ) arrangements of the marbles with each black ball adjacent to another, out of ( 20+4 4 ) equally likely arrangements total, so the required probability is ( 22 2 ) ( 24 4 ) = 1 46 . 021739 . . . ....
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This note was uploaded on 11/30/2010 for the course MATH 235/237 taught by Professor Wilkie during the Spring '10 term at Waterloo.
 Spring '10
 WILKIE
 Counting

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