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Unformatted text preview: such decimal digits. c) There are C (3, 2) =3 ways to choose the position of the two 5s. There are 9 ways (we cannot choose 5) to choose the other number. So totally, there are such decimals. Question 2: (4pts) How many numbers must be selected from the set {1, 3, 5, 7, 9, 11, 13, 15} to guarantee that at least one pair of these numbers add up to 16? Consider the 4 groups {1, 15}, {3, 13}, {5, 11} and {7, 9}. If there are 5 or numbers, then there must be two numbers from the same group. Thus, the sum of them is 16. If there are only 4 or less numbers, then there is a possibility that none of them are from the same group. Then the sum is not 16 for any pair. Therefore, 5 is sufficient and necessary....
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 Spring '08
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