Let the total possible arrangement for $n$ numbers be $f(n)$. Show that $f(n)=\sum_{i=0}^{n-1}f(i).f(n-1-i)$

$f(0)=1 \\ f(1)=1 \\ f(2)=2$

You can calculate $f(9)$ from here.

For the general case of calculating $f(n)$,

$f(n)=$ the coefficient of $x^{n-1}$ of  ${f(0)+f(1)x+f(2)x^2 + \cdot}^2$

Multiply the equation on the right and $x $  $f(n)$ becomes the coefficient of $x^n$.

Therefore,

$ x{f(0) + f(1)x+f(2)x^2+ \cdots}^2 = f(1)x+f(2)x^2+ \cdots $

Let $f(0)+f(1x)+ \cdots = P(x) $.

$ xP(x)^2=P(x)-1$

$\implies P(x)= \frac{1 \pm \sqrt{1-4x}}{2x}$

Determine the Maclaurin series of $\sqrt{1-4x} $ and put it in the equation.