What is the sum of all of the products of $3$ of the digits $1, \dots, 9$? Consider the numbers $1, 2, 3, \dots, 9$.  Take the product of any three of them.  What is the sum of all such products?
In other words, calculate $1 \cdot 2 \cdot3 + 1 \cdot 2 \cdot 4 + 1 \cdot 2 \cdot 5 + \dots + 7 \cdot 8 \cdot9$.
If we consider products of four numbers or more, what's the answer?

Use the Principle of Inclusion and exclusion. 
Your sum is equal to 
$$ \left[ (1 + 2 + \ldots + 9) ^3 - 3 \times ( 1^2 + 2^2 + \ldots + 9^2) \times (1 +2 + \ldots + 9 ) + 2 \times (1^3 + 2^3 + \ldots + 9^3) \right] \div 6$$
As an explanation, you can see if that $a\neq b, b\neq c, c\neq a$, then it will appear 6 times in the first time, 0 times in the second, 0 times in the third, hence appear $6/6 =1 $ time.
Terms of the form $aab$ will appear thrice in the first term, thrice in the second (which is subtracted), 0 times in the third, hence appear $(3 - 3)/6 = 0 $ times.
Terms of the form $aaa$ will appear once in the first term, thrice in the second, twice in the third, for a total of $(1 - 3 + 2)/6 = 0 $ times.
