Is this function increasing in $x$? Say we have $f(x)=\frac{x!}{(x-n)!n!}$ for any $n\geq1$ and integer and $x>n$ and integer.
How can we show whether it's increasing or decreasing in $x$?
I was thinking, if $x_1\leq x_1$, then $x_1!\leq x_2!$ and $(x_1-n)!\leq(x_2-n)!$ but $\frac{1}{(x_1-n)!}\geq \frac{1}{(x_2-n)!}$ so is it possible to show that $f(x_1)\leq f(x_2)$?
I'm more interested in learning about a general approach (if there's any) to tackle problems like this and some guidance will be much appreciated.
 A: Your approach is correct, now since we have $x$ in both the numerator and the denominator, we need to ask ourselves, which one of them grows faster with $x$, or if one of them cancels out with respect to the other.
In this case, the numerator $x!$ can be written as $x! = (x-n)!\cdot (x - n +1)\cdot...\cdot (x-1)\cdot x$.
Therefore
$\frac{x!}{(x-n)!} = \frac{(x-n)!\cdot (x -n +1)\cdot...\cdot (x-1)\cdot x}{(x-n)!} = (x -n +1)\cdot...\cdot (x-1) \cdot x$
Meaning that the function will grow with $x$.
A: Note that $x!=x\cdot(x-1)\dots(x-n+1)\cdot(x-n)!$ so $f(x)=\frac{x\cdot(x-1)\dots(x-n+1)}{n!}$. Let $k$ be a positive integer. Then
\begin{align}
f(x+k)&=\frac{(x+k)\dots (x+k-n+1)}{n!}\\
&>\frac{x\dots (x-n+1)}{n!}=f(x).
\end{align}
Of course, there's a physical interpretation of this formula. This formula is also known as $x\choose n$, the number of groups of size $n$ that can be constructed from a population of size $x$. Obviously, increasing the population increases the number of possible groups.
