<s>[INST] 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 stuck and would like to know how we would proceed with this.

Also how would we approach this problem if repetition was allowed ?

 [/INST] 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 $$(1+2+3)(64000 + 48\cdot 111) = 415968$$
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