This is actually a question involving the recurrence relation in binomial coefficient formula, which is $${n \choose k} = {n-1 \choose k-1} + {n-1 \choose k}$$.
I am trying to prove this formula by induction and I know the proof is valid. However, when I am trying to set up the base case, I am setting n and k to both 0. (I know in math natural number starts from 1, but here 0 is required to be in natural numbers due to restriction).
The problem I feel confused is that when I set n = k = 0, I know $${0 \choose 0} = 1$$
However, in this case, the RHS will be equal to 2 instead of 1, which causes the base case to fail.
I ask this question to my professor, my professor told me that: $${n-1 \choose -1} = 0$$
Why is this so? I feel so confused, I google that the factorial of -1 is equal to some complex number, which I don't know, I have no idea the above identity holds.
Can someone help figure out my confusion?
P.S. 0 must be in natural numbers.
Oftentimes things in probability are defined in such a way out of convenience, but this time you're just misinterpreting what Pascal's rule applies to. This identity is true for values of $1\leq k\leq n$, but $k=0$ is not included.