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Solutions from a student (dragged) 24

Solutions from a student (dragged) 24 - i players from n...

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this expression when n = 5, we need to evaluate H k ( n ) for k = 1 and k = 2. We have that H 1 ( n ) = n H 2 ( n ) = n j =1 H 1 ( j ) = n j =1 j = n ( n + 1) 2 H 3 ( n ) = n j =1 H 2 ( j ) = n j =1 j ( j + 1) 2 . Thus we can compute the first few values of H 2 ( · ) as H 2 (1) = 1 H 2 (2) = 3 H 2 (3) = 6 H 2 (4) = 10 H 2 (5) = 15 . So that we find that H 3 (5) = H 2 (1) + H 2 (2) + H 2 (3) + H 2 (4) + H 2 (5) = 1 + 3 + 6 + 10 + 15 = 35 . Problem 16 (the number of tied tournaments) Part (a): See Table 2 for the enumerations used in computing N (3). We have denoted A , B , and C by the people all in the first place. Part (b): To argue the given sum, we consider how many outcomes there are when i -players tie for last place. To determine this we have to choose the
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Unformatted text preview: i players from n that will tie (which can be done in ² n i ³ ways). We then have to distributed the remaining n-i players in winning combinations (with ties allowed). This can be done recursively in N ( n-i ) ways. Summing up all of these terms we Fnd that N ( n ) = n ± i =1 ² n i ³ N ( n-i ) . Part (c): In the above expression let j = n-i , then our limits on the sum above change as follows i = 1 → j = n-1 and i = n → j = 0 , so that the above sum for N ( n ) becomes N ( n ) = n-1 ± j =0 ² n j ³ N ( j ) ....
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