jalalam.ppt

1
Binomial Coefficients
CS/APMA 202
Rosen section 4.4
Aaron Bloomfield

2
Binomial Coefficients
It allows us to do a quick expansion of
(x+y)n
Why it’s really important:
It provides a good context to present
proofs
 Especially combinatorial proofs

3
Let n and r be non-negative integers with
r ≤ n. Then C(n,r) = C(n,n-r)
Or,
Proof (from last slide set):
 !
)
(
)!
(
!
)
,
(
r
n
n
r
n
n
r
n
n
C





)!
(
!
!
r
n
r
n


)!
(
!
!
)
,
(
r
n
r
n
r
n
C


Review: corollary 1
from section 4.3


















r
n
n
r
n

4
Review: combinatorial proof
A combinatorial proof is a proof that uses
counting arguments to prove a theorem,
rather than some other method such as
algebraic techniques
Essentially, show that both sides of the
proof manage to count the same objects
 Usually in the form of an English explanation
with supporting formulae

5
Polynomial expansion
Consider (x+y)3:
Rephrase it as:
When choosing x twice and y once, there are
C(3,2) = C(3,1) = 3 ways to choose where the x
comes from
When choosing x once and y twice, there are
C(3,2) = C(3,1) = 3 ways to choose where the y
comes from
3
2
2
3
3
3
3
)
( y
xy
y
x
x
y
x 




    3
2
2
2
2
2
2
3
)
)(
)(
( y
xy
xy
xy
y
x
y
x
y
x
x
y
x
y
x
y
x 











6
Consider
To obtain the x5 term
 Each time you multiple by (x+y), you select the x
 Thus, of the 5 choices, you choose x 5 times
C(5,5) = 1
 Alternatively, you choose y 0 times
C(5,0) = 1
To obtain the x4y term
 Four of the times you multiply by (x+y), you select the x
The other time you select the y
 Thus, of the 5 choices, you choose x 4 times
C(5,4) = 5
 Alternatively, you choose y 1 time
C(5,1) = 5
To obtain the x3y2 term
 C(5,3) = C(5,2) = 10
Etc…
5
4
3
2
2
3
4
5
5
5
10
10
5
)
( y
xy
y
x
y
x
y
x
x
y
x 







7
For (x+y)5
5
4
3
2
2
3
4
5
5
0
5
1
5
2
5
3
5
4
5
5
5
)
( y
xy
y
x
y
x
y
x
x
y
x 






















































5
4
3
2
2
3
4
5
5
5
10
10
5
)
( y
xy
y
x
y
x
y
x
x
y
x 







8
Polynomial expansion:
The binomial theorem
For (x+y)n
The book calls this Theorem 1
n
n
n
n
n
y
x
n
y
x
n
y
x
n
n
y
x
n
n
y
x 0
1
1
1
1
0
0
1
1
)
( 








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

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




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
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
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




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



 














n
j
j
j
n
y
x
j
n
0
n
n
n
n
y
x
n
n
y
x
n
n
y
x
n
y
x
n 0
1
1
1
1
0
1
1
0 












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
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







 



9
Examples
What is the coefficient of x12y13 in (x+y)25?
What is the coefficient of x12y13 in (2x-3y)25?
 Rephrase it as (2x+(-3y))25
 The coefficient occurs when j=13:
300
,
200
,
5
!
12
!
13
!
25
12
25
13
25



















  














25
0
25
25
)
3
(
)
2
(
25
)
3
(
2
j
j
j
y
x
j
y
x
00
,545,702,4
33,959,763
)
3
(
2
!
12
!
13
!
25
)
3
(
2
13
25 13
12
13
12














10
Rosen, section 4.4, question 4
Find the coefficient of x5y8 in (x+y)13
Answer:
1287
8
13
5
13



















11
Pascal’s triangle
0
1
2
3
4
5
6
7
8
n =

12
Pascal’s Identity
By Pascal’s identity: or 21=15+6
Let n and k be positive integers with n ≥ k.
Then
 or C(n+1,k) = C(n,k-1) + C(n,k)
We will prove this via two ways:
 Combinatorial proof
 Using the formula for


























 
k
n
k
n
k
n
1
1









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

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5
6
4
6
5
7








k
n

13
Combinatorial proof of Pascal’s
identity
Prove C(n+1,k) = C(n,k-1) + C(n,k)
Consider a set T of n+1 elements
 We want to choose a subset of k elements
 We will count the number of subsets of k elements via 2 methods
Method 1: There are C(n+1,k) ways to choose such a subset
Method 2: Let a be an element of set T
Two cases
 a is in such a subset
There are C(n,k-1) ways to choose such a subset
 a is not in such a subset
There are C(n,k) ways to choose such a subset
Thus, there are C(n,k-1) + C(n,k) ways to choose a subset of k
elements
Therefore, C(n+1,k) = C(n,k-1) + C(n,k)

14
Rosen, section 4.4, question 19:
algebraic proof of Pascal’s identity












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



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k
n
k
n
k
n
1
1
)!
)(
1
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)
1
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)
1
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1
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(
*
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1
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1
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k
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k
n
k
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n
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k
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1
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1
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1
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1
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1
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k
n
k
n
k
n
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1
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)(
1
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1
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n
Substitutions:

15
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16
Pascal’s triangle
0
1
2
3
4
5
6
7
8
n = 1
2
4
8
16
32
64
128
256
sum = = 2n

