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Unformatted text preview: MAD 6206 Combinatorics I, Fall 2011 CRN: 87401 Assignment 1 suggested solution Question 1 . (Multinomial coefficients) Let ℓ ≥ 1 and n,k 1 ,k 2 ,...,k ℓ be nonnegative integers satisfying k 1 + k 2 + ··· + k ℓ = n. Define the multinomial coefficient ( n k 1 ,k 2 ,...,k ℓ ) as the number of ordered partitions of [ n ] into ℓ (pairwsie disjoint but possibly empty) subsets A 1 ,A 2 ,...,A k where  A i  = k i , i = 1 , 2 ,...,ℓ . (a) Prove that parenleftbigg n k 1 ,k 2 ,...,k ℓ parenrightbigg = parenleftbigg n k 1 parenrightbiggparenleftbigg n − k 1 k 2 parenrightbiggparenleftbigg n − k 1 − k 2 k 3 parenrightbigg ··· parenleftbigg k ℓ k ℓ parenrightbigg . (b) Prove that parenleftbigg n k 1 ,k 2 ,...,k ℓ parenrightbigg = [ x k 1 1 x k 2 2 ··· x k ℓ ℓ ]( x 1 + x 2 + ··· + x ℓ ) n . That is, ( n k 1 ,k 2 ,...,k ℓ ) equals the coefficient of x k 1 1 x k 2 2 ··· x k ℓ ℓ in the expansion of ( x 1 + x 2 + ··· + x ℓ ) n . Solution . (a) There are ( n k 1 ) ways to choose A 1 , ( n k 1 k 2 ) ways to choose A 2 . This process continues until we get to the last set A ℓ , where we have k ℓ elements remaining, and we use all of them for A ℓ , giving ( k ℓ k ℓ ) = 1 way. The given formula now follows from the product principle. (b) Consider the expression ( x 1 + x 2 + ··· + x ℓ ) ··· ( x 1 + x 2 + ··· + x ℓ ) bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright n factors . The coefficient equals the number of ways to obtain a term involving x k 1 1 x k 2 2 ··· x k ℓ ℓ . To obtain such a term in the expansion, we need to choose from the n factors, k 1 of them for x 1 , k 2 of them for x 2 , ..., k ℓ of them for x ℓ . Thus there are ( n k 1 ,k 2 ,...,k ℓ ) ways to form a term involving x k 1 1 x k 2 2 ··· x k ℓ ℓ . Question 2 . The unimodal property of the binomial coefficients) Consider ( n k ) where n,k are nonnegative integers satisfying n ≥ k ≥ 0. (a) Let r k = ( n k +1 ) ( n k ) , k = 0 , 1 ,...,n − 1 . Prove that r k > 1 , if k < ( n − 1) / 2 = 1 , if k = ( n − 1) / 2 < 1 , if k > ( n − 1) / 2 . That is, the sequence { ( n k ) } ≤ k ≤ n is unimodal, with maximum at the center term(s). (b) Use Stirling’s formula to prove that parenleftbigg 2 n n parenrightbigg ∼ 2 2 n · ( πn ) 1 / 2 , as n → ∞ ....
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This note was uploaded on 01/02/2012 for the course MAD 6206 taught by Professor Suen during the Spring '11 term at University of South Florida.
 Spring '11
 Suen

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