Question

The sum of all five digit numbers that can be formed using the digits $1,2,3,4$ and $5$ is found.The first digit is (repetition of digits not allowed).

Answer 1

If $5$ is at the unit's place, the remaining digits $1,2,3$ and $4$ can be arranged among themselves in $4!$ ways. Hence, there are exactly $4!$ five-digit numbers using the given digits.

Similarly, there are exactly $4!$ five-digit numbers using the given digits. Similarly for other digits.

Hence the total sum of digits at unit's place for all numbers possible is $4!(5+4+3+2+1)$

Similarly, the total sum of digits at ten's place for all numbers possible is $10\times 4!(5+4+3+2+1)$

Similarly, for hundred's, thousand's and ten thousand's place the sum are ${10}^2\times 4!(5+4+3+2+1),{10}^3\times 4!(5+4+3+2+1),{10}^4\times 4!(5+4+3+2+1)$ respectively.

Hence, the required sum

$=\left(4\right)!(1+2+3+4+5)(1+10+{10}^2+{10}^3+{10}^4)$

$=24\times 15\times \left(\frac{{10}^5-1}{10-1}\right)$

$=24\times 15\times 11111$

$=3999960$

The first digit is $3$.