Minimize $\binom{m}{k}+\binom{n-m}{k}$ Fix integers $n, k,$
I want to find an integer $m$ that minimize $\binom{m}{k}+\binom{n-m}{k}$.
I have already tried to calculate the difference between the values for $m$ and $m+1$,
$$
\binom{m+1}{k}+\binom{n-m-1}{k}-\binom{m}{k}-\binom{n-m}{k} = \binom{m}{k} \frac{k}{m-k+1}-\binom{n-m-1}{k} \frac{k}{n-m-k}.$$
(Sorry for the typo)
It seems hard to determinate whether it is positive or negative.
Is there any strategy to deal with this problem?
 A: Hint1: What is the smallest posible value of (n m)?
Hint2: What kind of number can n - m be? You can't calculate binomial for every input...
EDIT:
I somehow missed that n and k must be fixed. Ideas of this answer will help you get solution if they are not.
A: I think the simplest way to think about it is in terms of choices:
$$
\begin{align*}
\binom{m}{k} \ & \colon \ \text{The total number of ways to pick $k$ objects from a total of $m$} \\ 
\binom{n-m}{k} \ & \colon \ \text{The total number of ways to pick $k$ objects from a total of $n-m$} 
\end{align*}
$$
It should be clear that as the total number of objects increases, so does the number of ways to pick $k$ objects from the total. In other words $\binom{m}{k}$ increases as $m$ increases, and $\binom{n-m}{k}$ decreases as $m$ increases. 
By symmetry (the first binomial factor increases as much as the last decreases) the minimum is obtained when both binomial factors are the same. This is obtained when 
$$ 
    m = n - m \,.
$$
Which you should be able to solve. Let 
$$
f(m) = \binom{m}{k} + \binom{n-m}{k}
$$
I will leave it to you showing that the function is symmetric about $n/2$, that is 
$f(n/2-m) = f(n/2+m)$
Thus extrema must occur at $m = n/2$. 

I was not able to give a good argument to how to prove that this is an minima without calculus. It is just logical to me.. Lock $k=2$
$$
h(m,n) = \binom{m}{2} + \binom{n-m}{2}
$$
We introduce the new variable $m = pn$, where $p$ can be anything. Then
$$
T(p) := \frac{h(pn,n)}{h(n/2,n)} = 2\frac{n-1}{n-2}+4\frac{pn(p-1)}{n-2)}
$$
Differentiation gives
$$
\frac{\partial T}{\partial p} = 4\frac{n(2p-1)}{n-2} 
\quad \text{and} \quad 
\frac{\partial^2 T}{\partial p^2} = \frac{8n}{n-2} 
$$
Thus proving that $T(p)$ is minimal for $p=1/2$, since the double-derivative is always positive. 

Let us look at $\binom{m}{n} + \binom{n-m}{k}$ when $m=n/2$ then we have
$$ g(n,k) := 2\binom{n/2}{k} $$
If $n$ is even it is clear that $g(n,k)$ produces. Thus assume that $n = 2p+1$ is odd. Then 
$$
g(2p+1,k) = \frac{2}{k!} \frac{(p+1/2)!}{(p+1/2-k)!} = \frac{2}{k!} \prod_{j=0}^{k-1} \left( p + \frac{1}{2} - j\right)\,,
$$
again just a nice simple fraction. 
