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.
 A: *This contsins some mistake.  Be careful
I try writing about my trying, but you may think that is  brute force. And it may have some mistake.
we should think the form of $2^a3^b5^c7^d(0\le a\le9,0\le b\le6,0\le c\le 3,0\le d\le3)$.
when $c=1,d=1$, then we can choose a number from$\{2,3,4,6,8,9\}$, so we get 6 numbers. 
And think the case that $c=1,d\neq1$. If $4\le a\le6$,we can get only $(a,b)=(4,0),(5,0),(6,0),(4,1)$(for example,$2\times8\times5,4\times8\times5,8\times8\times5,6\times8\times5$) because 8 or double $4$ is necessasry to make $a$ not less than $4$.If $3\le b\le4$, also we get $(a,b)=(0,3),(0,4),(1,3)$.
And in the case that $0\le a \le 3$ and $0\le b \le 2$, other than $(a,b)=(0,0),(1,0),(0,1)$, we can get such a product(plese think a bit). So, we can know the number in the case equles to $4\times3-3=9$.
So, we now know the number in the case that $c=1,d\neq1$ equals to $9+4+3=16$, and the number in the case that $c\neq1,d=1$. It is equal. So, we can know the number in the case that $c=1$ or $d=1$. That is  $16\times 2+6=38$
Let's think about the case that $c=0$ and $d=1$. The form is $2^a3^b(a\le a\le 9,0\le b\le6)$.
if $7\le a\le9$, we get only $(a,b)=(7,0),(7,1),(8,0),(9,0)$, and if $5\le b \le6$, we get only $(a,b)=(0,5),(1,5),(0,6)$.
When $0\le a\le6$ and $0\le b\le4$, then that is a little difficult.
we can choose three number,so we know $a+b\ge3$ And, except $8$, each number has only two or one prime factor. So, we can think that
\begin{equation}3\le a+b\le8(a=6)\\3\le a+b\le7(3\le a\le5)\\3\le a+b\le6(0\le a\le2) \end{equation}
and if we think in the case classification on the value of $a$, we know the full pairs of $a,b$ which meet the former equation can be gotten. The number is $3+3+4+5+4+4+4=27$, so we know the number in case that $c=0$ and $d=1$ is $27+4+3=34$. Then, we know the answer is $34+38=72$
former answer:"
Because of counting multisets, the number is less than $\binom {10}{3}=120$. But I can't get an accurate number. Sorry"
