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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?

What is the sum of all 6 digit numbers which can be formed using the digits 2, 3, 5, 6 and 9 exactly once? Correct Answer 319999680

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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