So today I was dealing with this factorial: $$ {m \choose n-1} $$
This expresses as:
$$ \frac{m!}{(n-1)!(m-n+1)!} $$ Normally when it's normal numbers and not factorials the signs doesnt matter but here if we change the signs and substitute with numbers will give 2 differents answers, I was wondering why this happen? does the order matters?
Why it has to be $$ \frac{m!}{(n-1)!(m-n+1)!} $$ Instep of $$ \frac{m!}{(n+1)!(m-n-1)!} $$
I still don't understand what you want exactly, but I'll give you some explanation on the binomial coefficient. ${n \choose k}$ is defined as $\frac{n!}{k! \cdot (n-k)!}$. So let's fill in some numbers. ${5 \choose 3} = \frac{5!}{3! \cdot (5-3)!}$. Now what happens, if i write 3 as 4-1? (that's of the form $n-1$ that you have, so your m now is 5 and your n is 4). ${5 \choose 4-1} = \frac{5!}{(4-1)! \cdot (5-(4-1))!}$ and what is -(4-1)? It's -4+1, so that's where your signs come from. To summarize, with m = 5 and n = 4 we get: ${5 \choose 4-1} = {m \choose n-1} = \frac{m!}{(n-1)! \cdot (m-(n-1))!} = \frac{5!}{(4-1)! \cdot (5-(4-1))!} = \frac{5!}{3! \cdot (5-3)!} = \frac{5!}{3! \cdot 2!} = {5 \choose 3}$