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Unformatted text preview: MATH 681 Final Exam Answer exactly four of the following six questions. Indicate which four you would like graded! Binomial coefficients, Stirling numbers, and arithmetic expressions need not be simplified in your answers. 1. (12 students attempted this problem) Answer the following questions about find ing the number of words a n of length n with letters A, B, and C using the letter A at least once. (a) (5 points) Find an exponential generating function for a n . The exponential generating function associated with a string of one or more As is: x + x 2 2 + x 3 6 + x 4 24 + = e x 1 The exponential generating function associated with a string of zero or more Bs (or zero or more Cs) is: 1 + x + x 2 2 + x 3 6 + x 4 24 + = e x Multiplying these three together, a mixed string of one or more As, zero or more Bs, and zero or more Cs is associated with the exponential generating function: ( e x 1)( e x )( e x ) = e 3 x e 2 x (b) (5 points) Develop an argument to show that a n satisfies the recurrence a n = 2 a n 1 + 3 n 1 with a = 0 . Clearly a = 0, since the unique string of length zero has no A in it. We may use a casewise argument on the first character (o last character, if you prefer) in a string of length n . If the first character is B or C, than the substring consisting of the n 1 subsequent characters must be a string of As, Bs, and Cs which contains at least one A: in other words, a string of length n 1 satisfying our original criterion. We know there are a n 1 such strings by definition, so there are 2 a n 1 strings of length n beginning with B or C. If, on the other hand, the first character is an A, then the n 1 subsequent characters are no longer restrained the string as a whole must contain an A, but it is guaranteed to do so! Thus, the following n 1 characters can be any string of A, B, and C, which we know can be produced in 3 n 1 ways. Putting these two cases together, we see that a n = 2 a n 1 + 3 n 1 . (c) (5 points) Using a method of your choice, find a closedform expression for a n . It is easy to derive this from the exponential generating function found in part (a), and only slightly harder to derive it from the recurrence relation found in part (b). There is also a direct enumeration using exclusion methods. Here the EGF approach is demonstrated. Page 1 of 6 December 12, 2009 MATH 681 Final Exam Expanding the exponential generating function we already discovered: X n =0 a n x n n ! = e 3 x e 2 x = X n =0 (3 x ) n n ! X n =0 (2 x ) n n ! = X n =0 (3 n 2 n ) x n n ! so a n = 3 n 2 n . 2. (6 students attempted this problem) Prove the following identities using combi natorial arguments, where F n represents the Fibonacci sequence indexed with F = 1 , F 1 = 1 , F 2 = 2 ....
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 Fall '09
 WILDSTROM
 Binomial

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