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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?

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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) 319999680

I had been asked this question in an online quiz.

I need to ask this question from Counting topic in chapter Counting of Discrete Mathematics

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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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