The sum of all the four-digit numbers that can be formed using all...

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

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

Numbers are

___3___3___3___2___2___2___2___2___2___1___1___1

Required sum is

$= \mathop{\mathop{\left[3 \times 3 + 2 \times 6 + 1 \times 3\right]}_{⏟}}_{unitdigit ' ssum} + \mathop{\mathop{\left[\left(3 \times 3 + 2 \times 6 + 1 \times 3\right) \times 10\right]}_{⏟}}_{ten ' splacesum} + \mathop{\mathop{\left[\left(3 \times 3 + 2 \times 6 + 1 \times 3\right) \times 100\right]}_{⏟}}_{hundred ' splacesum} + \mathop{\mathop{\left[\left(3 \times 3 + 2 \times 6 + 1 \times 3\right) \times 1000\right]}_{⏟}}_{thousand ' splacesum}$

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

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