1. Binomial Coefficients: Selected Exercises

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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 ).
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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?
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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.
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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.
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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 ).
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Exercise *30
Give a combinatorial argument to prove that
Σk=1,n kC( n, k )2 = nC( 2n – 1, n – 1 ).
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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.
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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 )
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10. Combinatorial Identities
Manipulation of the Binomial Theorem
“Committee” arguments
Block walking arguments – for identities involving sums
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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.
n2n – 1 = 1C( n, 1 ) + 2C( n, 2 ) + 3C( n, 3 ) nC( n, n ).
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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.
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13. 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 ) C( n, n )2 = C( 2n, n ).
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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
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15. 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
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16. 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
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17. 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
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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
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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
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20. 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)
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End
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*10
Give a formula for the coefficient of xk in the expansion of ( x + 1/x )100, where k is an even integer.
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*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 ).
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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.
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20 Solution
C( n – 1, k –1 )C( n, k + 1 )C( n + 1, k ) =
(n – 1)! n! (n+1)!
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(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 ).
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