The sum of all the four-digit numbers that can be formed using all the digits 2 ,   1 ,   2 ,…

Total $4$ digit numbers using the digits $1,2,2,3$ is

$=\frac{4!}{2!}=12$

Numbers are

$$\begin{gathered} \_ \_ \_ 3 \\ \_ \_ \_ 3 \\ \_ \_ \_ 3 \\ \_ \_ \_ 2 \\ \_ \_ \_ 2 \\ \_ \_ \_ 2 \\ \_ \_ \_ 2 \\ \_ \_ \_ 2 \\ \_ \_ \_ 2 \\ \_ \_ \_ 1 \\ \_ \_ \_ 1 \\ \_ \_ \_ 1 \end{gathered}$$

Required sum is

$=\underset{\mathrm{unit} \mathrm{digit}\text{'}\mathrm{s} \mathrm{sum}}{\underbrace{\left[3\times 3+2\times 6+1\times 3\right]}}+\underset{\mathrm{ten}\text{'}\mathrm{s} \mathrm{place} \mathrm{sum}}{\underbrace{\left[\left(3\times 3+2\times 6+1\times 3\right)\times 10\right]}}+\underset{\mathrm{hundred}\text{'}\mathrm{s} \mathrm{place} \mathrm{sum}}{\underbrace{\left[\left(3\times 3+2\times 6+1\times 3\right)\times 100\right]}}+\underset{\mathrm{thousand}\text{'}\mathrm{s} \mathrm{place} \mathrm{sum}}{\underbrace{\left[\left(3\times 3+2\times 6+1\times 3\right)\times 1000\right]}}$

$=\left[24\right]+\left[24\times 10\right]+\left[24\times 100\right]+\left[24\times 1000\right]$

$=24+240+2400+24000=26664$