A. Dold, B. Eckmann's Séminaire Bourbaki PDF

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Extra resources for Séminaire Bourbaki

Example text

N! (n−0)! n! n! n! = 1. (n−1)! = n and C(n, n − 1) = (n−1)! = n. (n−n)! n! n! (n−n+k)! (n−k)! = C(n, k). (c) We have n! n! (n − k + 1)! (n − k)! (n − k + 1)! (n − k + 1)! n! (n − k + 1)! (n + 1)! (n + 1 − k)! 1. 8 The Chess Club has six members. In how many ways (a) can all six members line up for a picture? (b) can they choose a president and a secretary? (c) can they choose three members to attend a regional tournament with no regard to order? 42 COUNTING AND COMBINATORICS Solution. (a) P (6, 6) = 6!

N! (n−0)! n! n! n! = 1. (n−1)! = n and C(n, n − 1) = (n−1)! = n. (n−n)! n! n! (n−n+k)! (n−k)! = C(n, k). (c) We have n! n! (n − k + 1)! (n − k)! (n − k + 1)! (n − k + 1)! n! (n − k + 1)! (n + 1)! (n + 1 − k)! 1. 8 The Chess Club has six members. In how many ways (a) can all six members line up for a picture? (b) can they choose a president and a secretary? (c) can they choose three members to attend a regional tournament with no regard to order? 42 COUNTING AND COMBINATORICS Solution. (a) P (6, 6) = 6!

The proof is by induction on n. Basis of induction: For n = 0 we have 0 C(0, k)x0−k y k = 1. (x + y)0 = k=0 Induction hypothesis: Suppose that the theorem is true up to n. That is, n (x + y)n = C(n, k)xn−k y k k=0 Induction step: Let us show that it is still true for n + 1. That is n+1 n+1 (x + y) C(n + 1, k)xn−k+1 y k . = k=0 4 PERMUTATIONS AND COMBINATIONS 43 Indeed, we have (x + y)n+1 =(x + y)(x + y)n = x(x + y)n + y(x + y)n n n C(n, k)xn−k y k + y =x k=0 n C(n, k)xn−k y k k=0 n C(n, k)xn−k+1 y k + = k=0 C(n, k)xn−k y k+1 k=0 =[C(n, 0)xn+1 + C(n, 1)xn y + C(n, 2)xn−1 y 2 + · · · + C(n, n)xy n ] +[C(n, 0)xn y + C(n, 1)xn−1 y 2 + · · · + C(n, n − 1)xy n + C(n, n)y n+1 ] =C(n + 1, 0)xn+1 + [C(n, 1) + C(n, 0)]xn y + · · · + [C(n, n) + C(n, n − 1)]xy n + C(n + 1, n + 1)y n+1 =C(n + 1, 0)xn+1 + C(n + 1, 1)xn y + C(n + 1, 2)xn−1 y 2 + · · · +C(n + 1, n)xy n + C(n + 1, n + 1)y n+1 n+1 C(n + 1, k)xn−k+1 y k .

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