Explanation The given identity is ∑ k = 0 n ( n k ) ( m − n n − k ) = ( m n ) . R.H.S: Here, ( m n ) is the number of arrangement of n object from m distinct object. For example, n balls are drawn from a box, which is contained m balls at the beginning. The balls are named from 1 to m and placing in a smaller box. L.H.S: Here, ( n k ) is the number of ways to select k objects from n elements. Also, ( m − n n − k ) is the number of ways to select n − k objects from m − n elements. By using the above example, k balls are drawn from a box and selecting balls are named with n ≤ m . After selecting a k balls, then select the extra n − k balls and the selected balls are named with numbers greater than n

Mathematical Statistics and Data Analysis

3rd Edition

ISBN: 9781111793715

Author: John A. Rice

Publisher: Cengage Learning

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Sequence and Series

A sequence is defined as a collection of things. Series is defined to sum the things one by one in the sequence. It was invented by German mathematician Carl Friedrich Gauss.

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Chapter 1.8, Problem 34P

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Prove the given identity: ∑k=0n(nk)(m−nn−k)=(mn).

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Chapter 1 Solutions

Mathematical Statistics and Data Analysis

Ch. 1.8 - Prob. 1P Ch. 1.8 - Prob. 2P Ch. 1.8 - Prob. 3P Ch. 1.8 - Prob. 4P Ch. 1.8 - Prob. 5P Ch. 1.8 - Prob. 6P Ch. 1.8 - Prob. 7P Ch. 1.8 - Prob. 8P Ch. 1.8 - Prob. 9P Ch. 1.8 - Prob. 10P

Ch. 1.8 - Prob. 11P Ch. 1.8 - Prob. 12P Ch. 1.8 - Prob. 13P Ch. 1.8 - Prob. 14P Ch. 1.8 - Prob. 15P Ch. 1.8 - Prob. 16P Ch. 1.8 - Prob. 17P Ch. 1.8 - Prob. 18P Ch. 1.8 - Prob. 19P Ch. 1.8 - Prob. 20P Ch. 1.8 - Prob. 21P Ch. 1.8 - Prob. 22P Ch. 1.8 - Prob. 23P Ch. 1.8 - Prob. 24P Ch. 1.8 - Prob. 25P Ch. 1.8 - Prob. 26P Ch. 1.8 - Prob. 27P Ch. 1.8 - Prob. 28P Ch. 1.8 - Prob. 29P Ch. 1.8 - Prob. 30P Ch. 1.8 - Prob. 31P Ch. 1.8 - Prob. 32P Ch. 1.8 - Prob. 33P Ch. 1.8 - Prob. 34P Ch. 1.8 - Prob. 35P Ch. 1.8 - Prob. 36P Ch. 1.8 - Prob. 37P Ch. 1.8 - Prob. 38P Ch. 1.8 - Prob. 39P Ch. 1.8 - Prob. 40P Ch. 1.8 - Prob. 41P Ch. 1.8 - Prob. 42P Ch. 1.8 - Prob. 43P Ch. 1.8 - Prob. 44P Ch. 1.8 - Prob. 45P Ch. 1.8 - Prob. 46P Ch. 1.8 - Prob. 47P Ch. 1.8 - Prob. 48P Ch. 1.8 - Prob. 49P Ch. 1.8 - Prob. 50P Ch. 1.8 - Prob. 51P Ch. 1.8 - Prob. 52P Ch. 1.8 - Prob. 53P Ch. 1.8 - Prob. 54P Ch. 1.8 - Prob. 55P Ch. 1.8 - Prob. 56P Ch. 1.8 - Prob. 57P Ch. 1.8 - Prob. 59P Ch. 1.8 - Prob. 60P Ch. 1.8 - Prob. 61P Ch. 1.8 - Prob. 62P Ch. 1.8 - Prob. 63P Ch. 1.8 - Prob. 64P Ch. 1.8 - Prob. 65P Ch. 1.8 - Prob. 66P Ch. 1.8 - Prob. 67P Ch. 1.8 - Prob. 68P Ch. 1.8 - Prob. 69P Ch. 1.8 - Prob. 70P Ch. 1.8 - Prob. 71P Ch. 1.8 - Prob. 72P Ch. 1.8 - Prob. 73P Ch. 1.8 - Prob. 74P Ch. 1.8 - Prob. 75P Ch. 1.8 - Prob. 77P Ch. 1.8 - Prob. 78P Ch. 1.8 - Prob. 79P Ch. 1.8 - Prob. 80P

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Prove the given identity: ∑ k = 0 n ( n k ) ( m − n n − k ) = ( m n ) .

Solution Summary: The author demonstrates that the given identity is displaystylek=0oversetnsum.

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Prove the given identity: ∑k=0n(nk)(m−nn−k)=(mn).

Chapter 1 Solutions