What happens to the formula of the binomial coefficient when k = 0?

Question

The formula for calculating a binomial coefficient is as follows : Binomial coefficient

My question is with the middle equation, when k = 0, the term n-k + 1 remains as n + 1 and then the numerator remains as n (n-1) ... (n + 1), I cannot understand why the term n + 1 remains. Also, this does not match the equation on the right side, for example (4 0) with the equation in the middle it remains as 4 * 5/0! but with the equation on the right it remains as 4! / (0!) (4-0)!

Answer 1

The middle equation is more of an intuitive tool than a hard rule, though it has use in defining the generalized binomial coefficient, and results from simplification of the right-hand formula when $k \ge 1$. However, the equation fails for $k < 1$. (Expand the factorials into products in order to see the simplification if you're at all curious.)

The "hard rule" to obey would be

$$\binom n k = \frac{n!}{k!(n-k)!}$$

This perfectly covers the case $k=0$ since $0! = 1$. However, $0!$ isn't defined by a product, unlike $1!$ or $2!$ or the higher factorials: it's defined through recursion. (That means, since $0!$ is not defined by a product, we can't be using the middle formula for $k=0$.) The recursion for the factorial is

$$a_n = na_{n-1} \text{ where } a_1 = 1$$

(though you can take $a_0 = 1$ instead but then $0!$ is still not defined through a product). Play around a bit: you'll see that $a_n = n!$. But we've only defined $n!$ for $n \ge 1$ in this way, which can be reduced to a product. We can't define $a_0 = 0!$ unless we reverse the recursion to have

$$a_{n-1} = \frac{a_n}{n}$$

and, if $n=1$, we see that $0! = a_0 = a_1/1 = 1$. (You can continue this "backwards recursion" again to show that $(-1)!$ wouldn't be well-defined.)

Answer 2

The combinatoric approach to this is asking: "Given $n$ items, how many ways can you select $k=0$ (that is, none) of them?"

There's only one way to select "no items", so that's why $\binom{n}{0} = 1$.

For good measure, the edges of Pascal's Triangle are always $1$'s, associated with this "choose-nothing" case for each $n$.