Sum of all $4$-digit numbers that can be formed with the digits $0,1,2,3$ What is the sum of all 4-digit numbers that can be formed with the digits 0,1,2,3 when repetition not allowed?

All I have been able to do is figure out that there will be $18$ such $4$ digit numbers ..but I am very much stuck and would like to know how we would proceed with this.

Also how would we approach this problem if repetition was allowed ?
We'll find sums at each place (units, tens, etc.) and add them.
At thousands place each of $1,2,3$ occurs $6$ times. Sum is  $6(1+2+3)\cdot 1000=36000$.
For hundreds place, we first look at numbers with thousands digit $1$. These are six numbers in which each of $2,3,0$ occurs twice. Hence in the $18$ nos, each of $1,2,3$ occurs $4$ times while $0$ occurs $18-3\cdot 4=6$ times as expected. Sum at hundreds place is $4(1+2+3)\cdot 100=2400$.
Sum at tens and units places are similar - $240$ and $24$ respectively.
Desired sum is $36000+2400+240+24=38664$.

The repetition case is exactly similar and probably easier.
We have $3\cdot 4^3$ nos in all, with $4^3$ starting with $1,2,3$ each. Sum at thousands place is $4^3(1+2+3)\cdot 1000$
For hundreds place, lets fix any digit as $$\square \, 0 \,  \square \, \square$$ There'll be $3\cdot 4^2=48$ occurences of the fixed digit. Thus sum at hundreds place is $48(0+1+2+3)\cdot 100$.
Similarly at tens and units place - $48(0+1+2+3)\cdot 10$ and $48(0+1+2+3)\cdot 1$ resp.
Desired sum is $$(1+2+3)(64000 + 48\cdot 111) = 415968$$
It would be easy to write down four blanks side by side, and then try to see how many possible choices could be filled in each block that would satisfy the condition.
      _____   ______  _____ _____

Check the different cases, for which repetition is allowed, and for repetition not being allowed.
For case 1, repetition being allowed, there are 3 choices for the first blank(0 not incuded), 4 for the next three.
So total number of choices, as per counting principle, total= 3 * 4 * 4 * 4=192
For case 2, repetition not being allowed, there are 3 choices for the first blank, 3 for the next, 2 for the third, and 1 for the last.
so total number of choices, total=3 * 3 * 2 * 1=18.