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Tucker, Applied Combinatorics, Section 5.5 Group G. Binomial Identities. Michael Duquette & Whitney Sherman. Consider the polynomial: Which can be expanded to: Expanding again gives: This equation can be written as:
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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. _________________________________________________ Tucker, Applied Combinatorics Section 4.2a
___________________________________________ • If we set a=1 we end up with: The Binomial Theorem Tucker, Applied Combinatorics Section 4.2a
Symmetry Identity ___________________________________________ • 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 _________________________________________________ • 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: _________________________________________________ • 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 _________________________________________________ • 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) _________________________________________________ Pascal’s Triangle and Interpretation • 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 _________________________________________________ • 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
_________________________________________________ Here if List of Identities Found on Pg 217 Tucker, Applied Combinatorics Section 4.2a
Example 2: Page 222 # 14 (b) _________________________________________________ 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) _________________________________________________ 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) _________________________________________________ Binomial coefficient: First Derivative: Second Derivative: Tucker, Applied Combinatorics Section 4.2a
Example 2: Page 222 # 14 (b) _________________________________________________ Compare the Second Derivative to our summation: So and Tucker, Applied Combinatorics Section 4.2a
Example 2: Page 222 # 14 (b) _________________________________________________ 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) _________________________________________________ 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) _________________________________________________ Solution: Tucker, Applied Combinatorics Section 4.2a