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What is the sum of all 6 digit numbers which can be formed using the digits 2, 3, 5, 6 and 9 exactly once?(a) 986546600(b) 25611866(c) 433338798(d) 319999680I had been asked this question in an online quiz.I need to ask this question from Counting topic in chapter Counting of Discrete Mathematics |
Answer» RIGHT answer is (d) 319999680 For EXPLANATION: Note that sum of all possible NUMBERS = (n-1)!(sum of the digits involved)(1111…n times), where n is the number of digits. Here n = 6, we have (6-1)!(2+3+5+6+9)(111111) = 5!*(24)*(111111) = 319999680. |
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