figuring out an integer function $f(1) = 1\\
f(2) = 2\\
f(3) = 6\\
f(4) = 20\\
f(5) = 70\\
f(6) = 252\\
f(7) = 924\\
f(8) = 3432\\
f(9) = 12870$
Then what is $f(n)$ (where $n > 0$)?
I though about many many possibilities but still cannot figure out the expression.

You could approach this kind of question in this way. You'd like to find what is the relation between $f_n$ and $f_{n+1}$ (you believe that there is some)  so that's why most of the time it's a good idea to analyze expressions like $\frac{f_{n+1}}{f_n}$ or $f_{n+1}-f_n$. Let's look at $\frac{f_{n+1}}{f_n}$. You may notice that $\frac{f_{n+1}}{f_{n}}\sim 4$ as the sequence grows, more precisely:
$$
\begin{array}{ccl}
 \frac{f_2}{f_1}& = & 2 \\
 \frac{f_3}{f_2} & = & 3 \\
 \frac{f_4}{f_3} & = &3.(3) \\
 \frac{f_5}{f_4} & = & 3.5 \\
 \frac{f_6}{f_5} & = & 3.6 \\
 \frac{f_7}{f_6} & = & 3.(6)\\
 \frac{f_8}{f_7} & = & 3.(714285) \\
 \frac{f_9}{f_8} & = & 3.75 \\
\end{array}
$$
So you might expect that $\frac{f_{n+1}}{f_{n}} = 4 - g_n$ where $g_n \to 0$ as $n\to +\infty$. Upon more precise calculation we get: 
$$
\begin{array}{ccl}
 \frac{f_2}{f_1}& = & 4-\frac{2}{1} \\
 \frac{f_3}{f_2} & = & 4-\frac{2}{2} \\
 \frac{f_4}{f_3} & = &4-\frac{2}{3} \\
 \frac{f_5}{f_4} & = &4-\frac{2}{4} \\
 \frac{f_6}{f_5} & = &4-\frac{2}{5} \\
 \frac{f_7}{f_6} & = & 4-\frac{2}{6}\\
 \frac{f_8}{f_7} & = & 4-\frac{2}{7} \\
 \frac{f_9}{f_8} & = & 4-\frac{2}{8} \\
\end{array}
$$
So we could see that $\frac{f_{n+1}}{f_n} = 4-\frac{2}{n} = \frac{2(2n-1)}{n}$. From here 
$$\frac{f_{n+1}}{1} =\frac{f_{n+1}}{ f_n } \times \frac{f_{n}}{f_{n-1}} \times \dots \times \frac{f_2}{f_1} = \frac{2(2n-1)}{n}\times \frac{2(2n-3)}{n-1} \dots  = \frac{2^n(2n-1)!!}{n!}$$
which is $\binom{2n}{n}$ as already mentioned
