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Binomial Coefficients: Selected Exercises 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 2 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 3 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 4 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 5 Let n & k be positive integers, with n > k. C( n, k ) = C( n - 1, k – 1 ) + C( n - 1, k ). 1. The left hand side (LHS) counts the number of subsets of size k from a set of n elements. 2. The RHS counts these same subsets using the sum rule: Partition the subsets into 2 parts: 1. 2. Subsets of k elements that include element 1: 1. Pick element 1: 1 2. 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 6 Exercise *30 Give a combinatorial argument to prove that Σk=1,n kC( n, k )2 = nC( 2n – 1, n – 1 ). Copyright © Peter Cappello 7 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 8 Exercise *30 Solution Σk=1,n kC( n, k )2 = nC( 2n – 1, n – 1 ) The RHS counts the # of such committees: 1. Pick the chair from the n math professors: n 2. 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, 1. Pick the k math professor members: C( n, k ) 2. Pick the committee chair: k 3. Pick the n - k CS professor members: C( n, n – k ) = C( n, k ) Copyright © Peter Cappello 9 Combinatorial Identities • Manipulation of the Binomial Theorem • “Committee” arguments • Block walking arguments – for identities involving sums Copyright © Peter Cappello 10 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, 1. Manipulate the binomial theorem algebraically; 2. Evaluate the resulting equation for values of x & y, producing the desired result. Prove that n2n – 1 = 1C( n, 1 ) + 2C( n, 2 ) + 3C( n, 3 ) + . . . + nC( n, n ). Copyright © Peter Cappello 11 Committee Arguments Show that 1. n2n – 1 = 1C( n, 1 ) + 2C( n, 2 ) + 3C( n, 3 ) + . . . + nC( n, n ). Hint: committees of any size, 1 of whom is chair. 2. C( n, k )C( k, m ) = C( n, m )C( n – m, k – m ). Hint: committees of k people, m of whom are leaders. 3. 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 12 Block-Walking Arguments 1. Draw Pascal’s triangle. 2. Interpret a node in the triangle as the # of ways to walk from the apex to the node, always going down. Show that 1. C( n, k ) = C( n – 1, k ) + C( n – 1, k – 1 ) 2. C( n, 0 )2 + C( n,1 )2 + . . . + C( n, n )2 = C( 2n, n ). Copyright © Peter Cappello 13 Pascal’s Triangle kth number in row n is nCk: k=0 1 n=0 n=1 1 n=2 1 1 n=3 n=4 k=1 1 2 3 4 k=2 1 k=3 1 3 6 Copyright © Peter Cappello k=4 1 4 1 14 Displaying Pascal’s Identity k=0 0C0 n=0 1C0 n=1 2C0 n=2 n=3 n=4 3C0 4C0 k=1 1C1 2C1 3C1 4C1 4C2 k=2 2C2 3C2 4C3 Copyright © Peter Cappello k=3 3C3 k=4 4C4 15 Block-Walking Interpretation nCk = # strings of n Ls & Rs with k Rs. 0C0 1C0 n=1 2C0 n=2 n=4 = # ways to get to corner n, k starting from 0, 0 k=0 n=0 n=3 nCk 3C0 4C0 k=1 1C1 2C1 3C1 4C1 4C2 k=2 2C2 3C2 4C3 Copyright © Peter Cappello k=3 3C3 k=4 4C4 16 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 1C0 n=1 2C0 n=2 n=3 n=4 1C1 2C1 3C0 4C0 k=1 3C1 4C1 4C2 k=2 2C2 3C2 k=3 3C3 4C3 Copyright © Peter Cappello k=4 4C4 17 2+ C 2+ C 2+ … + C 2 = C n 0 n 1 n 2 n n 2nCn k=0 0C0 n=0 1C0 n=1 2C0 n=2 n=3 n=4 3C0 4C0 k=1 1C1 2C1 3C1 4C1 4C2 k=2 2C2 3C2 4C3 Copyright © Peter Cappello k=3 3C3 k=4 4C4 18 2+ C 2+ C 2+ … + C 2 = C n 0 n 1 n 2 n n 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 19 Give a Committee Argument nC0 2+ nC1 2+ nC2 2+ … + 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? Copyright © Peter Cappello 20 End Copyright © Peter Cappello 21 *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 22 *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 23 20 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 24 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 25