Solutions from a student (dragged) 27

Solutions from a student (dragged) 27 - ² r k ³² n-1...

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lower bound on x i ) r ± i =1 ( x i - m i )= n - r ± i =1 m i . This expression motivates us to defne v i = x i - m i .Th en v i 0sowearelook ingForthe number oF solutions to the equation r ± i =1 v i = n - r ± i =1 m i , where v i must be greater than or equal to zero. This number is given by ² n - r i =1 m i + r - 1 r - 1 ³ . Problem 21 ( k zeros in an integer equation ) To fnd the number oF solutions to x 1 + x 2 + ··· + x r = n, where exactly k oF the x r ’s are zero, we can select k oF the x i ’s to be zero in ² r k ³ ways and then count the number oF solutions with positive (greater than or equal to one solutions) For the remaining r - k variables. The number oF solutions to the remaining equationi s ² n - 1 r - k -
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Unformatted text preview: ² r k ³² n-1 r-k-1 ³ . Problem 22 (the number of partial derivatives) Let n i be the number oF derivatives taken oF the x i th variable. Then a total order oF n derivatives requires that these componentwise derivatives satisFy ∑ n i =1 n i = n , with n i ≥ 0. The number oF such is given by ² n + n-1 n-1 ³ = ² 2 n-1 n-1 ³ . Problem 23 (counting discrete wedges) We require that x i ≥ 1 and that they sum to a value less than k , i.e. n ± i =1 x i ≤ k ....
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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 Penn State.

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