Number of possible products If I have set (2,3,4,5,6,7,8,9), how many possible products are there of any three numbers from the set. Repeats are allowed e.g. 2*2*2, 2*3*3, etc.
I don't know how to solve this other than brute force. Thx a lot.
The number of products consisting of $3$ elements with repetition from the $8$ element set 
  \begin{align*}
\{2,3,4,5,6,7,8,9\}=\{2,2^2,2^3,2\cdot 3, 3,3^2,5,7\}
\end{align*}
  is
  \begin{align*}
\binom{8+3-1}{3}=\binom{10}{3}=\color{blue}{120}
\end{align*}


Here the strategy is to count  the products which  occur more than once manually and then subtract this number from $120$.  


We can see at a glance  that  products having two elements from $\{5,7\}$ cannot have two different representations. Therefore candidates have either a factor $5$ or $7$ or none of them.


We need some kind of systematic way to find all possible products with multiple representations. In order to do so we arrange all candidates  according  to  two  criterias:


The product contains either a  factor $5$ or  $7$ or none  of them
We sort candidates by increasing  powers of  $2$



  Products containing  $5$ or $7$:
  
  We only need to list two factors, since the third one is either $5$ or $7$. We obtain
  \begin{align*}
\text{factor }2\ :&amp;\qquad 2\cdot 3^2=(2\cdot 3)\cdot 3&amp;(90,126)\\
\text{factor }2^2:&amp;\qquad  2^2\cdot3=2\cdot(2\cdot 3)&amp;(60,84)\\
&amp;\qquad 2^2\cdot  3^2=(2\cdot  3)(2\cdot 3)&amp;(180,252)\\
\text{factor }2^3:&amp;\qquad  2^3\cdot 3=2^2\cdot (2\cdot 3)&amp;(120, 168)\\
\text{factor }2^4:&amp;\qquad 2\cdot 2^3=2^2\cdot 2^2&amp;(80,112)\\
\end{align*}
  
  We see there are $5$ possibilities  to  write   a product  with two different representations. Since the third factor is either $5$ or  $7$ we have a total of $\color{blue}{10}$  possibilities. The bracketed values on the right of each line give the number with factor  $5$ resp. $7$.
  
  Products containing neither $5$ nor $7$:
  
  \begin{align*}
\text{factor }2\ :&amp;\qquad   2\cdot  3\cdot  3^2=(2\cdot 3)\cdot 3\cdot 3=54\\
&amp;\qquad 2\cdot 3^2\cdot 3^2=(2\cdot 3)\cdot   3\cdot 3^2=162\\
&amp;\qquad \rightarrow 2\text{  multiples}\\
\text{factor  }2^2:&amp;\qquad 2\cdot 2\cdot 3^2=2^2\cdot 3\cdot 3=2\cdot (2\cdot 3)\cdot 3=36\\
&amp;\qquad 2^2\cdot 3\cdot 3^2=2\cdot (2\cdot 3)\cdot 3^2=(2\cdot 3)(2\cdot 3)\cdot 3=108\\
&amp;\qquad 2^2\cdot 3^2\cdot 3^2=(2\cdot 3)(2\cdot 3)\cdot 3^2=324\\
&amp;\qquad \rightarrow 5\text{  multiples}\\
\text{factor  }2^3:&amp;\qquad 2\cdot 2^2\cdot 3=2\cdot 2\cdot (2\cdot 3)=24\\
&amp;\qquad 2\cdot 2^2 \cdot  3^2=2^2\cdot(2\cdot3)\cdot      3=2^3\cdot 3\cdot  3=2\cdot(2\cdot3)(2\cdot 3)=72\\
&amp;\qquad 2^3\cdot 3\cdot   3^2=2^2\cdot (2\cdot  3)\cdot 3^2=(2\cdot  3)(2\cdot 3)(2\cdot 3)=216\\
&amp;\qquad \rightarrow 6\text{  multiples}\\
\text{factor }2^4:&amp;\qquad 2\cdot2^3\cdot    3=2^2\cdot2^2\cdot    3=2\cdot 2^2\cdot (2\cdot 3)=48\\
&amp;\qquad 2\cdot 2^3\cdot  3^2=2^2\cdot2^2\cdot   3^2=2^3\cdot(2\cdot3)\cdot3=2^2\cdot   (2\cdot      3)(2\cdot  3)=144\\
&amp;\qquad \rightarrow 5\text{  multiples}\\
\text{factor  }2^5:&amp;\qquad   2\cdot 2\cdot2^3=2\cdot2^2\cdot2^2=32\\
&amp;\qquad   2^2\cdot   2^3\cdot 3=2\cdot  2^3\cdot (2\cdot  3)=2^2\cdot   2^2\cdot   (2\cdot  3)=96\\
&amp;\qquad       2^2\cdot  2^3\cdot 3^2=2^3\cdot   (2\cdot    3)(2\cdot   3)=288\\
&amp;\qquad \rightarrow 4\text{  multiples}\\
\text{factor }2^6:&amp;\qquad     2\cdot  2^2\cdot 2^3=2^2\cdot    2^2\cdot 2^2=64\\
&amp;\qquad 2^3\cdot 2^3\cdot 3=2^2\cdot  2^3\cdot (2\cdot 3)=192\\
&amp;\qquad \rightarrow 2\text{  multiples}\\
\text{factor }2^7:&amp;\qquad 2\cdot 2^3\cdot 2^3=2^2\cdot 2^2\cdot 2^3=128\\
&amp;\qquad \rightarrow 1\text{  multiple}\\
\end{align*}
  giving a total of $2+5+6+5+4+2+1=\color{blue}{25}$ multiple product representations.
  
  We conclude the number of different products is $$120-10-25=\color{blue}{85}.$$
$8^3$ (3 slots, 8 possibilities).

EDIT: This doesn't account for product degeneracy.