$C(n,p) = C(n,q) => p + q= n$ How to prove the following theorem?
Theorem:
If the binomial numbers $C(n,p)$ and $C(n,q)$ are equal, with $p$ different from $q$, then $p+q = n$.
($n$, $p$ and $q$ are natural numbers)
 A: The binomial coefficients are Unimodal.
A: It is suffices to show that $q! (n-q)! = p! (n-p)!$ implies $p+q=n$ ($p \ne q$). Assume without loss of generality that $p > q$. Then, we need to show that $p!/q! = (n-q)!/(n-p)!$ implies $p+q=n$, or $(q+1) (q+2) \cdots p = (n-p+1) (n-p+2) \cdots (n-q)$ implies $p+q=n$. Since both sides of the last equality have identical number of terms ($p-q$), we must have $q+1 = n-p+1$, that is $p+q=n$.
A: Mike Spivey correctly points out that unimodality does not suffice. Here's a quick way to fill the gap:  Consider the set $S = \{p,q,n-p,n-q\}$. Now $n\choose x$ assumes only one value for $x\in S$, by hypothesis and the fact that ${n\choose p}={n\choose n-p}$. But $p\neq q$ and $p+q\neq n$ together imply that $S$ has at least three distinct values.  This plus unimodality gives a contradiction.
A: HINT $\ $ In order to prove that $\rm\ \binom{n}{p}\ =\ \binom{n}{q}\ \:\Rightarrow\ \  \ p+q = n\ \ \ if\ \ \ p \ne q\ \ $ for $\rm\ \ p,q\in [0, n]$
consider how to prove that $\rm\ \ \ sin(p)\ =\ sin(q)\ \Rightarrow\ p+q = \pi\ \ \ if\ \ \ p \ne q\ \ $ for $\rm\ \ p,q\in [0,\pi]$
MORAL $\ $ A proper choice of $\rm \:sin\:$ can get you "over the hump".
A: Suppose that for a fixed $n$ we have the equality $\binom{n}{p} = \binom{n}{q}$ with $p \neq q$. Observe that
\begin{eqnarray}
0 = \frac{\pi}{2^{n-1}} \left[ \binom{n}{p} - \binom{n}{q} \right] = \int_{-\pi}^{\pi} \left( \cos (2p - n)x - \cos (2q - n)x \right) \ \cos^{n} x \ dx.
\end{eqnarray}
The cosine difference identity
\begin{eqnarray}
\cos A - \cos B = -2 \sin (\tfrac{A + B}{2}) \ \sin (\tfrac{A - B}{2})
\end{eqnarray}
implies 
\begin{eqnarray}
\int_{-\pi}^{\pi} \left( \sin (p + q - n)x \  \sin (p- q)x \right) \ \cos^{n} x \ dx = 0 .
\end{eqnarray}
I'll leave it as an exercise to finish the proof and conclude the result.
