# X n e 1 2 n x so for all k the egf is e x 1 k the

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+ _+ x n! +_ = e -1 2 n x so for all k the e.g.f. is (e x - 1) k . The coefficient of n x n! in (e x - 1) k is precisely k! S(n,k); we can compute it as follows: i=0 k i (k-i)x x k k i (-1) e = (e -1) = H(x) . Substitute (k - i)x for x in the usual series for e x = n 0 n x n! to get H(x) = i=0 k i n 0 n k i (-1) 1 n! (k- i) xSUPn = n 0 n i=0 k i n x n! (-1) k i (k- i)

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July 30, 1996 12 k! S(n,k) = i=0 k i n (-1) k i (k- i) S(n,k) = 1 (-1 k (k- i i=0 k i n Exercise Find the number of r-digit quatenary requences (digits 0,1,2,3) with an even number of zeros. Off number of 1's. Egf for 0's egf for 1's = = = 1 4 4 = 4 if r > 0 r r-1 Simple form combinatorial argument exists. Can you find one?
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• Fall '06
• miller
• Generating function, xk, e.g.f.

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