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,weneedtoevaluate H k ( n )for k =1and k =2 .Wehavethat 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 Frst 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 Fnd that H 3 (5) = H 2 (1) + H 2 (2) + H 2 (3) + H 2 (4) + H 2 (5) =1+ 3+ 6+ 1 0+ 1 5= 3 5 . Problem 16 (the number of tied tournaments) Part (a): See Table 2 for the enumerations used in computing N (3). We have denoted A ,
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Unformatted text preview: i players from n that will tie (which can be done in n i 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 n i 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 n j N ( j ) ....
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This note was uploaded on 02/25/2011 for the course STAT 418 taught by Professor G.jogeshbabu during the Winter '08 term at Pennsylvania State University, University Park.

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