What is the sum of digits in the unit place of all numbers

Question

What is the sum of digits in the unit place of all numbers formed using 1,2,3,4,5,6 taken all at a time without repeating any of them?

Answer 1

Sum of digits in the unit place $$=(6-1)!(1+2+3+4+5+6)$$ $$=5! ~\times~21~=~120~\times~21~=~2520$$

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Explanation : Let us first choose the unit place as $~6~$, then the possible choice for other places are $~5\times 4\times 3\times 2\times 1~=~5!~=~120$.

Now if we choose the unit place as $~5~$, then the possible choice for other places are again $~5\times 4\times 3\times 2\times 1~=~5!~=~120$.

Similarly, the others.

So the total sum of digits in the unit place of all numbers formed using $~1,~2,~3,~4,~5,~6~$ taken all at a time without repeating any of them $$~=(~6~\times~5!~)+(~5~\times~5!~)+(~4~\times~5!~)+(~3~\times ~5!~)+(~2~\times~5!~)+(~1~\times ~5!~)$$ $$=~5!~\times~(6+5+4+3+2+1)~$$$$=~120~\times~21~$$$$=~2520$$