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By Ross S., Weatherwax J.L.

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This can be done in n select one of the n! orderings of the remaining n players to form matches with. Finally, we must select winners of each match in 2n ways. In total we would then conclude that we have 2n n · n! · 2n = (2n)! 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 .

Problem 3 (counting presidents) Part (a): With no restrictions we must select three people from ten. This can be done in 10 ways. Then with these three people there are 3! ways to specify which person is the 3 president, the treasurer, etc. Thus in total we have 10 3 · 3! = 10! = 720 , 7! possible choices. Part (b): If A and B will not searve together we can construct the total number of choices by considering clubs consisting of instances with A included but no B, B included by no A, and finally neither A or B included.

Then the probability a random child comes from a family with i children is 16 8 5 4 , P2 = 48 , P3 = 15 , P4 = 48 , and P5 = 48 . given by (and denoted by Pi ) is P1 = 48 48 1 0 0 0 0 0 0 1 2 3 4 5 6 2 1 0 0 0 0 0 3 1 1 0 0 0 0 4 1 1 1 0 0 0 5 1 1 1 1 0 0 6 1 1 1 1 1 0 Table 3: The elements of the sample space where the second die is strictly larger in value than the first. Problem 22 (shuffling a deck of cards) To have the ordering exactly the same we must have k heads in a row (which leave the first k cards unmoved) followed by n − k tails in a row (which will move the cards k + 1, k + 2, .

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A solution manual for A first course in probability by Ross S., Weatherwax J.L.

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