Why is the binomial coefficient $\binom00$ equal to 1? I can understand why the generalized binomial coefficient $\binom a0$ equals $1$ when $a$ is not zero.
Here $a$ can be any real number.
Why? Well, we have:
$$\binom {a}{n} = \frac{a(a-1)\dots(a-(n-1))}{1.2.3 \dots n}$$
$$\Rightarrow \binom {a}{n+1} = \binom {a}{n} \cdot \frac{(a-n)}{n+1}$$
Now if we plug $n=0$ in the above, that gives us:
$$\binom {a}{1} = \binom {a}{0} \cdot a$$
which is the same as:
$$a = \binom {a}{0} \cdot a$$
So when $a$ is non-zero, we cancel the $a$ term. So we must have
$$\binom {a}{0} = 1$$
That is clear.
But why is $\binom {a}{0}$ also equal to $1$ when $a$ is zero?
At least that is what WolframAlpha says, that it's $1$.
But I find it somewhat counter intuitive. Why is it not zero, for example?
 A: How many subsets of size $4$ are of a set of size $7$?
$$7\choose{4}.$$
How many subsets of size $0$ (i.e., empty sets) are there in a set of size $0$?
$$0\choose 0$$
which is $1$, because the empty set has exactly $1$ subset of size zero, namely itself.
A: One reason might be from the binomial theorem:
$$(a+b)^n=\sum_{k=0}^n\binom nka^kb^{n-k},\quad n\in \mathbb N$$
When $n=0$, $$(a+b)^0=1=\sum_{k=0}^0\binom0ka^kb^{0-k}=\binom00a^0b^0=\binom00,\quad a\ne -b$$
A: You can see it in two ways:

*

*$\binom{a}{b}$ corresponds to the number of subsets of a set $E$ with $a$ elements, that have exactly $b$ elements. That means that $\binom{0}{0}$ corresponds to the number of subsets of the empty set, that have no elements. There is one set corresponding to that description, the empty set itself, which means: $\binom{0}{0} = 1$


*With the formula for the binomial coefficient: We know that $\binom{n}{k} = \frac{n!}{k!(n-k)!}$, from which we can deduce that $$\binom{0}{0} = \frac{0!}{0!(0-0)!} = \frac{1}{1 \times 1} = 1$$
Knowing that $0! = 1$
A: If you want to stick to a combinatorial interpretation, $\binom{n}{k}$ is the number of length-$n$ binary strings with exactly $k$ ones. There is exactly $1$ length-$0$ binary string with exactly $0$ ones.
But it is even possible to extend in a natural way to negative $n$, namely to define $C : ℤ×ℕ→ℤ$ such that:

*

*$C(n,0) = 1$ for every $n∈ℤ$.

*$C(0,k+1) = 0$ for every $k∈ℕ$.

*$C(n+1,k+1) = C(n,k)+C(n,k+1)$ for every $n∈ℤ$ and $k∈ℕ$.

It can be shown that such a $C$ exists, and this yields the correct extension of binomial coefficients that works with the Newton series.
It must be emphasized that you can define anything in any way you like, but whether or not they have nice properties is another matter. If you want the Newton series to have binomial coefficients as the basis, and the forward-difference operator to be just a shift of the coefficients, then you have no choice but to use this definition.
