$The sum of all the four digit distinct numbers that can be formed with the digits 1,2,2, and 3 is$

$\begin{array}{l}1\text{\hspace{0.17em}at unit place }\Rightarrow \text{\hspace{0.17em}number of such numbers is }\frac{3!}{2!}=3\\ 2\text{\hspace{0.17em}at unit place }\Rightarrow \text{\hspace{0.17em}number of such numbers is }3!=6\\ 3\text{\hspace{0.17em}at unit place }\Rightarrow \text{\hspace{0.17em}number of such numbers is }\frac{3!}{2!}=3\\ \text{Hence the sum of the digits in the unit place is }3\times 1+6\times 2+3\times 3=24\\ \text{The sum of all 4 digit numbers is 1111}\times 24=26664\\ \end{array}$