A%20Probability%20Approach%20For%20Solving%20Counting%20Problems

E 777 p11 110110110 11000 p1 110003 31000

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Unformatted text preview: N = 1,000 When 7 appears three times (i.e. 777) P11 = 1/10*1/10*1/10 = 1/1000 P1 = 1/1000*3 = 3/1000 (since we are counting the appearances of sevens!) When 7 appears twice (i.e. X77 or 77X or 7X7) P21 = 1/10*1/10*9/10 = 9/1000 P22 = 1/10*1/10*9/10 = 9/1000 P23 = 1/10*1/10*9/10 = 9/1000 1/10 is for picking 7 out of 10 digits and the 9/10 is for picking any other digit except 7. P2 = 27/1000*2 = 54/1000 (since we are counting the appearances of sevens!) P31 = 1/10*9/10*9/10 = 81/1000 P32 = 9/10*1/10*1/10 = 81/1000 P33 = 9/10*9/10*1/10 = 81/1000 P3 = 243/1000*1 = 243/1000 (since we are counting the appearances of sevens!) When 7 appears once (i.e. XX7 or 7XX or X7X) SP = 3/1000+54/1000+243/1000 = 300/1000 N*P = 1000*300/1000 = 300 The answer is (D) Of the three-digit positive integers that have no digits equal to zero, how many have two digits that are equal to each other and the remai...
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