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Tucker, Applied Combinatorics, Section 5.5 Group G Binomial Identities Michael Duquette & Whitney Sherman 5/24/2017 Tucker, Applied Combinatorics Section 4.2a 1 •Consider the polynomial: •Which can be expanded to: ( a x )3 (a x)(a x)(a x) •Expanding again gives: aaa aax axa axx xaa xax xxa xxx •This equation can be written as: a 3 3a 2 x 3ax 2 x 3 There are 2 choices for each term in this example and 3 terms So 2 2 2 8 formal products. 10 So in (a x)10 there would be 2 1024 formal products. The question now becomes… How many formal products in 3 the expansion of (a x) contain k x ' s and (3 k ) a ' s ? 5/24/2017 Tucker, Applied Combinatorics Section 4.2a 2 This question is the same as asking for the coefficient of a3 (3 k ) x k in a 3 3a 2 x 3ax 2 x3. How many three letter sequences with k x ' s and (3 k ) a ' s are there? _________________________________________________ • The answer is C (3, k ) (3 spaces, choose k of them to be x’s) and so the (a x)3 can be written 3 3 3 3 0 3 a 1 2 a x 2 2 ax 3 x 3 • We see that the coefficient of a n k x k in (a x)n will be equal to the number of n-letter sequences formed by k x’s and (n-k) a’s so C (n, k ) i.e. 5/24/2017 n n k k (a x) a x k n Tucker, Applied Combinatorics Section 4.2a 3 n n k k (a x) a x k ___________________________________________ n • If we set a=1 we end up with: The Binomial Theorem n (1 x) 0 n 5/24/2017 n 1 n x 2 2 n x ... n Tucker, Applied Combinatorics Section 4.2a n x 4 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. n k 5/24/2017 n n! k ! n k ! n k Tucker, Applied Combinatorics Section 4.2a 5 Another Fundamental Identity _________________________________________________ n k n 1 n 1 k k 1 PROOF: • There are C (n, k )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 C (n 1, k )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 C (n 1, k 1) ways. Thus C (n, k ) C (n 1, k ) C (n 1, k 1) 5/24/2017 Tucker, Applied Combinatorics Section 4.2a 6 Example 1 _________________________________________________ Show that: n k k m n m n m k m •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: 5/24/2017 n k k n 1 n n or k 1 k Tucker, Applied Combinatorics Section 4.2a n n 1 k k 1 7 Pascal’s Triangle _________________________________________________ n 1 n 1 k k 1 •Using n k and C (n, 0) C (n, n) 1 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 n=4 2 1 n=3 1 1 3 4 1 3 6 K=4 1 4 1 Table of binomial coefficients: kth number in row n is 5/24/2017 Tucker, Applied Combinatorics Section 4.2a n k 8 Pascal’s Triangle and Interpretation _________________________________________________ •Look at the map of the streets and label the start (0,0). •Label each additional vertex with a ( n, k ) where n indicates the number of blocks traversed from (0,0) and k is the number of times the person chose the ‘right’ branch. (0,0) (1,0) •Any route to corner ( n, k )can be written as a list of the branches (left or right) chosen at the successive corners on the path from (0,0) to ( n, k ) . (2,0) (3,1) (4,2) (5,3) (6,3) •This list is just a sequence of k Rights and (n-k) Lefts. •Our route from (0,0) shows the sequence LLRRRL. •Let s (n, k ) be the number of possible routes from the start to corner ( n, k ) . •This is the number of sequences of k R’s and (n-k) L’s and so s (n, k ) C (n, k ) 5/24/2017 Tucker, Applied Combinatorics Section 4.2a 9 n n 1 n 1 k k k 1 _________________________________________________ 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 s(n, k ) s(n 1, k ) s(n 1, k 1) (0,0) s (6,3) s(5,3) s(5, 2) (1,0) (2,0) (3,1) (4,2) (5,2) (5,3) (6,3) 5/24/2017 Tucker, Applied Combinatorics Section 4.2a 10 List of Identities Found on Pg 217 _________________________________________________ Here C (n, r ) 0 if 0 n n 0 r n 0 n ... n r 1 r 2 n ... r r r 2 n 1 5/24/2017 n 2 n 2 n 1 n 2 n r n r 1 ... 1 2 r r n 0 r r n 1 r k 0 m k 0 m k 0 2 n 2 m k m k 2 n ... n n 1 r 1 2 2n n n m n r k r n m n r k m r m k n k m n 1 r s r s 1 Tucker, Applied Combinatorics Section 4.2a 11 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: n k k 1 k k 0 n Solution: n n n n n k k 1 0 0 1 1(1 1) 2(2 1) ... n( n 1) k 0 1 2 n k 0 n n n n 2 6 12 ... n( n 1) 2 3 4 n n 5/24/2017 Tucker, Applied Combinatorics Section 4.2a 12 Example 2: Page 222 # 14 (b) _________________________________________________ Compare this summation with the binomial coefficient. n n n n n k k 1 2 6 12 ... n(n 1) k 2 3 4 n k 0 n Binomial coefficient: n n n 2 n n n n k n 1 x x x ... x x k 0 k 0 1 2 n Notice that our summation starts with the third term, therefore we need to take the second derivative of the binomial coefficient. 5/24/2017 Tucker, Applied Combinatorics Section 4.2a 13 Example 2: Page 222 # 14 (b) _________________________________________________ n n n k Binomial coefficient: 1 x x k 0 k First Derivative: n 1 x n 1 n k 1 k x k k 0 n Second Derivative: n n 11 x 5/24/2017 n2 n k 2 k k 1 x k k 0 n Tucker, Applied Combinatorics Section 4.2a 14 Example 2: Page 222 # 14 (b) _________________________________________________ Compare the Second Derivative to our summation: n n k 2 k k 1 x k k 0 n n n n n k k 1 2 6 12 ... n(n 1) k 2 3 4 n k 0 n n n n k 2 k k 1 k k 1 1 So k k 0 k k 0 n n n2 n2 k k 1 n n 1 1 1 n n 1 2 and k k 0 n 5/24/2017 Tucker, Applied Combinatorics Section 4.2a 15 Example 2: Page 222 # 14 (b) _________________________________________________ To Check to see whether our equation works, plug in a specific n. Let n= 3 n(n 1)2n2 3(3 1)232 12 3 3 2 6 6 6 12 2 3 5/24/2017 Tucker, Applied Combinatorics Section 4.2a 16 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: n 1 k 0 k n k Binomial coefficient: 1 x 5/24/2017 n n n n 2 n n n n k x x ... x x 0 1 2 n k 0 k Tucker, Applied Combinatorics Section 4.2a 17 Problem for Class to Try: Page 222 # 14 (a) _________________________________________________ Solution: n n n n n 1 2 n 1 1 1 ... 1 k 0 k 0 1 2 n n n n n n n ... 1 0 1 2 3 n n k 1 x with x 1 n 1 1 0 n 5/24/2017 Tucker, Applied Combinatorics Section 4.2a 18