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.
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