We make 4 digit codes and each digit of the code form from 1, 2, 3, an

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

--> (1,000*6+2,000*6+3,000*6+4,000*6)+(100*6+200*6+300*6+400*6)
+(10*6+20*6+30*6+40*6)+(1*6+2*6+3*6+4*6)
=10,000*6+1,000*6+100*6+10*6
=66,660

Therefore, the answer is D.

Numbers to be used: 1 , 2 , 3 , 4
total number of possibilites= 4 . 3 . 2 . 1 = 24 (no repetition of any number)
We need to calculate sum of these 24 combinations of numbers.

Since there are 4 numbers (1,2,3,4) and 24 combinations of numbers, then each number will be used 24/4 =6 times in the total sum.
xxxx
yyyy (x+y+z) = 6 x 4 + 6 x 3 + 6 x 2 + 6 x 1 = 60
+zzzz at units place = 0 (6 will be carried)
at tens place = 0+6 = 6 ( 6 is from previous carry, and 6 again will be carried)
at 100s place = 6 with 6 carried forward
at 1000s place =6 with 6 carried forward
at 10,000s place = 6 from previous carry

ANS = 66660

I feel it is a very tough question that cannot be solved within 2 min.
Has it appeared on the GMAT or is it that only I am finding it difficult.

formula =1111*sum(a+b+c+d)*n-1! (in this case n-3)

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

Can somebody please tell me if my approach is correct?

total combinations possible - 4! = 24; with one number appearing only once, I can say that we will get 12 complementing pairs (eg. 1234 and 4321)

we add 1234 and 4321 to get 5555 and multiply it by 12, we get 66660.

Please confirm my understanding.

Each digit can appear in a given spot 3!=6 ways since the other 3 digits can be arranged that way.

So each spot gets weighted

6*4+6*3+6*2+6=6*10=60 times

So 60*1000+60*100+60*10+60=

60*(1000+100+10+1)=
60*1111= 66,660

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So I believe that there is a simple formula for such question :
Where it goes like you have to calculate the sum of various combinations of N things in our case (1,2,3,4) and we cannot repeat the digits.

So fix one digit and then we have (n-1)! or 3! in our case possible combinations.
multiply this to the sum of the digits (1+2+3+4) = 10
Then multiply the same to 111...till N digits (in our case 1111).

so we finally get 3! * 10 * 1111 = 6 * 10 * 1111 = 66660.