Why compare f(n)/f(n-1) = 1 to solve for the maxima of a discrete function? I am aware of a general strategy where you have a discrete function, e.g.,
$f = \dfrac{{10 \choose 5}{n-10 \choose 15}}{n \choose 20}$
And in order to find the maximum, you solve $\frac{f(n)}{f(n-1)} = 1$.
I may be mistaken in this method applying to discrete functions only... But the bottom line is that I am trying to understand why this method works, and then understand how broadly this strategy can be used (in what situations).
I appreciate any insight. I feel like there must be something intuitive I am missing.
 A: It's the discrete equivalence of finding the point(s) for which the derivative is $0$ in the differentiable case, i.e. finding local extrema. 
One definition of the derivative is
$$f'(a) = \lim_{h\to 0}\frac{f(a+h)-f(a)}{h}$$
A discrete counterpart would be to remove the limit and set $h$ to the smallest increment available, i.e. 1. Then you get
$$\frac{f(n+1)-f(n)}1$$
Setting this property to $0$ will give you 
$$f(n+1)-f(n) = 0 \implies \frac{f(n)}{f(n+1)}=1$$
Is this a maximum? You can't be sure, you only know it is a local extremum. Equivalent to the continuous case, you'll have to check the sign of the second "derivative" to decide if it's a maximum or minimum.
A: The idea is that (if $f(n) \gt 0$), $\frac {f(n)}{f(n-1)} \gt 1 \implies f(n) \gt f(n-1)$, so if I can find an $n$ such that $\frac {f(n)}{f(n-1)} \gt 1 $ and $\frac {f(n)}{f(n+1)} \gt 1 $ then $n$ is a (local) maximum of $f(n)$.  Note that I have used inequalities where you have equality.  We are looking for the point where the inequality changes sense.
In your example, you should be able to convince yourself that $\frac {f(n)}{f(n-1)}=\frac {(n-10)(n-20)}{(n-25)n}$, which equals $1$ at $n=40$.  This gives us $f(39)=f(40)$, then the functions starts decreasing.
