Solutions from a student (dragged) 23

Solutions from a student (dragged) 23 - Part (c): Consider...

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Unformatted text preview: Part (c): Consider the following manipulations of a binomial like sum n j =i n j j i n xj −i y n−i−(j −i) = j =i n i n n−i j−i n−i j−i n−i j n−i j xj −i y n−j xj −i y n−j xj y n−(j +i) xj y n−i−j = n i n i n i n i j =i n−i j =0 n−i j =0 = = = In summary we have shown that n (x + y ) n − i . j =i n j j i xj −i y n−j = n i (x + y ) n − i for i ≤ n Now let x = 1 and y = −1 so that x + y = 0 and using these values in the above we have n j =i n j j i (−1)n−j = 0 for i ≤ n . Problem 15 (the number of ordered vectors) As stated in the problem we will let Hk (n) be the number of vectors with components x1 , x2 , · · · , xk for which each xi is a positive integer such that 1 ≤ xi ≤ n and the xi are ordered i.e. x1 ≤ x2 ≤ x3 ≤ · · · ≤ xn Part (a): Now H1 (n) is the number of vectors with one component (with the restriction on its value of 1 ≤ x1 ≤ n). Thus there are n choices for x1 so H1 (n) = n. We can compute Hk (n) by considering how many vectors there can be when the last component i.e. xk has value of j . This would be the expression Hk−1(j ), since we know the value of the k -th component. Since j can range from 1 to n the total number of vectors with k components (i.e. Hk (n)) is given by the sum of all the previous Hk−1 (j ). That is n Hk (n) = j =1 H k − 1 (j ) . Part (b): We desire to compute H3 (5). To do so we first note that from the formula above the points at level k (the subscript) depends on the values of H at level k − 1. To evaluate ...
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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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