jmc

Note that there are $4!=24$ numbers ending in 1, since we have 4 choices for the 10s digit, 3 choices for the 100s digit, 2 choices for the 1000s digit, and 1 choice for the remaining digit. Thus there are also 24 numbers ending in each of 3, 4, 5, 9, and the total contribution of ones digits to the sum is $24(1+3+4+5+9)=528$. But we can make a similar argument about the contribution of the digits in the other places (10s, 100s, etc.), so our total sum is $528+5280+\dots +5280000=528(1+10+\dots +10000)=528\cdot 11,111=\boxed{5,866,608}$.