Why is this binomial equation not an identity between polynomials?

Question

In the book Concrete Mathematics, the following identity appears: $$(r-k){r\choose k} = r {r-1\choose k}$$ It is followed by a proof that it holds for all real r. This proof makes use of the property that

both sides of [the identity] are polynomials in r of degree k+1.

So far so good. However, the passage goes on by claiming that the following identity cannot be extended to real values for r this way, because it is not and identity between polynomials. $${n \choose k} = {n \choose n-k}$$

How can the factors (r-k) and r make the first identity polynomial when the second is not; am I missing some other difference that is critical?

Answer 1

We recall the definition (5.1) of binomial coefficients stated in chapter 5 Binomial coefficients with $\alpha\in\mathbb{C}$ and integer values $p$

\begin{align*} \binom{\alpha}{p}= \begin{cases} \frac{\alpha(\alpha-1)\cdots(\alpha-p+1)}{p!}&p\geq 0\\ 0&p<0 \end{cases}\tag{1} \end{align*}

The identity $(r-k)\binom{r}{k}=r\binom{r-1}{k}$ has a representation according to (1) as \begin{align*} \frac{1}{k!}(r-k)r(r-1)\cdots(r-k+1)=\frac{1}{k!}r(r-1)(r-2)\cdots(r-k) \end{align*} Both sides coincide and are polynomials in $r$ of degree $k+1$. On the other hand the identity $\binom{n}{k}=\binom{n}{n-k}$ has a representation according to (1) as \begin{align*} \frac{1}{k!}n(n-1)\cdots(n-k+1)=\frac{n(n-1)\cdots(n-(n-k)+1)}{(\color{blue}{n}-k)(\color{blue}{n}-k-1)\cdots (\color{blue}{n}-k-(n-k)+1)}\tag{2} \end{align*}

In (2) we see the right hand side is a rational function with numerator and denominator being polynomials in $n$ of degree $n-k$.

Since the argument stated in the book is valid for polynomials only but not for rational functions in general, it cannot be applied to (2).