An exercise in chapter 2 of Spivak's Calculus (4th ed.) talks about how Pascal's triangle gives the binomial coefficients. It explains this by saying that the relation $\binom{n+1}{k} = \binom{n}{k-1}+\binom{n}{k}$. I'm having trouble seeing how this equation gives rise to Pascal's triangle, so any explanation of what's really going on would be helpful, thanks.
[Math] Why does Pascal’s Triangle give the Binomial Coefficients
binomial-coefficientscombinatorics
Related Solutions
For the non-trivial interpretation, you're looking for non-negative solutions of $a + b + c = n$ (each of these corresponds to a term $x^a y^b z^c$). Code each of these solutions as $1^a 0 1^b 0 1^c$, for example $(2,3,5)$ would be coded as $$110111011111.$$ Now it should be easy to see why the answer is $\binom{n+2}{n}$.
To your first question, yes, it does make sense to define $P(n,r)=\prod_{i=0}^{r-1}(n-i)$, thus extending domain of $n$ to all of $\mathbb C$ (or really any commutative ring with identity). Then $P(n,r)$ is also called the falling factorial, sometimes denoted $n^{\underline r}\,$.
Just like $C(n,k)$ counts sets of size $k$ for $n>0$, there is a nice combinatorial interpretation $|C(-n,k)|$; it counts multisets of size $k$. There is a similar duality for $P(n,k)$. Whereas $P(n,k)$ counts ordered lists without repetition, $|P(-n,k)|$ counts the number of ways to place $k$ distinct flags on $n$ distinct flagpoles, where the order of the stack of flags on each pole matters. Note that $|P(-n,k)|=n(n+1)\cdots (n+k-1)$ is the rising factorial, denoted $n^{\overline k}$. This is sort of like an ordered version of a multiset. You can think of the flagpoles as the objects being chosen, and putting a flag on it an object represents choosing it. You can put multiple flags on a pole, so you can choose an object multiple times. Since the flags are distinct, order matters.
Finally, the reason that Wolfram Alpha cannot compute $P(n,r)$ when $n$ is a negative integer is because $(-n)!$ is undefined. Specifically, the Gamma function has a simple pole at each nonnegative integer. However, we can still calculate $P(n,r)$ using $n!/(n-r)!=\Gamma(n+1)/\Gamma(n-r+1)$. Since we have a simple pole in the numerator and denominator, they effectively cancel. The ratio now has a removable singularity at each nonnegative integer, and we can define $P(n,r)$ as the limit for $n$ in the neighborhood of the negative integer. In this case, we recover the expected $P(n,r)=\prod_{i=0}^{r-1}(n-i)$ result.
Best Answer
If $(n,k)$ is the $k$th entry of the $n$th row of Pascal's triangle, then we have the following equation from the way Pascal's triangle is built:
$$(n+1,k)=(n,k-1)+(n,k)$$
Notice the similarity with the binomial coefficient identity you mention. Now, since $(n,1)=(n,n)=1$, and since $\binom{n}{1}=\binom{n}{n}=1$, by induction it follows that $(n,k)=\binom{n}{k}$.
Explicitly, if $(n,k)=\binom{n}{k}$ for all $k=1,\ldots,n$, then we have
$$(n+1,k)=(n,k-1)+(n,k)=\binom{n}{k-1}+\binom{n}{k}=\binom{n+1}{k}$$
for all $k=1,\ldots,n$, and by definition, we already have $(n+1,n+1)=1=\binom{n+1}{n+1}$