1. Michael Duquette & Whitney Sherman
Tucker, Applied Combinatorics, Section 5.5 Group G
Binomial Identities
Michael Duquette & Whitney Sherman
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Tucker, Applied Combinatorics Section 4.2a

2. 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 ?
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Tucker, Applied Combinatorics Section 4.2a

3. 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 k x’s and (n-k) a’s so
i.e.
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4. Tucker, Applied Combinatorics Section 4.2a
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If we set a=1 we end up with: The Binomial Theorem
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Tucker, Applied Combinatorics Section 4.2a

5. Tucker, Applied Combinatorics Section 4.2a
Symmetry Identity
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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.
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Tucker, Applied Combinatorics Section 4.2a

6. 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
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Tucker, Applied Combinatorics Section 4.2a

7. 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:
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Tucker, Applied Combinatorics Section 4.2a

8. 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
n=2 1 2 1
n=3 1 3 3 1
K=4
n=4 1 4 6 4 1
Table of binomial coefficients: kth number in row n is
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Tucker, Applied Combinatorics Section 4.2a

9. Pascal’s Triangle and Interpretation
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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
(0,0)
(1,0)
(2,0)
(3,1)
(4,2)
(5,3)
(6,3)
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10. Tucker, Applied Combinatorics Section 4.2a
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
(0,0)
(1,0)
(2,0)
(3,1)
(4,2)
(5,3)
(6,3)
(5,2)
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11. List of Identities Found on Pg 217
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Here if
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12. 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:
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13. 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.
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14. Tucker, Applied Combinatorics Section 4.2a
Example 2:
Page 222 # 14 (b)
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Binomial coefficient:
First Derivative:
Second Derivative:
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15. 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
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16. 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
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17. 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:
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Tucker, Applied Combinatorics Section 4.2a

18. Tucker, Applied Combinatorics Section 4.2a
Problem for Class to Try:
Page 222 # 14 (a)
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Solution:
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Tucker, Applied Combinatorics Section 4.2a