3. What is the sum of all four-digit numbers written using the digits $1,2,3$, 4, 5, 6, 7, 8 and 9 such that the digits do not repeat?
First method:

In those numbers, each of those digits appears in each decimal place

$8 \cdot 7 \cdot 6=336$ times.

Therefore, the sum of the digits in each decimal place is equal to

$336 \cdot 9+336 \cdot 8+336 \cdot 7+336 \cdot 6+336 \cdot 5+336 \cdot 4+336 \cdot 3+336 \cdot 2+336 \cdot 1=$

$=336 \cdot(1+2+3+4+5+6+7+8+9)=336 \cdot 45=15120$.

Then the sum of all those numbers is

$15120 \cdot 1000+15120 \cdot 100+15120 \cdot 10+15120 \cdot 1=15120 \cdot(1000+100+10+1)=$

$=15120 \cdot 1111=16798320$.