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Bunuel
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Bunuel
What is the sum of all the four-digit numbers formed by digits 3, 5, 5, 6, using each digit once ?

A. 65297

B. 64427

C. 63327

D. 43521

E. 43519
 
Hi Bunuel, I tried using the formula you mentioned in a different post:

"1. Sum of all the numbers which can be formed by using the n digits without repetition is: (n−1)!∗(sum of the digits)∗(111... n times)

2. Sum of all the numbers which can be formed by using the n digits (repetition being allowed) is: n−1∗(sum of the digits)∗(111... n times)"

But I am getting the wrong answer. Can you please help. Thanks­
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I use below formula for this type of question which is useful in both cases with all distinct digits or repeated digits

{(number of rearrangement of digits) / (number of total digits)} * (sum of all digits) * (1111)

so incase of ( 3, 5, 5, 6)

total number of rearrangements = 4!/2! =12

sum of all 4 digit numbers = (12/4)*(3+5+5+6)(1111) = 3*19*1111 = 63327

Bunuel please validate if my approach is correct
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So just to make sure, the reason we get 4!/2! is because the digit 5 can only be used once?

so I first put A, B, C, D - > 3, 5, 5, 6, respectfully. Thinking that the second digit 5 was treated differently..
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