(B)
Consider the digits $1,2,3$.
The possible numbers are $\{123,132,213,231,312,321\}$, count = $3! = 6$.
- Consider the cases where the digit $3$ is at the unit position. The number of such numbers (fix $3$ at unit, permute the rest) will be $(3-1)! = 2!$. In all such cases, the $3$ digit will contribute a $3$ to the final sum. Total = $2! \times 3$
- Consider the cases where the digit $3$ is at the decimal position. The number of such numbers (fix $3$ at decimal, permute the rest) will be $(3-1)! = 2!$. In all such cases, the $3$ digit will contribute a $30$ to the final sum. Total = $2! \times 30$
- Consider the cases where the digit $3$ is at the hundreds position. The number of such numbers (fix $3$ at hundreds, permute the rest) will be $(3-1)! = 2!$. In all such cases, the $3$ digit will contribute a $300$ to the final sum. Total = $2! \times 300$
In total, the digit $3$ contributes $2! \times (3+30+300) = 2! \times 333$ to the final sum.
The same happens for all other digits.
Hence, the net sum (for the original question) will be:
$(5-1)! \times (11111 + 33333 + 55555 + 77777 + 99999)$
Note: $(5-1)!$ = the number of permutations after fixing $1$ digit. $11111 \ldots$ because the digit $1$ contributes a $1, 10, 100, 1000, 10000$ to the final sum.
Required Answer = $6666600$.
Hence, option B is correct.