BTGmoderatorLU wrote:We make 4 digit codes and each digit of the code form from 1, 2, 3, and 4. If none of each digit use more than once, eg, 1234 can be a code but 1124 cannot be a code, what is the sum of all the possible codes?

A. 56,660
B. 58,660
C. 60,660
D. 66,660
E. 68,660

Since we are forming a code using 4 distinct digits, the number of codes that can be formed is 4! = 24. However, for each digit at a certain place value, it can be at that place value 6 times. For example, at the thousands place, 1 can only appear 6 times. That is because 2, 3 and 4 will also appear 6 times each at the the thousands place so that there are a total of 24 numbers. Therefore, the following must be true:

1, 2, 3 and 4 each appears:

i) at the thousands place 6 times with a total value of 6 x (1+2+3+4) x 1000 = 60 x 1000 = 60,000.

ii) at the hundreds place 6 times with a total value of 6 x (1+2+3+4) x 100 = 60 x 100 = 6,000.

iii) at the tens place 6 times with a total value of 6 x (1+2+3+4) x 10 = 60 x 10 = 600.

iv) at the ones place 6 times with a total value of 6 x (1+2+3+4) x 1 = 60 x 1 = 60.

So the sum of these 24 codes (or numbers) must be 60,000 + 6,000 + 600 + 60 = 66,660.

Answer: D