Number of 4 digits that can be formed using the digits a, b, b, c = 4!/2! = 12

--> Out of these 12, 'a' occurs in each places (thousands, hundreds, tens and units) 3 times
--> Sum of all values of a's places = 3*1000a + 3*100a + 3*10a + 3*a = 3333*a

Same is for c
--> Sum of all values of c's places = 3333*c

Note that, b occurs 6 times each in thousands, hundreds, tens and units places
--> Sum of all values of b's places = 6*1000a + 6*100a + 6*10a + 6*a = 6666*a

--> Sum of all digits places = 3333*a + 6666*b + 3333*c
= 3333*(a + 2b + c)
= 3333*(3 + 10 + 6)
= 63327

Option C

Total possible numbers = 4!/2! = 12

5 will appear 6 times at each digit place and 3 and 6 appear 3 times at each digit place.

Sum of all 4 digits numbers = 1111 * (6*5 + 3*3+3*6) = 1111 * 57 = 63327

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

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

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..