sum of numbers formed by permutations

Question

I have digits 2,3,4,5. I have been asked to find the sum of all 4 digits the numbers that can be formed using these digits without repetition such that all are included in the number.

Can someone help me solve this quickly?

The answer is 93324.

Sorry guys, I have corrected the question. Can someone please help me now? — user6037890, May 03 '16 at 08:17

Aniket Bhattacharyea · Answer 1 · 2016-05-03T08:19:22.993

1

Fixing 5 in the units digit, you get $3!=6$ numbers. So 5 comes 6 times in the units digit. Same goes for 2,3,4

hence sum of the units digits = $6(2+3+4+5)=84$

Same argument goes for ten's place, hundreds place and so on. hence the sum=

$84 \times 1000+84 \times 100+84 \times 10+84=84 \times 1111=93324$

edited May 03 '16 at 08:19

answered May 03 '16 at 08:14

Aniket Bhattacharyea

991

Seems that someone has down-voted each and every answer here. – barak manos May 07 '16 at 11:44
1

Maybe some troll – Aniket Bhattacharyea May 07 '16 at 15:45
I hope it is not the OP. – Peter May 07 '16 at 21:02
@Peter: I don't think that OP has the privilege of voting. In any case, since he or she has not referred to any of the answers, I'm deleting mine (as you did). – barak manos May 08 '16 at 06:19

score -1 · Answer 2 · answered May 03 '16 at 08:23

-1

For the 2 you get:
$2000 \cdot 3! + 200 \cdot 3! + 20 \cdot 3! + 2 \cdot 3!$
The 1st term corresponds to 2 being at position 1,
the 2nd term corresponds to 2 being at position 2 and so on.

Then you have the same pattern for 3,4,5.

So you get: $(2222 + 3333 + 4444 + 5555) \cdot 6 = 93324$

answered May 03 '16 at 08:23

peter.petrov

12,568

Seems that someone has down-voted each and every answer here. – barak manos May 07 '16 at 11:44

sum of numbers formed by permutations

2 Answers2