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hw4solns

# hw4solns - Discrete Mathematics Spring 2004 Homework 4...

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Discrete Mathematics, Spring 2004 Homework 4 Sample Solutions 4.2 #77. Let s n,k denote the number of ways to seat n persons at k round tables, with at least one person at each table. (The numbers s n,k are called Stirling numbers of the first kind .) The ordering of the tables is not taken into account. The seating arrangement at a table is taken into account, except for rotations. (a) Show that s n,k = 0 if k > n . (b) Show that s n,n = 1 for all n 1. (c) Show that s n, 1 = ( n 1)! for all n 1. (d) Show that s n,n - 1 = C ( n, 2) for all n 2. (e) Show that s n, 2 = ( n 1)! parenleftbigg 1 + 1 2 + 1 3 + · · · + 1 n 1 parenrightbigg for all n 2. (f) Show that n k =1 s n,k = n ! for all n 1 . (g) Find a formula for s n,n - 2 , n 3, and prove it. Solution . (a) If k < n , there are no ways to place n people at k tables without at least one of the tables being empty. (b) There is exactly one way to place n people at n tables; each table has a single person seated at it. (c) Denote the individuals by x 1 , . . . , x n ; we wish to seat them at a single table. Since rotations are not taken into account, given any ordering of the x i in a line with x 1 in the first position, we can associate to it a unique seating around the table; simply proceed counterclockwise around the table, starting with x 1 . Clearly we can reverse this process, so that given a seating we can obtain an ordering of the x i in a line with x 1 in the first position. 1

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But given that x 1 must be in the first position of the line, there are exactly ( n 1)! ways to order the remaining elements, and hence s n, 1 = ( n 1)!. (d) With notation as in (c), the choice of any unordered pair { x i , x j } determines a unique seating of the n people at n 1 tables; this pair sits at a single table (there is only one way to seat them there), and each of the remaining tables has a single person seated there. So the number of possible seatings is C ( n, 2). (e) We determine a seating of x 1 , . . . , x n at two tables in the following process: (1) Select a k -person subset of { x 2 , . . . , x n } to sit at the same table with x 1 . The remaining individuals will, of course, sit at the other table. (Notice that k cannot equal n 1, or else no one will be sitting at the other table.) (2) Order the people sitting at the same table as x 1 . (3) Order the people at the other table. For a fixed value of k , there are C ( n 1 , k ) ways to perform step 1. By the same argument as in (c), there are k ! ways to perform step 2, and there are ( n k 2)! ways to perform step 3 (note that one performs this step after already having chosen the set of people to sit at this table, so fix one of them and then use the same argument as in (c)). Summing over the possible values of k , it follows that the total number of ways to seat x 1 , . . . , x n at two tables is given by n - 2 k =0 C ( n 1 , k ) · k !( n k 2)! = n - 2 k =0 ( n 1)!
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hw4solns - Discrete Mathematics Spring 2004 Homework 4...

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