Lecture on Permutation Combination

# Thereare subsetsofk elementsthatcontainasincethereare

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Unformatted text preview: of the form xn jyj for j = , , ,…,n. To form the term xn jyj, it is necessary to choose n j xs from the n sums. Therefore, the coefficient of xn jyj is which equals . 17 Using the Binomial Theorem Example: What is the coefficient of x12y13 in the expansion of ( x y)25? Solution: We view the expression as ( x +( y))25. By the binomial theorem x12y13 18 A Useful Identity Corollary 1: With n 0, Proof (using binomial theorem): With x = 1 and y = 1, from the binomial theorem we see that: Proof (combinatorial): Consider the subsets of a set with n elements. There are subsets with zero elements, with one element, with two elements, …, and with n elements. Therefore the total is Since, we know that a set with n elements has 2n subsets, we conclude: 19 Blaise Pascal (1623‐1662) Pascal’s Identity Pascal’s Identity: If n and k are integers with n k 0, then Proof (combinatorial): Let T be a set where |T| = n + 1, a T, and S = T {a}. There are subsets of T containing k elements. Each of these subsets either: contains a with k 1 other elements, or contains k elements of S and not a. There are subsets of k elements that contain a, since there are subsets of k 1 elements of S, subsets of k elements of T that do not contain a, because there are subsets of k elements of S. Hence, See Exercise 19 for an algebraic proof. 20 Pascal’s Triangle The nth row in the triangle consists of the binomial coefficients , k = 0,1,….,n. By Pascal’s identity, adding two adjacent bionomial coefficients results is the binomial coefficient in the next row between these two coefficients. 21 Section 22 Section Summary Permutations with Repetition Combinations with Repetition Permutations with Indistinguishable Objects Distributing Objects into Boxes 23 Permutations with Repetition Theorem 1: The number of r‐permutations of a set of n objects with repetition allowed is nr. Proof: There are n ways to select an element of the set for each of the r positions in the r‐permutation when repetition is allowed. Hence, by the product rule there are nr r‐permutations with repetition. Example: How...
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## This document was uploaded on 03/29/2014 for the course COT 3100h at University of Central Florida.

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