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If $G$ be a set and $f:G \times G \longmapsto G$ be a binary operation on G then $f$ is called associative on $G$ if $a(bc)=(ab)c$ for all a,b in $G$.

My question is that for a finite number of elements $a_1,a_2,...,a_n$ how many ways are there to operate the element for a particular ordering of them .

Let's call a function $N(m)$ for $m$ elements defined by the above condition . Recall that by associative law $a_1(a_2a_3)$ and$(a_1a_2)a_3$ are same and this are the only ways to operate elemets$a_1a_2a_3$ so $N(3)=2$

Now consider four elements $a_1,a_2,a_3,a_4$.The only way to operate this elements in this order is $((a_1a_2)a_3)a_4$,$(a_1a_2)(a_3a_4)$,$(a_1(a_2a_3))a_4$,$a_1((a_2a_3)a_4)$ and $a_1(a_2(a_3a_4))$ thus $N(4)=5$

So can we generalize thus process to get the value $N(m)$ in general?

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You want to count certain binary trees. The result is given by Catalan numbers. With $n$ denoting that number of leaves, this can be expressed as$$\frac1{n}{2(n-1)\choose n-1} =\frac{(2n-2)!}{(n-1)!\,n!}.$$

For example, $n=3$ leads to $\frac{4!}{3!2!}=2$ and $n=4$ leads to $\frac{6!}{4!3!}=5$.

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