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Question: By using the digits 1, 2, 3, 4 and 5, 5-digit numbers are formed such that every digit is used at most once, then find the sum of all such possible numbers.

Attempt: I found the answer 3,999,960. The sum of digits at unit place will be

$$s = 24 * ( 5+4+3+2+1) $$

Now adding s for every place - unit, tenth, hundredth, thousandth, and ten-thousandth.

But there must some other ways also to approach this problem. I would like to understand about them. Thanks.

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  • $\begingroup$ I think that your approach is good. You could formalize it or make it a little more general, but this is the approach that I would use. Potentially, you could reorganize it as $24(55555+44444+33333+22222+11111)$. $\endgroup$ – Michael Burr Jan 18 '18 at 13:20
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In your method, you did $s = 24*(5+4+3+2+1)$, and multiplied 1, 10, 100, 1000, and 10000 accordingly. Instead, you could have just taken $24*(55555+44444+33333+22222+11111)=3999960$.

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