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By Ross S., Weatherwax J.L.

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N ·2 , n! total first round results. The problem with this is that it will double count the total number of pairings. It will count the pairs AB and BA as distinct. To remove this over counting we need to divide by the total number of ordered n pairs. This number is 2n . When we divide by this we find that the total number of first round results is given by (2n)! n! Problem 12 (selecting committees) Since we must select a total of six people consisting of at least three women and two men, we could select a committee with four women and two mean or a committee with three woman and three men.

Problem 14 (counting vectors that sum to less than k) We want to evaluate the number of solutions to ni=1 xi ≤ k for k ≥ n, and xi a positive integer. Now since the smallest value that ni=1 xi can be under these conditions is given when xi = 1 for all i and gives a resulting sum of n. Now we note that for this problem the sum ni=1 xi take on any value greater than n up to and including k. Consider the number of solutions to ni=1 xi = j when j is fixed such that n ≤ j ≤ k. This number is given by j−1 .

Thus the number of solutions to the first equation r−1 8 4+5−1 = 70, the number of solutions to the second = above is given by 4 5−1 9 5+5−1 = 126, and finally the number of solutions to = equation is given by 4 5−1 6+5−1 10 the third equation is given by = = 210. Thus the total number of 5−1 4 solutions is given by the product of these three numbers. m for these calculations. Problem 14 (counting vectors that sum to less than k) We want to evaluate the number of solutions to ni=1 xi ≤ k for k ≥ n, and xi a positive integer.

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