How many answers can be created using the elementary arithmetic operators? If I gave you an amount of $n$ numbers, how many anwswer will you be able to create using the elementary arithmetic operators ($+, -, \times, /$)? 
These are the rules:


*

*All numbers $\in\mathbb{Q}_{0>}$.

*All numbers are different ($a\neq b \neq c \neq \cdots$).

*All numbers in the answer must be used and can only be used once.

*An operator can be used several times or zero times.

*Use of parentheses is allowed.


Lets's have a look at some examples:
We have 2 numbers called $a$ and $b$;


*

*$a + b$ or $b+a$

*$a-b$

*$b-a$

*$b\times a$ or $a\times b$

*$b/a$

*$a/b$


Not allowed: 


*

*$a\times a+b$


We have 3 numbers called $a, b$ and $c$;


*

*$a+b+c$ or ...

*$a\times b\times c$ or ...

*$a-b-c$

*$(a+b)\times c$

*$(a/b)/c$

*etc..


Please edit if necessary!
 A: If we ignore all "coincidences", including associativity and commutativity where it applies, we can enumerate all valid exprsssions by


*

*arranging the $n$ numbers in one of $n!$ orders

*insertingof parentheses among them in $C(n-1)$$=\frac{(2n-2)!}{(n-1)!n!}$ ways

*inserting operators among the subexpressions in $4^{n-1}$ ways


That gives us a total of $$\tag1\frac{4^{n-1}(2n-2)!}{(n-1)!} $$
expressions.
If one wants to regard commutativity, each occurance of $+$ or $\times$ leads to double-counting, that is: For each $k$, $0\le k\le n-1$, the number $\frac{2^{n-1}(2n-2)!}{(n-1)!}{n-1\choose k}$ expressions with $k$ commutative operators should be divided by $2^k$. This give us 
$$\tag2\begin{align}\sum_{k=0}^{n-1} 2^{-k}\frac{2^{n-1}(2n-2)!}{(n-1)!}{n-1\choose k}&=\frac{2^{n-1}(2n-2)!}{(n-1)!}\left(1+\frac12\right)^{n-1}\\&=\frac{3^{n-1}(2n-2)!}{(n-1)!}.\end{align}$$
The next goal would be to fully regard associativity. However it doesn't lend itself to simple investigations; one should leave the realm of binary trees and instead consider trees of arbitrary degree with twocoloured edges and I have no idea how to count these efficiently.
