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Find the sum of all distinct four digit numbers that can be formed using the digits 1, 2, 3, 4, 5, each digit appearing at most once. My Answer For each $x\in$ {$1,2,3,4,5$}, there are $4!$ such numbers whose last digit is $x$. Thus the digits in the unit place of all the $120$ numbers add up to $4!(1+2+3+4+5)$. Similarly the numbers at ten’s place add up to $360$ and so on. Thus the sum is $360$($1+10+100+1000$) is this correct

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  • $\begingroup$ Yes, this is correct. $\endgroup$ – Wolfram Jan 29 '17 at 17:58
  • $\begingroup$ thnx for checking $\endgroup$ – Upstart Jan 29 '17 at 18:00

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