# a01a - MAD 6206 Combinatorics I Fall 2011 CRN 87401...

This preview shows pages 1–2. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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 → ∞ ....
View Full Document

## 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.

### Page1 / 4

a01a - MAD 6206 Combinatorics I Fall 2011 CRN 87401...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online