Digits are $1, 2, 2, 3$

total distinct numbers $\frac{4!}{2!}=12.$

total numbers when

$1$ at unit place is $3.$
$2$ at unit place is $6$
$3$ at unit place is $3.$

So, sum $=(3+12+9)\left({10}^3+{10}^2+10+1\right)$

$=\left(1111\right)\times 24$

$=26664$