Smriti Shashikumar wrote:Find the sum of all numbers that can be formed by taking all the digits at a time from 1,2,4 and 6 without repetition.

A. 86568
B. 86658
C. 85866
D. 88665
E. 86656

There are four positions in the integer: the thousands digit, the hundreds digit, the tens digit, and the units digit.
For each position, there are four digit options: 1, 2, 4 and 6.

The number of ways to arrange the four digit options = 4! = 24.
Each of the four digit options will appear in each position an EQUAL number of times.
Thus, each digit option will appear in each position = 24/4 = 6 times.

The sum of the digit options = 1+2+4+6 = 13.
Since each digit option will appear in each position 6 times, the sum of the digits in each position = 6*13 = 78.

If the four digits are A, B, C and D, the integer can be represented as follows:
1000A + 100B + 10C + D.
Since the sum of the digits in each position is 78, we get:
78(1000) + 78(100) + 78(10) + 78 = 78(1000+100+10+1) = 78(1111) = 86658.

The correct answer is B.

well, you don't need to do the entire calculation. As you know that when you add the all the digits at a particular face value of these 24 numbers, you are going to get 78, you can simplify by:

unit digit is 8, carrying over 7 to tens place.
sum of tens place will again be 78. Add the carry over of 7 to it and you get 85. So, put 85 in tens place.
So, the number ends with 58. You can shortlist the answer!!!

Thanks a lot! That was super helpful!