You have a multiset containing one $3$, two $4$’s, two $5$’s, one $6$, and three $7$’s, for a total of $9$ elements. You must pick at least $2$ of these $9$ elements to form your product, and the question is how many different products you can form in this way.
A set of $9$ elements has $2^9$ subsets; one of them is empty, and $9$ have only one member each, so you’re down to $2^9-10$ subsets containing at least $2$ of these $9$ numbers. However, some of these subsets generate the same product. For instance, there are $\binom32=3$ ways to pick two of the $7$’s, so there are $3$ ways to get the product $49$. There are $2\cdot2=4$ ways to get the product $20$, since there are two $4$’s and two $5$’s. A case by case analysis like this looks very messy, however.
Notice that the only prime factors available are $2,3,5$, and $7$, so every product must have the form $2^a3^b5^c7^d$ for some non-negative integers $a,b,c,d$. It’s also clear that $a\le 5,b\le 2,c\le 2$, and $d\le 3$. One way to approach the problem is to count the $4$-tuples $\langle a,b,c,d\rangle$ that can actually be formed.
Suppose that we use the $6$, so that $a\ge 1$ and $b\ge 1$. The possible values of $a$ are then $1,3$, and $5$, depending on whether we use $0,1$ or $2$ of the $4$’s. The possible values of $b$ are $1$ and $2$. The possible values of $c$ are $0,1$, and $2$, and the possible values of $d$ are $0,1,2$, and $3$. That’s a total of $$3\cdot2\cdot3\cdot4=72$$ combinations, but one of them is $a=1=b$ and $c=d=0$, which uses only one of the $9$ numbers, namely, the $6$. Thus, there are $71$ different products that can be formed using the $6$.
Can you now count the products that can be made without using the $6$?