Binomial Formula conversion I saw an answer in stackoverflow about binomial here
The answerer provide some nice explanation how can binomial can be calculated by pascal triangle.
But I'm still not sure how to convert this
       N!                   N!               N!      (N-K)
---------------- = -------------------- = -------- x -----
 (N-K-1)! (K+1)!   (N-K)!/(N-K) K!(K+1)   (N-K)!K!   (K+1)

And then the correlation with this
(N choose K+1) = (N choose K) * (N-K)/(K+1)

and this
 N!
---- = 1
N!0!

with pascal triangle? 
Thanks!
 A: The question you refer to shows how to compute a function which computes $\binom{a}{b}$ recursively instead of using the definition of $\binom{a}{b}$.
Instead of using the definition
$$
\binom{n}{k+1} = \frac{n!}{(n-(k+1))!(k+1)!}
$$
one can do
$$
\binom{n}{k+1} = \binom{n}{k} \frac{n-(k+1)+1}{k+1}
$$
so one must calculate the same function but for $k$ instead of $k+1$. One can apply the same formula replacing $k+1$ by $k$
$$
\binom{n}{k+1} = \binom{n}{k} \frac{n-(k+1)+1}{k+1}
 = \bigg( \binom{n}{k-1} \frac{n-k+1}{k}\bigg) \frac{n-(k+1)+1}{k+1}
$$
and now you get $k-1$, so every time you get the same function but with one of the variables decreasing by one.
This is easy to give to a computer, but it must stop somewhere. To do that, one has that by definition
$$
\binom{n}{k-k}=\binom{n}{0}=\frac{n!}{(n-0)!0!}=\frac{n!}{n!1}
$$
so $\binom{n}{0}$ does not depend on $\binom{n}{-1}$.
To put it short, Pascal's triangle is an array where the $n$-th floor has $\binom{n}{0}$ up to $\binom{n}{n}$
