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Exponential Generating Functions Book Problems 22. Determine the exponential generating function for the sequence of factorials 0!, 1!, 2!, . . . , n!, . . . ∞ X k=0 xk = ∞ X k! k=0 1 xk = k! 1−x 28. Determine the number of n-digit numbers with all digits at least 4, such that 4 and 6 occur an even number of times, and 5 and 7 each occur at least once, there being no restrictions on the digits 8 and 9. 2 2 2 x2 x4 x6 x2 x3 x2 x3 1+ + + + ... x+ + + ... 1+x+ + + ... 2! 4! 6! 2! 3! 2! 3! In closed form, this is 1 x e + e−x 2 The answer we seek will be the coefficient of 2 (ex − 1)2 e2x xn n! . Additional Problems 1. For each of the following sequences {an }, find a simple, closed form expression for the exponential generating function. (a) (5, 5, 5, . . .) 5ex (b) ak = 3k e3x (c) (1, 0, 0, 1, 1, . . .) ex − x − x2 2! (d) (0, 0, 1, 1, . . .) ex − 1 − x (e) (1, 0, 1, 0, 1, . . .) 1 x e + e−x 2 (f) (2, 1, 2, 1, 2, 1, . . .) 1 x 3 1 ex + e−x + e − e−x = ex + e−x 2 2 2 2. How many 10-letter words are there in which each letter e,n,r and s occur (a) At most one? 22 x2 1+x+ + ... (1 + x)4 2! We need the coefficient of x10 10! . To find this, we will use the closed form and some algebra. (1 + 4x + 6x2 + 4x3 + x4 )e22x = e22x + 4xe22x + 6x2 e22x + 4x3 e22x + x4 e22x We have to factor in that there is a shift, so 10! will not be associated with x10 . We have 10 to use the ‘multiplication’ trick to get x10! . • For e22x , the coefficient is 2210 10 • For 4xe22x , we have x9! , so we need to multiply by 10 10 to get the right term. So, the coefficient would therefore be 4 · 10 · 229 = 40 · 229 . 10 9·10 • For 6x2 e22x , we have x8! , so we need to multiply by 9·10 to get the right term. So, 8 the coefficient would therefore be 6 · 9 · 10 · 22 = 540 · 228 . 10 • For 4x3 e22x , we have x7! , so we need to multiply by 8·9·10 8·9·10 to get the right term. So, 7 the coefficient would therefore be 4 · 8 · 9 · 10 · 22 = 2880 · 227 . 10 • For x4 e22x , we have x6! . So, we need to multiply by 7·8·9·10 7·8·9·10 , so the coefficient would 6 6 therefore be 7 · 8 · 9 · 10 · 22 = 5040 · 22 . So, the coefficient of x10 10! will be 2210 + 40 · 229 + 540 · 228 + 2880 · 227 + 5040 · 226 (b) At least once? 22 4 x2 x2 x3 1+x+ + ... x+ + + ... 2! 2! 3! We need the coefficient of x10 10! . We again will use closed forms and algebra. e22x (ex − 1)4 = e22x (e4x − 4e3x + 6e2x − 4ex + 1) = e26x − 4e25x + 6e24x − 4e23x + e22x So, the coefficient of x10 10! will be 2610 − 4 · 2510 + 6 · 2410 − 4 · 2310 + 2210