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Copyright © Peter Cappello 2
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 3
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 4
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 5
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 6
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 7
Exercise *30
Give a combinatorial argument to prove that
Σk=1,n kC( n, k )2 = nC( 2n – 1, n – 1 ).
Copyright © Peter Cappello 8
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 9
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 10
Combinatorial Identities
• Manipulation of the Binomial Theorem
• “Committee” arguments
• Block walking arguments – for identities involving sums
Copyright © Peter Cappello 11
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.
Prove that
n2n – 1 = 1C( n, 1 ) + 2C( n, 2 ) + 3C( n, 3 ) + . . . + nC( n, n ).
Copyright © Peter Cappello 12
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.
– 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 13
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 14
Pascal’s Trianglekth number in row n is nCk:
1
1 1
1 2 1
1 3 3 1
n = 4
n = 3
n = 2
n = 1
n = 0
1 4 6 4 1
k = 0
k = 1
k = 2
k = 3
k = 4
Copyright © Peter Cappello 15
Displaying Pascal’s Identity
0C0
n = 4
n = 3
n = 2
n = 1
n = 0
k = 0
k = 1
k = 2
k = 3
k = 4
1C0 1C1
2C0 2C1 2C2
3C0 3C1 3C2 3C3
4C0 4C1 4C2 4C3 4C4
Copyright © Peter Cappello 16
Block-Walking Interpretation
0C0
n = 4
n = 3
n = 2
n = 1
n = 0
k = 0
k = 1
k = 2
k = 3
k = 4
1C0 1C1
2C0 2C1 2C2
3C0 3C1 3C2 3C3
4C0 4C1 4C2 4C3 4C4
nCk = # ways toget to corner n, kstarting from 0, 0
nCk = # strings of n Ls & Rs with k Rs.
Copyright © Peter Cappello 17
Pascal’s Identity via Block-Walking
0C0
n = 4
n = 3
n = 2
n = 1
n = 0
k = 0
k = 1
k = 2
k = 3
k = 4
1C0 1C1
2C0 2C1 2C2
3C0 3C1 3C2 3C3
4C0 4C1 4C2 4C3 4C4
# routes to corner n, k = # routes thru corner n-1, k + # routes thru corner n-1, k-1
Copyright © Peter Cappello 18
nC02 + nC1
2 + nC22 + … + nCn
2 = 2nCn
0C0
n = 4
n = 3
n = 2
n = 1
n = 0
k = 0
k = 1
k = 2
k = 3
k = 4
1C0 1C1
2C0 2C1 2C2
3C0 3C1 3C2 3C3
4C0 4C1 4C2 4C3 4C4
Copyright © Peter Cappello 19
nC02 + nC1
2 + nC22 + … + nCn
2 = 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 20
Give a Committee Argument
nC02 + nC1
2 + nC22 + … + nCn
2 = 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 2011 22
*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 23
*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 24
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.