key3fa02

key3fa02 - COT3100C-01, Fall 2002 Oct. 15, 2002, correction...

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COT3100C-01, Fall 2002 Oct. 15, 2002, correction on S. Lang Solution Keys to Assignment #3 (40 pts.) 3(c) in red added 10/18, a simpler solution to 2(b) in blue added 10/28 1. (12 pts.) Count the number of 7-digit integers under each of the following restrictions: (a) All digits are distinct. (Thus, integers such as 1234567 and 3456980 are counted, but integers such as 2345 and 2233445 are not counted.) Answer: 9 * P(9, 6) = 9*9*8*7*6*5*4, where the first factor 9 counts the choices of a non-zero digit (1 through 9) for the leftmost digit of the 7-digit integer; the factor P(9, 6) counts the number of permutations for the remaining 6 digits of the 7-digit integer. (b) The digits may be repeated (such as 2233456) but the integer must be divisible by 5. (Hint: what would be the last, rightmost digit for an integer to be divisible by 5?) Answer: 9 * 10 5 * 2, where the first factor 9 counts the 9 choices (1 through 9) for the leftmost digit; each of the middle 5 digits can be any of the 10 choices (0 through 9); finally, the rightmost digit must be either 0 or 5, for two choices, for the integer to be divisible by 5. (c) The digits may be repeated but digit 3 must be used (as part of the integer) at least twice. (Hint: consider the cases: digit 3 used twice, three times, etc., until 7 times, then count each case or group separately. Alternatively, consider the complementary question, i.e, if digit 3 is not used or used exactly once, and count them separately.) Answer: We consider the complementary question. First, we count those 7-digit integers that don’t use digit 3. This count is 8 * 9 6 because the first digit can have 8 choices (1, 2, 4 through 9), each of the remaining 6 digits has 9 choices (0 though 9 excluding 3). We then count those 7-digit integers that contain digit 3 exactly once.
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key3fa02 - COT3100C-01, Fall 2002 Oct. 15, 2002, correction...

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