How many unique combination of $n$ numbers are there when multiplying $x$ numbers? If you have $n$ numbers and you multiply them in every combination
$$A\cdot A,\  A\cdot B,\ B\cdot A,\ B\cdot B$$
and so on, how many times do you get the same answer? Or, how many times do you get a unique answer? That is just for a product of two numbers, what if you did product of $3, 4,... n$ numbers. Is there a formula that could calculate this? 
An example would be, let's say you have numbers $2,3,5,7$. You can do
$$2\cdot 2,\; 2\cdot 3,\; 3\cdot 2, \;3\cdot 3, \;2\cdot 5, \;5\cdot 2, \;3\cdot 5, \; 5\cdot 3,\; 5\cdot 5,\; 2\cdot 7,\; 7\cdot 2, \;3\cdot 7, \;7\cdot 3, \;5\cdot 7,\; 7\cdot 5, \;7\cdot 7$$
With that you get 16 combinations and only 10 unique answers.
 A: We assume that we are given a set $S=\{s_1,s_2,\dots,s_n\}$ of $n$ distinct numbers, and want to count the number of numerically distinct products  $a_1a_2\cdots a_k$, where the $a_i$ range over $S$.  
The answer very much depends on numerical properties of the numbers in the set $S$. We will find the maximum possible number of numerically distinct products. We get this maximum if, for example, the elements of $S$ are distinct primes, and in many other cases. 
Let $P$ be a product of $k$ elements of $S$. Let $x_1$ be the number of copies of $s_1$ that we used to form $P$. Let $x_2$  be the number of copies of $s_2$ that we used, and so on.  
Then the $x_i$ are non-negative integers, and $x_1+\cdots+x_n=k$. Furthermore, under our assumptions any solution of $x_1+\cdots +x_n=k$ in non-negative integers gives a numerically distinct product.
Thus by Stars and Bars (please see Wikipedia) the number of numerically distinct products, under our assumptions, is $\binom{n+k-1}{n-1}$ or equivalently $\binom{n+k-1}{k}$.  
Remark: The case $n=4$, $k=2$ of the OP does not require the above machinery, and neither does $n$ arbitrary, $k=2$. But note that for the case $k=2$, the general formula above gives the answer $\binom{n+1}{2}$.
