Finding the sum of numbers formed by digits

Question

I know how to find the sum of all numbers formed by digits 1,2,3,4,5 taken all at a time without repetitions. How can I find the sum when digit repetitions are allowed?

Answer 1

Let us focus on the last digit in the ten-thousands place. Every digit there has to be multiplied by $10^4$. Each number $1,2,3,4,5$ can occur $4^5$ times. This is because each other $4$ digits places can be occupied by any of the $5$ numbers. Hence multipliying all this gives:

$$S=10^4\times 4^5\times (1+2+3+4+5)$$ $$S=10^4\times 4^5\times 15$$

Now we simply extend this for every digit's place. The final sum will be $$S= 4^5\times 15\times 11111$$