(n – 1)! - UCSB Computer Science

Transcript (n – 1)! - UCSB Computer Science

Binomial Coefficients: Selected Exercises

Preliminaries

What is the coefficient of x 2 y 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 = x 3 + 3x 2 y + 3xy 2 + y 3 .

The answer thus is 3 .

There are 2 3 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 y s?

• This is the

coefficient

of x n 3 y 3 in ( x + y ) n .

• How many formal terms have exactly

y s?

The Binomial Theorem

Let

be variables, and



Partition the set of 2 n terms of the formal expansion of ( x + y ) n into n + 1 classes according to the # of y s in the term: (

)

= Σ j=0 to n C(

)

x n-j y j

= C(

, 0 )

x n y 0

+ C(

,1 )

x n-1 y 1

…

+ C(

)

x n-j y j

…

+ C(

)

x 0 y n

Pascal

’

s Identity

Let

be positive integers, with

n > k

. Give a

combinatorial argument

to show that C(

) = C(

n - 1

– 1 ) + C(

n - 1

combinatorial argument

proves that the equation ’ s LHS & RHS are different ways to count the elements of the same set.

Let

be positive integers, with

n > k

. C(

) = C(

n - 1

– 1 ) + C(

n - 1

The left hand side (LHS) counts the number of subsets of size

from a set of

elements.

The RHS counts these same subsets using the sum rule :

Partition

the subsets into 2 parts: Subsets of

elements that

include

element 1 : 2.

Pick element 1 : 1 Pick the remaining

k – 1

subset elements from the remaining

n - 1

set elements: C(

n - 1

– 1 ).

Subsets of

elements that

exclude

element 1 : Pick the

elements from the n - 1 remaining elements: C(

n - 1

Exercise *30

Give a

combinatorial argument

to prove that Σ

k=1,n k

) 2 =

2n – 1

n – 1

Give a

combinatorial argument

that Σ

k=1,n k

) 2 =

2n – 1

n – 1

The set of committees with

members from a group of

math professors &

computer science professors, such that the committee chair is a mathematics professor.

Exercise *30 Solution

k=1,n k

) 2 =

2n – 1

n – 1

) The RHS counts the # of such committees: 1.

Pick the chair from the

math professors:

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

, the # of math professors on the committee.

For each

, 1. Pick the

math professor members: C(

) 2. Pick the committee chair:

3. Pick the

n - k

CS professor members: C(

n, n – k

) = C(

n, k

Combinatorial Identities

Manipulation of the Binomial Theorem

(

)

= Σ j=0 to n C(

)

x n-j y j

= C(

, 0 )

x n y 0

+ C(

,1 )

x n-1 y 1

…

+ C(

)

x n-j y j

…

+ C(

)

x 0 y n

Prove that C( n, 0 ) + C( n, 1 ) +

. . .

+ C( n, n ) = 2 n .

In general, 1.

Manipulate Evaluate the binomial theorem algebraically; the resulting equation for values of x & y , producing the desired result.

Prove that n2 n – 1 = 1C( n, 1 ) + 2C( n, 2 ) + 3C( n, 3 ) +

. . .

+ nC( n, n ) .

Committee Arguments

Show that 1.

n2 n – 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.

Σ k = 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

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 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 Triangle

Displaying Pascal

’

s Identity

Block-Walking Interpretation

C

= # strings of Ls & Rs with

k n

Rs.

n = 0 0 C 0 k = 0 k = 1 n

C

= # ways to get to corner

,

starting from 0, 0

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 n = 0 0 C 0 k = 1 n = 1 1 C 0 1 C 1 k = 2 n = 2 n = 3 3 C 0 2 C 0 3 C 1 2 C 1 3 C 2 2 C 2 3 C 3 k = 3 k = 4 n = 4 4 C 0 4 C 1 4 C 2 4 C 3 Copyright © Peter Cappello 4 C 4 17

C

0 2

+

C

1 2

+

C

2 2

+ … +

C

n 2

=

C

n n = 0 n = 1 n = 2 1 C 0 0 C 0 k = 0 k = 1 1 C 1 k = 2 3 C 0

2 C 0

3 C 1

2 C 1

3 C 2

2 C 2

3 C 3 k = 3 k = 4 n = 3 n = 4 4 C 0 4 C 1

4 C 2

C

0 2

+

C

1 2

+

C

2 2

+ … +

C

n 2

=

C

n • RHS = all routes to corner 4,2 • LHS partitions routes to 4,2 into those that: – go thru corner 2,0 : 2 C 0  2 C 2 – go thru corner 2,1 : 2 C 1  2 C 1 – go thru corner 2,2 : 2 C 2  2 C 0 • The identity generalizes this argument: – # routes to 2n, n that go thru n,k = n C k  n C n-k – Sum over k = 0 to n Copyright © Peter Cappello 19

Give a Committee Argument

n C 0 2 + n C 1 2 + n C 2 2 + … + n C n 2 = 2n C n 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 = ( Σ j=0 to n C(

)

x n-j y j

) ( Σ j=0 to n C(

)

x n-j y j

) Evaluate this identity at x = 1 : (1 + y) 2n = (1 + y) n (1 + y) n = ( Σ j=0 to n C(

)

y j

) ( Σ j=0 to n C(

)

y j

) What is the coefficient of

y n

in the above polynomial product?

End

*10

Give a formula for the coefficient of

x k

in the expansion of

( x + 1/x ) 100

, where

is an

even

integer.

*10 Solution

• By the Binomial Theorem,

(x + 1/x) 100 =

Σ j=0 to 100 C(

100

)

x 100-j (1/x) j =

Σ j=0 to 100 C(

100

)

x 100-2j .

• We want the

coefficient

x 100-2j

, where k = 100 – 2j  j = (100 – k)/2 .

• The coefficient we seek is C(100, (100 – k)/2 ).

20

Suppose that

are integers with



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

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

(n – 1)! - UCSB Computer Science

Transcript (n – 1)! - UCSB Computer Science

Binomial Coefficients: Selected Exercises

Preliminaries

Preliminaries

The Binomial Theorem

Pascal

s Identity

Exercise *30

Exercise *30 Solution

Combinatorial Identities

Manipulation of the Binomial Theorem

Committee Arguments

Block-Walking Arguments

Pascal

s Triangle

Displaying Pascal

s Identity

Block-Walking Interpretation

C

= # strings of Ls & Rs with

Rs.

C

= # ways to get to corner

,

starting from 0, 0

Pascal

s Identity via Block-Walking

C

+

C

+

C

+ … +

C

=

C

C

+

C

+

C

+ … +

C

=

C

Give a Committee Argument

End

*10

*10 Solution

20

20 Solution

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