The sum of all numbers greater than 1000 formed by using the digits 0, 1, 2, 3, no digit being repeated in any number, is

All the numbers are of four digits and they do not have 0 in thousand’s place.

The number of numbers having 0 in unit’s place = 3! (∵ the other three places are to be filled by 1, 2 or 3).

The number of numbers having 1 in unit’s place ${=}^2{\mathrm{P}}_1{\times }^2{\mathrm{P}}_2$

(∵ thousand’s place can be filled by one of 2, 3 and the remaining two places can be filled by the remaining two digits).

Similarly, the number of numbers having 2 or 3 in unit’s place

${=}^2{\mathrm{P}}_1{\times }^2{\mathrm{P}}_2$ in each case.

Thus, the sum of the digits in unit’s place for all the numbers

$\begin{array}{l}=3!\times 0{+}^2{\mathrm{P}}_1{\times }^2{\mathrm{P}}_2\times 1{+}^2{\mathrm{P}}_1{\times }^2{\mathrm{P}}_2\times 2{+}^2{\mathrm{P}}_1{\times }^2{\mathrm{P}}_2\times 3\\ =4+8+12=24\end{array}$

Similarly, the sum of the digits in ten’s and hundred’s places = 24 each.

Now, the thousand’s place can have only 1 or 2 or 3.

Number of numbers having 1 in thousand’s place = 3!, (for the other three places will be filled by 0, 2, 3).

Similarly, the number of numbers having 2 or 3 in thousands’s place is 3! in each case.

∴ Sum of the digits in the thousands’s place for all the numbers = 3! × 1 + 3! × 2 + 3! × 3 = 6 + 12 + 18 = 36.

∴ Required sum of all the numbers

= 36 × 1000 + 24 × 100 + 24 × 10 + 24 × 1

= 36000 + 2400 + 240 + 24 = 38664.