Binomial Coefficients: Selected Exercises

Copyright © Peter Cappello Preliminaries What is the coefficient of x2y in ( x + y )3? ( x + y )3 = ( x + y )( x + y )( x + y ) = ( xx + xy + yx + yy )( x + y ) = xxx + xyx + yxx + yyx + xxy + xyy + yxy + yyy = x3 + 3x2y + 3xy2 + y3. The answer thus is 3. There are 23 terms in the formal expansion. Answer: The # of ways to pick the y position in the formal expansion: C( 3, 1 ). Copyright © Peter Cappello

Copyright © Peter Cappello Preliminaries How many terms are there in the formal expansion of ( x + y )n? How many formal terms have exactly 3 ys? This is the coefficient of xn-3y3 in ( x + y )n. How many formal terms have exactly j ys? Copyright © Peter Cappello

Copyright © Peter Cappello The Binomial Theorem Let x & y be variables, and n  N. Partition the set of 2n terms of the formal expansion of ( x + y )n into n + 1 classes according to the # of ys in the term: ( x + y )n = Σj=0 to n C( n, j )xn-jyj = C( n, 0 )xny0 + C( n,1 )xn-1y1 + … + C( n, j )xn-jyj + … + C( n, n )x0yn. Copyright © Peter Cappello

Copyright © Peter Cappello Pascal’s Identity Let n & k be positive integers, with n > k. Give a combinatorial argument to show that C( n, k ) = C( n - 1, k – 1 ) + C( n - 1, k ). A combinatorial argument proves that the equation’s LHS & RHS are different ways to count the elements of the same set. Copyright © Peter Cappello

Copyright © Peter Cappello Let n & k be positive integers, with n > k. C( n, k ) = C( n - 1, k – 1 ) + C( n - 1, k ). The left hand side (LHS) counts the number of subsets of size k from a set of n elements. The RHS counts these same subsets using the sum rule: Partition the subsets into 2 parts: Subsets of k elements that include element 1: Pick element 1: 1 Pick the remaining k – 1 subset elements from the remaining n - 1 set elements: C( n - 1, k – 1 ). Subsets of k elements that exclude element 1: Pick the k elements from the n - 1 remaining elements: C( n - 1, k ). Copyright © Peter Cappello

Copyright © Peter Cappello Exercise *30 Give a combinatorial argument to prove that Σk=1,n kC( n, k )2 = nC( 2n – 1, n – 1 ). Copyright © Peter Cappello

Copyright © Peter Cappello Give a combinatorial argument that Σk=1,n kC( n, k )2 = nC( 2n – 1, n – 1 ). The set of committees with n members from a group of n math professors & n computer science professors, such that the committee chair is a mathematics professor. Copyright © Peter Cappello

Copyright © Peter Cappello Exercise *30 Solution Σk=1,n kC( n, k )2 = nC( 2n – 1, n – 1 ) The RHS counts the # of such committees: Pick the chair from the n math professors: n Pick the remaining n – 1 members from the remaining 2n – 1 professors: C( 2n – 1, n – 1 ) The LHS counts the committees: Partition the set of such committees into subsets, according to k, the # of math professors on the committee. For each k, Pick the k math professor members: C( n, k ) Pick the committee chair: k Pick the n - k CS professor members: C( n, n – k ) = C( n, k ) Copyright © Peter Cappello

Combinatorial Identities Manipulation of the Binomial Theorem “Committee” arguments Block walking arguments – for identities involving sums Copyright © Peter Cappello

Manipulation of the Binomial Theorem ( x + y )n = Σj=0 to n C( n, j )xn-jyj = C( n, 0 )xny0 + C( n,1 )xn-1y1 + … + C( n, j )xn-jyj + … + C( n, n )x0yn. Prove that C( n, 0 ) + C( n, 1 ) + . . . + C( n, n ) = 2n. In general, Manipulate the binomial theorem algebraically; Evaluate the resulting equation for values of x & y, producing the desired result. n2n – 1 = 1C( n, 1 ) + 2C( n, 2 ) + 3C( n, 3 ) + . . . + nC( n, n ). Copyright © Peter Cappello

Copyright © Peter Cappello Committee Arguments Show that n2n – 1 = 1C( n, 1 ) + 2C( n, 2 ) + 3C( n, 3 ) + . . . + nC( n, n ). Hint: committees of any size, 1 of whom is chair. C( n, k )C( k, m ) = C( n, m )C( n – m, k – m ). Hint: committees of k people, m of whom are leaders. Σk = 0 to r C( m, k )C( w, r – k ) = C( m + w, r ). Hint: committees of r people taken from m men & w women. Copyright © Peter Cappello

