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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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