17
Proof practice: corollary 1
Let n be a non-negative integer. Then
Algebraic proof











n
k
n
k
n
0
2












n
k
k
n
k
k
n
0
1
1
n
n
)
1
1
(
2 












n
k k
n
0













n
j
j
j
n
n
y
x
j
n
y
x
0
)
(

18
Combinatorial proof
 A set with n elements has 2n subsets
By definition of power set
 Each subset has either 0 or 1 or 2 or … or n elements
There are subsets with 0 elements, subsets with 1
element, … and subsets with n elements
Thus, the total number of subsets is
 Thus,











n
k
n
k
n
0
2
n
n
k k
n
2
0



















0
n










n
k k
n
0








1
n








n
n

20
Pascal’s triangle
0
1
2
3
4
5
6
7
8
n =

21
Let n be a positive integer. Then
Algebraic proof
This implies that
n
0
0 












n
k
k
k
n
0
0
)
1
(

 






















































5
3
1
4
2
0
n
n
n
n
n
n













n
k
k
n
k
k
n
0
1
)
1
(
 n
1
)
1
( 


k
n
k k
n
)
1
(
0









 


22
Algebraic proof












n
k
k
k
n
k
n
0
2
1











n
k
n
k
k
n
0
3
2
n
n
)
2
1
(
3 












n
k
k
k
n
0
2

23
Vandermonde’s identity
Let m, n, and r be non-negative integers
with r not exceeding either m or n. Then



























  r
k k
n
k
r
m
r
n
m
0

24
Combinatorial proof of
Vandermonde’s identity
Consider two sets, one with m items and one with n
items
 Then there are ways to choose r items from the union of
those two sets
Next, we’ll find that value via a different means
 Pick k elements from the set with n elements
 Pick the remaining r-k elements from the set with m elements
 Via the product rule, there are ways to do that for EACH
value of k
 Lastly, consider this for all values of k:
Thus, 


























  r
k k
n
k
r
m
r
n
m
0







 
r
n
m
















 k
n
k
r
m



















r
k k
n
k
r
m
0

25
Review of Rosen, section
4.3, question 11 (a)
How many bit strings of length 10 contain
exactly four 1’s?
 Find the positions of the four 1’s
 The order of those positions does not matter
Positions 2, 3, 5, 7 is the same as positions 7, 5, 3, 2
 Thus, the answer is C(10,4) = 210
Generalization of this result:
 There are C(n,r) possibilities of bit strings of length n
containing r ones

26
Yet another combinatorial proof
Let n and r be non-negative integers with r ≤ n.
Then
 The book calls this Theorem 4
We will do the combinatorial proof by showing
that both sides show the ways to count bit
strings of length n+1 with r+1 ones
From previous slide: achieves this




















 n
r
j r
j
r
n
1
1










1
1
r
n

27
Yet another combinatorial proof
Next, show the right side counts the same objects
The final one must occur at position r+1 or r+2 or … or
n+1
Assume that it occurs at the kth bit, where r+1 ≤ k ≤ n+1
 Thus, there must be r ones in the first k-1 positions
 Thus, there are such strings of length k-1
As k can be any value from r+1 to n+1, the total number
of possibilities is
Thus,







 
r
k 1











 
1
1
1
n
r
k r
k











n
r
k r
k











n
r
j r
j




















 n
r
j r
j
r
n
1
1

28
Show that if p is a prime and k is an integer such that
1 ≤ k ≤ p-1, then p divides
We know that
p divides the numerator (p!) once only
 Because p is prime, it does not have any factors less than p
We need to show that it does NOT divide the
denominator
 Otherwise the p factor would cancel out
Since k < p (it was given that k ≤ p-1), p cannot divide k!
Since k ≥ 1, we know that p-k < p, and thus p cannot
divide (p-k)!
Thus, p divides the numerator but not the denominator
Thus, p divides








k
p
)!
(
!
!
k
p
k
p
k
p


















k
p

29
Give a combinatorial proof that if n is positive
integer then
Provided hint: show that both sides count the
ways to select a subset of a set of n elements
together with two not necessarily distinct
elements from the subset
Following the other provided hint, we express
the right side as follows:
2
0
2
2
)
1
( 











 n
n
k
n
n
k
n
k
1
2
0
2
2
2
)
1
( 













 n
n
n
k
n
n
n
k
n
k

30
Show the left side properly counts the
desired property












n
k k
n
k
0
2
Choosing a subset of k
elements from a set of
n elements
Consider each
of the possible
subset sizes k
Choosing one of
the k elements in
the subset twice

31
Two cases to show the right side: n(n-1)2n-2+n2n-1
 Pick the same element from the subset
Pick that one element from the set of n elements: total of n possibilities
Pick the rest of the subset
 As there are n-1 elements left, there are a total of 2n-1 possibilities to pick a given
subset
We have to do both
 Thus, by the product rule, the total possibilities is the product of the two
 Thus, the total possibilities is n*2n-1
 Pick different elements from the subset
Pick the first element from the set of n elements: total of n possibilities
Pick the next element from the set of n-1 elements: total of n-1 possibilities
Pick the rest of the subset
 As there are n-2 elements left, there are a total of 2n-2 possibilities to pick a given
subset
We have to do all three
 Thus, by the product rule, the total possibilities is the product of the three
 Thus, the total possibilities is n*(n-1)*2n-2
 We do one or the other
Thus, via the sum rule, the total possibilities is the sum of the two
Or n*2n-1+n*(n-1)*2n-2

32
Quick survey
 I felt I understood the material in this slide set…
a) Very well
b) With some review, I’ll be good
c) Not really
d) Not at all

33
Quick survey
 The pace of the lecture for this slide set was…
a) Fast
b) About right
c) A little slow
d) Too slow

34
Quick survey
 How interesting was the material in this slide
set? Be honest!
a) Wow! That was SOOOOOO cool!
b) Somewhat interesting
c) Rather borting
d) Zzzzzzzzzzz

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