Block-Walking Arguments Draw Pascal’s triangle. Interpret a node in the triangle as the # of ways to walk from the apex to the node, always going down. Show that C( n, k ) = C( n – 1, k ) + C( n – 1, k – 1 ) C( n, 0 )2 + C( n,1 )2 + . . . + C( n, n )2 = C( 2n, n ). Copyright © Peter Cappello

Copyright © Peter Cappello Pascal’s Triangle kth number in row n is nCk: k = 0 1 n = 0 k = 1 k = 2 n = 1 1 1 k = 3 n = 2 1 2 1 k = 4 1 3 3 1 n = 3 n = 4 1 4 6 4 1 Copyright © Peter Cappello

Displaying Pascal’s Identity k = 0 0C0 n = 0 k = 1 1C0 1C1 k = 2 n = 1 2C0 2C1 2C2 k = 3 n = 2 k = 4 3C0 3C1 3C2 3C3 n = 3 4C0 4C1 4C2 4C3 4C4 n = 4 Copyright © Peter Cappello

Block-Walking Interpretation nCk = # strings of n Ls & Rs with k Rs. nCk = # ways to get to corner n, k starting from 0, 0 k = 0 0C0 n = 0 k = 1 1C0 1C1 k = 2 n = 1 2C0 2C1 2C2 k = 3 n = 2 k = 4 3C0 3C1 3C2 3C3 n = 3 4C0 4C1 4C2 4C3 4C4 n = 4 Copyright © Peter Cappello

Pascal’s Identity via Block-Walking # routes to corner n, k = # routes thru corner n-1, k + # routes thru corner n-1, k-1 k = 0 0C0 n = 0 k = 1 1C0 1C1 k = 2 n = 1 2C0 2C1 2C2 k = 3 n = 2 k = 4 3C0 3C1 3C2 3C3 n = 3 4C0 4C1 4C2 4C3 4C4 n = 4 Copyright © Peter Cappello

Copyright © Peter Cappello nC02 + nC12 + nC22 + … + nCn2 = 2nCn k = 0 0C0 n = 0 k = 1 1C0 1C1 k = 2 n = 1 2C0 2C1 2C2 k = 3 n = 2 k = 4 3C0 3C1 3C2 3C3 n = 3 4C0 4C1 4C2 4C3 4C4 n = 4 Copyright © Peter Cappello

Copyright © Peter Cappello nC02 + nC12 + nC22 + … + nCn2 = 2nCn RHS = all routes to corner 4,2 LHS partitions routes to 4,2 into those that: go thru corner 2,0: 2C0  2C2 go thru corner 2,1: 2C1  2C1 go thru corner 2,2: 2C2  2C0 The identity generalizes this argument: # routes to 2n, n that go thru n,k = nCk  nCn-k Sum over k = 0 to n Copyright © Peter Cappello

Give a Committee Argument nC02 + nC12 + nC22 + … + nCn2 = 2nCn Hint: Number of committees of size n from a set of n men and n women. Challenge question: Derive this identity via the Binomial Theorem Use the algebraic fact: (x + y)2n = (x + y)n (x + y)n = (Σj=0 to n C( n, j )xn-jyj ) (Σj=0 to n C( n, j )xn-jyj ) Evaluate this identity at x = 1: (1 + y)2n = (1 + y)n (1 + y)n = ( Σj=0 to n C( n, j )yj ) ( Σj=0 to n C( n, j )yj ) What is the coefficient of yn in the above polynomial product? (Convolution) Copyright © Peter Cappello

Copyright © Peter Cappello End Copyright © Peter Cappello

Copyright © Peter Cappello 2011 *10 Give a formula for the coefficient of xk in the expansion of ( x + 1/x )100, where k is an even integer. Copyright © Peter Cappello 2011

Copyright © Peter Cappello 2011 *10 Solution By the Binomial Theorem, (x + 1/x)100 = Σj=0 to 100 C(100, j)x100-j(1/x)j = Σj=0 to 100 C(100, j)x100-2j. We want the coefficient of x100-2j, where k = 100 – 2j  j = (100 – k)/2. The coefficient we seek is C(100, (100 – k)/2 ). Copyright © Peter Cappello 2011

Copyright © Peter Cappello 2011 Suppose that k & n are integers with 1  k < n. Prove the hexagon identity C(n - 1, k –1)C(n, k + 1)C(n + 1, k) = C(n-1, k)C(n, k-1)C(n+1, k+1), which relates terms in Pascal’s triangle that form a hexagon. Hint: Use straight algebra. Copyright © Peter Cappello 2011

Copyright © Peter Cappello 2011 20 Solution C( n – 1, k –1 )C( n, k + 1 )C( n + 1, k ) = (n – 1)! n! (n+1)! _______________________________________ (k – 1)!(n – k)! (k + 1)!(n – k – 1)! k! (n + 1 – k)! = C( n – 1, k )C( n, k – 1 )C( n + 1, k + 1 ). Copyright © Peter Cappello 2011