Every 4-digit number that uses each of the digits 0, 1, 2 and 3 exactly once is written down. What is the total of all these numbers?

Allowing 0 to lead, each digit appears $3!=6$ times in each place: $6\times (0+1+2+3)\times 1111=39996$. Remove the 'numbers' with 0 in front (3-digit numbers from 1, 2, 3): each digit appears twice per place, $2\times 6\times 111=1332$. Sum $=39996-1332=38664$.