Binomial Theorem. Combinations WITH repetition. PERMUTATIONS AND COMBINATIONS Pascal’s TRIANGLE The Binomial Theorem Given a set with n elements. Combinations WITH REPETITION PERMUTATIONS WITH The number of permutations of the elements the set: REPETITION P(n) = n! = n (n 1) (n 2) ... 1 · − · − · · The number of r-permutations of the set: n! P(n, r) = = n (n 1) (n 2) ... (n r + 1) (n r)! | · − · − {z· · − } − product of r numbers The number of unordered r-combinations (“n choose r”): n P(n, r) n! = = r P(r) (n r)! r! − p. 2 n What ARE THE PROPERTIES OF r ? Pascal’s TRIANGLE The Binomial Theorem n Combinations WITH REPETITION How does r change with r? PERMUTATIONS WITH REPETITION 0 0! 1. 0 = 0! 0! = 1 1! 1 1! 1, 1. 0 = 1! 0! = 1 = 0! 1! = 2 2! 2 2! 2 2! 1, 2, 1. 0 = 2! 0! = 1 = 1! 1! = 2 = 0! 2! = 3 3! 3 3! 3 3! 3 3! 1, 3, 3, 1. 0 = 3! 0! = 1 = 2! 1! = 2 = 1! 2! = 3 = 0! 3! = p. 3 Pascal’s TRIANGLE Pascal’s TRIANGLE The Binomial Theorem Combinations WITH REPETITION 0 1 PERMUTATIONS WITH 0 REPETITION 1 1 11 0 1 2 2 2 121 0 1 2 3 3 3 3 0 1 2 3 1331 4 4 4 4 4 0 1 2 3 4 14641 5 5 5 5 5 5 0 1 2 3 4 5 15 10 1051 p. 4 Pascal’s TRIANGLE Pascal’s TRIANGLE The Binomial The numbers in Pascal’s Triangle are the coefficients of the polyno- Theorem n Combinations WITH mials of the form (x + y) : REPETITION PERMUTATIONS WITH 0 REPETITION (x + y) = 1 1 (x + y) = 1x + 1y 2 2 2 (x + y) = 1x + 2xy + 1y 3 3 2 2 3 (x + y) = 1x + 3x y + 3xy + 1y 4 4 3 2 2 3 4 (x + y) = 1x + 4x y + 6x y + 4xy + 1y n n n n n 1 n n 2 2 n n (x + y) = x + x − y + x − y + ... + y . 0 · 1 · 2 · n · n X n x y n x n k y k ( + ) = k − k=0 p. 5 Pascal’s TRIANGLE Pascal’s TRIANGLE The Binomial The numbers in Pascal’s Triangle are the coefficients of the polyno- Theorem n Combinations WITH mials of the form (x + y) : REPETITION PERMUTATIONS WITH 0 REPETITION (x + y) = 1 1 (x + y) = 1x + 1y 2 2 2 (x + y) = 1x + 2xy + 1y 3 3 2 2 3 (x + y) = 1x + 3x y + 3xy + 1y 4 4 3 2 2 3 4 (x + y) = 1x + 4x y + 6x y + 4xy + 1y n n n n n 1 n n 2 2 n n (x + y) = x + x − y + x − y + ... + y . 0 · 1 · 2 · n · n X n x y n x n k y k ( + ) = k − k=0 p. 6 Pascal’s TRIANGLE Pascal’s TRIANGLE The Binomial The numbers in Pascal’s Triangle are the coefficients of the polyno- Theorem n Combinations WITH mials of the form (x + y) : REPETITION PERMUTATIONS WITH 0 REPETITION (x + y) = 1 1 (x + y) = 1x + 1y 2 2 2 (x + y) = 1x + 2xy + 1y 3 3 2 2 3 (x + y) = 1x + 3x y + 3xy + 1y 4 4 3 2 2 3 4 (x + y) = 1x + 4x y + 6x y + 4xy + 1y n n n n n 1 n n 2 2 n n (x + y) = x + x − y + x − y + ... + y . 0 · 1 · 2 · n · n X n x y n x n k y k ( + ) = k − k=0 p. 7 Pascal’s TRIANGLE Pascal’s TRIANGLE The Binomial The numbers in Pascal’s Triangle are the coefficients of the polyno- Theorem n Combinations WITH mials of the form (x + y) : REPETITION PERMUTATIONS WITH 0 REPETITION (x + y) = 1 1 (x + y) = 1x + 1y 2 2 2 (x + y) = 1x + 2xy + 1y 3 3 2 2 3 (x + y) = 1x + 3x y + 3xy + 1y 4 4 3 2 2 3 4 (x + y) = 1x + 4x y + 6x y + 4xy + 1y n n n n n 1 n n 2 2 n n (x + y) = x + x − y + x − y + ... + y . 0 · 1 · 2 · n · n X n x y n x n k y k ( + ) = k − k=0 p. 8 n CoeffiCIENTS OF (x + y) Pascal’s TRIANGLE The Binomial Theorem Let’s prove Combinations WITH REPETITION PERMUTATIONS WITH n n n n n 1 n n 2 2 n n REPETITION (x + y) = x + x − y + x − y + ... + y . 