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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- Winter '08
- Probability, HK, HMS H1, 1 j, 1 J