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We make 4 digit codes and each digit of the code form from

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by BTGmoderatorLU » Tue Jul 03, 2018 2:03 pm

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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

The OA is D.

Please, can someone assist me with this PS question? I'm not sure how can I solve it. Thanks!
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by Shahrukh_mbabreakspace » Wed Jul 04, 2018 2:43 am
Out of all the possible lock codes,
1 will come at 1000th position in 3! ways
Similarly 2,3 and 4 will also come 3! times each.

Similarly 1 will come at 100th position in 3! ways
and 2,3 and 4 will also come 3! times each.

Similarly 1 will come at 10th position in 3! ways
and 2,3 and 4 will also come 3! times each.

Similarly 1 will come at unit position in 3! ways
and 2,3 and 4 will also come 3! times each.

So, when we take sum of all the codes, we multiply the digit with no. of times it will come in that position and also multiply place value of it

=>(1+2+3+4)*6*1000 + (1+2+3+4)*6*100 + (1+2+3+4)*6*10 + (1+2+3+4)*6*1
= 66660

________________________________________

Shahrukh Moin Khan
QA Mentor at Breakspace
https://www.mbabreakspace.com
PGDM: IIM Calcutta
B.Tech IIT Roorkee
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by Scott@TargetTestPrep » Mon Jul 09, 2018 7:02 pm
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

Scott Woodbury-Stewart
Founder and CEO
[email protected]

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