Binomial Coefficients: Selected Exercises. Copyright © Peter Cappello2 Preliminaries What is the coefficient of x 2 y in ( x + y ) 3 ? ( x + y ) 3 = (

Binomial Coefficients: Selected Exercises

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 ).


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?


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.


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.


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 ).


Exercise *30

Give a combinatorial argument to prove that

Σk=1,n kC( n, k )2 = nC( 2n – 1, n – 1 ).


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.


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 )


Combinatorial Identities

• Manipulation of the Binomial Theorem

• “Committee” arguments

• Block walking arguments – for identities involving sums


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 ).


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.


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 ).


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


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


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.


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


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


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



• The identity generalizes this argument:

– # routes to 2n, n that go thru n,k = nCk nCn-k

– Sum over k = 0 to n


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?


End

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.


*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 ).


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.


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 ).

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Binomial Coefficients: Selected Exercises. Copyright © Peter Cappello2 Preliminaries What is the coefficient of x 2 y in ( x + y ) 3 ? ( x + y ) 3 = (