0 · 1 · 2 · n · Example: 3 (x + y) =(x + y)(x + y)(x + y) = x x x+ x x y + x y x + y x x+ x y y + y x y + y y x+ y y y n 3 2 = 2 = 8 terms in total. The same as the number of the bit strings of length 3. p. 9 n CoeffiCIENTS OF (x + y) Pascal’s TRIANGLE The Binomial Theorem Combinations WITH REPETITION n (x + y) = (x + y) (x + y) ... (x + y) = ? PERMUTATIONS WITH | · {z · · } REPETITION n times What is happening when we multiply (x + y) n times? We get the sum of p. 10 n CoeffiCIENTS OF (x + y) Pascal’s TRIANGLE The Binomial Theorem Combinations WITH REPETITION n (x + y) = (x + y) (x + y) ... (x + y) = ? PERMUTATIONS WITH | · {z · · } REPETITION n times What is happening when we multiply (x + y) n times? We get the sum of x x x x x ... x+ yxxxx ... x+ xyxxx ... x+ yyxxx ... x+ x xyxx ... x+ ... + y y y y y ... y p. 11 n CoeffiCIENTS OF (x + y) Pascal’s TRIANGLE The Binomial Theorem Combinations WITH REPETITION PERMUTATIONS WITH REPETITION n (x + y) = (x + y) (x + y) ... (x + y) = | · {z · · } n times x x ... x + | {z } · =·x n · (x ... x) y + ... + y (x ... x)+ | {z } · x n 1· · · · · = − (x ... x) (y y) + ... + (y y) (x ... x)+ | {z } | {z } · x n 2· · · · · · x n 2· = − = − ... + y y ... y | · {z· · } =y n p. 12 n CoeffiCIENTS OF (x + y) Pascal’s TRIANGLE The Binomial Theorem Combinations WITH REPETITION PERMUTATIONS WITH REPETITION n (x + y) = (x + y) (x + y) ... (x + y) = | · {z · · } n times n 1 x + · n 1 n x − y + · n n 2 2 x − y + 2 · ... + 1 y n · p. 13 The Binomial Theorem Pascal’s TRIANGLE The Binomial Theorem Combinations WITH REPETITION PERMUTATIONS WITH REPETITION n (x + y) = (x + y) (x + y) ... (x + y) = | · {z · · } n times n n n n 1 n n 2 2 n n x + x − y + x − y + ... + y . 0 · 1 · 2 · n · Shorter notation for the same thing: n X n x y n x n k y k. ( + ) = k − k=0 This result is called The Binomial Theorem, and this is why the co- n efficients, k , are also called the binomial coefficients. p. 14 The Binomial Theorem Pascal’s TRIANGLE The Binomial Theorem Combinations WITH REPETITION PERMUTATIONS WITH Let’s prove that REPETITION n n n ... 2n. 0 + 1 + + n = n n n X n X n X n 1 1 n 1n k1k 1 . ( + ) = k − = k = k k=0 k=0 · k=0 Therefore, n X n 2n. k = k=0 p. 15 The Binomial Theorem Pascal’s TRIANGLE The Binomial Theorem Combinations WITH REPETITION PERMUTATIONS WITH Let’s prove that REPETITION n n n ... 2n. 0 + 1 + + n = n n n X n X n X n 1 1 n 1n k1k 1 . ( + ) = k − = k = k k=0 k=0 · k=0 Therefore, n X n 2n. k = k=0 p. 16 The Binomial Theorem Pascal’s TRIANGLE The Binomial Theorem Combinations WITH REPETITION n n n n + + ... + = 2 . PERMUTATIONS WITH 0 1 n REPETITION Results like this are not very obvious. Recall that n n! = k (n k)! k! So, − n! n! n! n! n + + + ... + = 2 . n! 0! (n 1)! 1! (n 2)! 2! 0! n! − − In this form, the result seems to be much harder to prove. However, can you think of another way to prove this identity? (You can try to use double counting) p. 17 The Binomial Theorem Pascal’s TRIANGLE The Binomial Theorem Combinations WITH REPETITION PERMUTATIONS WITH Using the binomial theorem REPETITION n X n x y n x n k y k, ( + ) = k − k=0 prove that n X n 2k k = k=0 n n n n 2 22 ... 2n 3n. 0 + 1 + 2 + + n = p. 18 Counting ROUTES Pascal’s TRIANGLE The Binomial Theorem Combinations WITH REPETITION PERMUTATIONS WITH REPETITION You can go only North and East. 5+3 Count the number of paths from (0, 0) to (5, 3). Answer: 3 . p. 19 Counting ROUTES Pascal’s TRIANGLE The Binomial Theorem Combinations WITH REPETITION PERMUTATIONS WITH REPETITION You can go only North and East. 5+3 Count the number of paths from (0, 0) to (5, 3). Answer: 3 . p. 20 Pascal’s TRIANGLE Again Pascal’s TRIANGLE The Binomial Theorem Combinations WITH REPETITION 0 1 PERMUTATIONS WITH 0 REPETITION 1 1 11 0 1 2 2 2 121 0 1 2 3 3 3 3 0 1 2 3 1331 4 4 4 4 4 0 1 2 3 4 14641 5 5 5 5 5 5 0 1 2 3 4 5 15 10 1051 p. 21 Pascal’s Identity Pascal’s TRIANGLE The Binomial Theorem Combinations WITH REPETITION PERMUTATIONS WITH REPETITION n + 1 n n k = k 1 + k − Every number in Pascal’s triangle is equal to the sum of the two numbers that are immediately above. p. 22 Another Identity Pascal’s TRIANGLE The Binomial Theorem Combinations WITH REPETITION PERMUTATIONS WITH REPETITION Prove that for r n and r m: ≤ ≤ r m + n X m n r = r k k k 0 = − p.