notes-090908 - MATH 681 Notes Combinatorics and Graph...

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Unformatted text preview: MATH 681 Notes Combinatorics and Graph Theory I 1 Fun with the combination statistic Of the 12 enumerative functions weve seen, the one with the most obvious utility is the combination statistic , a.k.a. the binomial coefficient : n k = n ! ( n- k )! k ! = n ( n- 1)( n- 2) ... ( n- k + 2)( n- k + 1) k ! That this is even guaranteed to be an integer is slightly surprising, but how useful it is is even more surprising. Today were going to look at a few of those uses and segue into a powerful combinatorial tool. 1.1 The multinomial/multicombinations The binomial was useful for enumerating the ways that two different types of item could be arranged: recall our items-and-dividers arguments, for instance. By selecting k elements of type 1 from a set of size n , we also select n- k elements of type 2. So even though a binomial appears to be a selection of a single class of item, it is implicitly a division of an ordered list between two classes. But what do we do when were dividing a list into more than 2 classes? Question 1: How many anagrams are there of the word MISSISSIPPI? Answer 1: Here we want to divide the positions of an 11-letter word into 4 classes of specified sizes: 4 positions for I, 4 positions for S, 2 for P, and 1 for M. We may start by choosing 4 of our ten positions as I locations: there are ( 11 4 ) ways to do that. Now we have 7 remaining locations, so there are ( 7 4 ) ways to choose positions for S, now of the remaining 3 positions, 2 must be P and there are ( 3 2 ) ways to do that; finally, the position of the M is forced, so in total there are ( 11 4 )( 7 4 )( 3 2 ) = 34650 ways to do this. Note that if we had assigned items to classes in a different order, wed have different enumeration statistics, e.g., if we chose a position for M, then the two Ps, and then the four Is, wed have ( 11 1 )( 10 2 )( 8 4 ) , which would have the same value. In fact, this would give us a free combinatorial identity: Proposition 1. ( n k )( n- k r ) = ( n r )( n- r k ) This identity could also be proven algebraically with ease, and combinatorially based on the above logic. Because multiple-class selections are useful enough, we actually have a generalization of the combination statistic: Definition 1. The multinomial coefficient is defined as follows, for n = k 1 + k 2 + + k r : n k 1 ,k 2 ,...,k r = n k 1 n- k 1 k 2 n- k 1- k 2 k 3 k n- 1 + k n k n- 1 k n k n = n ! k 1 ! k 2 ! k 3 ! k n ! Note that the combination statistic is, as mentioned above, a division into clases of size k and n- k , so ( n k ) = ( n k,n- k ) . The permutation statistic n P k can be considered as choosing n- k non- selected elements, and then selecting each of the remaining elements for a distinguished class, so n P k = ( n n- k, 1 , 1 , 1 ,..., 1 ) ....
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This note was uploaded on 01/12/2012 for the course MATH 681 taught by Professor Wildstrom during the Fall '09 term at University of Louisville.

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notes-090908 - MATH 681 Notes Combinatorics and Graph...

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