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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 ni players in winning combinations (with ties allowed). This can be done recursively in N ( ni ) ways. Summing up all of these terms we Fnd that N ( n ) = n i =1 n i N ( ni ) . Part (c): In the above expression let j = ni , then our limits on the sum above change as follows i = 1 j = n1 and i = n j = 0 , so that the above sum for N ( n ) becomes N ( n ) = n1 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.
 Winter '08
 G.JOGESHBABU
 Probability

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