The sum of all the 4-digit distinct numbers that can be formed with the digits 1, 2, 2 and 3 is :

Total possible numbers using 1, 2, 2 and 3 is <br><br>= ${{4!} \over {2!}}$ = 12 <br><br>When unit place is 1, the total possible numbers using remaining 2, 2 and 3 are <br><br>= ${{3!} \over {2!}}$ = 3 <br><br>When unit place is 2, the total possible numbers using remaining 1, 2 and 3 are <br><br>= 3! = 6 <br><br>When unit place is 3, the total possible numbers using remaining 1, 2 and 2 are <br><br>= ${{3!} \over {2!}}$ = 3 <br><br>$\therefore$ Sum of unit places of all (3 + 6 + 3) 12 numbers is <br><br>= ( 1$\times$3 + 2$\times$6 + 3$\times$3) <br><br>Similarly, <br><br>When 10^th place is 1, the total possible numbers using remaining 2, 2 and 3 are <br><br>= ${{3!} \over {2!}}$ = 3 <br><br>When 10^th place is 2, the total possible numbers using remaining 1, 2 and 3 are <br><br>= 3! = 6 <br><br>When 10^th place is 3, the total possible numbers using remaining 1, 2 and 2 are <br><br>= ${{3!} \over {2!}}$ = 3 <br><br>$\therefore$ Sum of 10^th places of all (3 + 6 + 3) 12 numbers is <br><br>= ( 1$\times$3 + 2$\times$6 + 3$\times$3) $\times$ 10 <br><br>Similarly, <br><br>Sum of 100^th places of all (3 + 6 + 3) 12 numbers is <br><br>= ( 1$\times$3 + 2$\times$6 + 3$\times$3) $\times$ 100 <br><br>and Sum of 1000^th places of all (3 + 6 + 3) 12 numbers is <br><br>= ( 1$\times$3 + 2$\times$6 + 3$\times$3) $\times$ 1000 <br><br>$\therefore$ Total sum = ( 1$\times$3 + 2$\times$6 + 3$\times$3) + ( 1$\times$3 + 2$\times$6 + 3$\times$3) $\times$ 10 <br><br>+ ( 1$\times$3 + 2$\times$6 + 3$\times$3) $\times$ 100 + ( 1$\times$3 + 2$\times$6 + 3$\times$3) $\times$ 1000 <br><br>= (3 + 12 + 9) (1 + 10 + 100 + 1000) = 1111 $\times$ 24 = 26664