Math 55, Second Midterm Exam SOLUTIONS (1) Part (a) is problem # 21 (d) on page 361. The desired permutations of the seven letters A,B,C,D,E,F and G are in bijection with the permutations of the four letters X,Y,F and G , where X represents the string ABC and Y represents the string DE . The number of permutations of any set with four elements is 4! = 24. Hence the answer is 24 . For part (b) we observe that there are 7! = 5040 permutations in total, and in half of them the letter A precedes the letter B , and in the others the letter B precedes A . Switching A and B de±nes a bijection between these two subsets. The answer is 2520 . (2) This is problem # 25 on page 440. First consider the case n = 1. If one fair coin is ²ipped, then X 1 (heads) =-1, X 1 (tails) = 1, the expected value is E ( X 1 ) = (1 / 2)(-1) + (1 / 2)(1) = 0, and the variance is V ( X 1 ) = E ( X 2 1 )-E ( X 1 ) 2 = 1-0 = 1. By linearity of expectation (Theorem 3 on page 429), we have E ( X n ) = n · E ( X 1 ) = 0. By linearity of
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This note was uploaded on 10/10/2009 for the course MATH 55 taught by Professor Strain during the Spring '08 term at University of California, Berkeley.