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Binomial Identities

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Tucker, Applied Combinatorics, Section 5.5 Group G

Binomial Identities

Michael Duquette & Whitney Sherman

Tucker, Applied Combinatorics Section 4.2a

- Consider the polynomial:
- Which can be expanded to:
- Expanding again gives:
- This equation can be written as:
- There are 2 choices for each term in this example and 3 terms So formal products.
- So in there would be formal products.

The question now becomes… How many formal products in the expansion of contain and ?

Tucker, Applied Combinatorics Section 4.2a

- This question is the same as asking for the coefficient
- of in .
- How many three letter sequences with and are there?
- The answer is (3 spaces, choose k of them to be x’s) and so the can be written
- We see that the coefficient of in will be equal to the number of n-letter sequences formed by kx’s and (n-k) a’s so
- i.e.

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Tucker, Applied Combinatorics Section 4.2a

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- If we set a=1 we end up with: The Binomial Theorem

Tucker, Applied Combinatorics Section 4.2a

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- Says that the number of ways to select a subset of k objects out of n objects is equal to the number of ways to select n-k of the objects to set aside.

Tucker, Applied Combinatorics Section 4.2a

Another Fundamental Identity

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- PROOF:
- There are committees formed out of k people chosen from a set of n people.
- They are put into two categories, depending on whether or not the committee contains a person p.
- If p is not part of the committee, there are ways to form the committee from the other n-1 people.
- On the other hand if p is on the committee, the problem reduces to choosing the k-1 remaining members of the committee from the other n-1 people.
- This can be done ways. Thus

Tucker, Applied Combinatorics Section 4.2a

Example 1

Show that:

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- The left hand side of the equation counts the ways to select a group of k people chosen from a set of n people, and then to select a subset of m leaders within the group of k people.
- The right hand side first selects a subset of m leaders from n people, and then selects the remaining k-m members of the group from the remaining n-m people.
- Note that when m=1:

Tucker, Applied Combinatorics Section 4.2a

Pascal’s Triangle

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- Using and for all
- nonnegative n, we can recursively build successive rows in the table of binomial coefficients: Pascal’s triangle.
- Each number in the table, except for the last numbers in a row, is the sum of the two neighboring numbers in the preceding row.

K=0

K=1

n=0

1

K=2

n=1

1

1

K=3

1

2

n=2

1

n=3

1

3

3

K=4

1

n=4

1

4

6

4

1

Table of binomial coefficients: kth number in row n is

Tucker, Applied Combinatorics Section 4.2a

(0,0)

(1,0)

(2,0)

(3,1)

(4,2)

(5,3)

(6,3)

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- Look at the map of the streets and label the start (0,0).
- Label each additional vertex with a where n indicates the number of blocks traversed from (0,0) and k is the number of times the person chose the ‘right’ branch.
- Any route to corner can be written as a list of the branches (left or right) chosen at the successive corners on the path from (0,0) to .
- This list is just a sequence of k Rights and (n-k) Lefts.
- Our route from (0,0) shows the sequence LLRRRL.
- Let be the number of possible routes from the start to corner .
- This is the number of sequences of k R’s and (n-k) L’s and so

Tucker, Applied Combinatorics Section 4.2a

(1,0)

(2,0)

(0,0)

(3,1)

(4,2)

(5,3)

(6,3)

(5,2)

Proof #2 of

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- Use our ‘block-walking’ method to prove this by example:
- To get to corner (6,3), the person needs to walk left from either (5,3) or walk right from (5,2)
- Thus showing that

Tucker, Applied Combinatorics Section 4.2a

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Here if

List of Identities Found on Pg 217

Tucker, Applied Combinatorics Section 4.2a

Example 2:

Page 222 # 14 (b)

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Question: By setting x equal to the appropriate values in the binomial expansion (or one of its derivatives, etc. ) evaluate:

Solution:

Tucker, Applied Combinatorics Section 4.2a

Example 2:

Page 222 # 14 (b)

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Compare this summation with the binomial coefficient.

Binomial coefficient:

Notice that our summation starts with the third term, therefore we need to take the second derivative of the binomial coefficient.

Tucker, Applied Combinatorics Section 4.2a

Example 2:

Page 222 # 14 (b)

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Binomial coefficient:

First Derivative:

Second Derivative:

Tucker, Applied Combinatorics Section 4.2a

Example 2:

Page 222 # 14 (b)

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Compare the Second Derivative to our summation:

So

and

Tucker, Applied Combinatorics Section 4.2a

Example 2:

Page 222 # 14 (b)

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To Check to see whether our equation works, plug in a specific n.

Let n= 3

Tucker, Applied Combinatorics Section 4.2a

Problem for Class to Try:

Page 222 # 14 (a)

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Question: By setting x equal to the appropriate values in the binomial expansion (or one of its derivatives, etc. ) evaluate:

Binomial coefficient:

Tucker, Applied Combinatorics Section 4.2a

Problem for Class to Try:

Page 222 # 14 (a)

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Solution:

Tucker, Applied Combinatorics Section 4.2a