
The sum of all the numbers that can be formed with any given set of four different figures is always 6,666 multiplied by the sum of the four figures. Thus, 1, 2, 3, 4 add up 10, and ten times 6,666 is 66,660. Now, there are thirtyfive different ways of selecting four figures from the seven on the dice—remembering the 6 and 9 trick. The figures of all these thirtyfive groups add up to 600. Therefore 6,666 multiplied by 600 gives us 3,999,600 as the correct answer.
A different solution, though with the same final result has been sent
in by Craig, one of our visitors. It is provided below.
If we consider the first digit in all its permutations, we have a sum
of 30 (1+2+3+4+5+6+9). The remaining dice are limited in choice to 6
options, then 5 options, then 4 options (avoiding duplicates), so
there are 120 (6x5x4) combinations times 30 = 3600, which is the sum
for each die over all permutations. The dice have place values of
1000, 100, 10, and 1, so the total sum of all combinations is 3600 x
1111 = 3,999,600.
We want to thank Craig for this!

