Prove Pascal's formula by substituting the values of the binomials coefficient as given in the equation $$\binom nk= \frac{n!}{k!\,(n-k)!} = \frac{n(n-1)\cdots(n-k+1)}{k(k-1)\cdots1}.$$
I guess I just don't know what values I'm supposed to be substituting in.
$$ {n\choose k}=\frac{n(n-1)\dots(n-k+1)}{k!}=\frac{(n-1)\dots(n-k+1)}{(k-1)!}\frac{n}{k}\\ =(\frac{k}{k}+\frac{n-k}{k})\frac{(n-1)\dots(n-k+1)}{(k-1)!}\\ =\frac{(n-1)\dots(n-k+1)}{(k-1)!}+\frac{(n-1)\dots(n-k+1)(n-k)}{k!}\\ ={n-1\choose k-1}+{n-1\choose k} $$
$$\binom nk=\frac{n!}{k!(n-k)!}=\frac{n!}{(n-k)!(n-(n-k))!}=\binom{n}{n-k}....[1]$$ $$\binom nk=\frac{n(n-1)!}{k(k-1)!(n-k)!}=\frac{ n}{k}\frac{(n-1)!}{(k-1)!((n-1)-(k-1))!}=\frac{n}{k}\binom {n-1}{k-1}...[2]$$ from [1] and [2] we have $$\binom {n}{k}=\binom {n}{n-k}=\frac {n}{n-k}\binom {n-1}{n-k-1}=\frac{n-k+k}{n-k} \binom {n-1}{(n-1)-(n-k-1)}=$$ $$=\left(1+\frac{ k}{n-k}\right) \binom {n-1}{k}=\binom {n-1}{k}+\frac{ k}{n-k}\binom {n-1}{k}=$$ $$=\binom {n-1}{k}+\frac{ k}{n-k}\frac{n-1}{k}\binom {n-2}{k-1}=\binom {n-1}{k}+\frac{n-1}{n-k}\binom {n-2}{(n-2)-(k-1)}=$$ $$=\binom {n-1}{k}+\frac{n-1}{n-k}\binom {n-2}{n-k-1}=\binom {n-1}{k}+\binom {n-1}{n-k}=$$ $$=\binom {n-1}{k}+\binom {n-1}{(n-1)-(n-k)}=\binom {n-1}{k}+\binom {n-1}{k-1}$$
I see you didn't accept any of the answers here, who do prove Pascal's formula by substituting the values. So, it seems like you're looking for a more elaborative answer:
[Credit: Brilliant.org]
Every row, and every number in that row, are numbered $0$ to infinity. You may wondering why I said they both go to infinity. The Pascal's triangle sits in a sea of zeros. They are mostly trivial, as they do nothing value-wise, $n + 0 = n$, but they are important conceptually. Let's call the row's number $n$, and the number's number $k$. That is, $k$ denotes the placement of the number, not the number's value.
So, why is any non-zero number in Pascal's triangle equal to $\displaystyle \binom{\text{row's number}}{\text{number's number}}= \binom nk$?
There are three facts, all necessary, that makes this true:
Firstly, $\binom nk = \binom{n-1}{k-1} + \binom{n-1}{k}$.
Secondly, the $0$th and $1$st numbers in the $1$st row are both binomial coefficients of $1$: $\binom{1}{0}$ and $\binom{1}{1}$.
Thirdly, every number in Pascal's triangle is the sum of the two above it, or precisely: the $\ $ $k$th number of the $n$th row is the sum of the $k$th and $(k-1)$th numbers of the $(n-1)$th row. Example:
The second number, $k=1$, of the third row, $n= 2$, is equal to $1+1 = \binom{1}{0}$ + $\binom{1}{1} = \binom{n-1}{k-1} + \binom{n-1}{k} = 2$.
So, given that this based on three facts, we must show that they are indeed facts. The second fact is evident upon observation, and the third fact is true by definition of Pascal's triangle. That leaves us with the first fact, which is proven through algebra. I think the users that already answered this question before had good proofs for this, but my answer will not be complete without one. So, here it is:
Statement: $$ \binom nk = \frac{n!}{k!(n-k)!} = \binom{n-1}{k-1} + \binom{n-1}{k} = \frac{(n-1)!}{(k-1)!(n-k)!} + \frac{(n-1)!}{k!(n-1-k)!}$$ $ $
Proof: $$\frac{(n-1)! \times k}{(k-1)!(n-k)! \times k} + \frac{(n-1)! \times (n-k)}{k!(n-1-k)! \times (n-k)} = \frac{(n-1)!k}{k!(n-k)!} + \frac{(n-1)!(n-k)}{k!(n-k)!} $$
$ $
$$\frac{(n-1)!k + (n-1)!(n-k)}{k!(n-k)!} = \frac{(k+(n-k))(n-1)!}{k!(n-k)!} $$
$ $
$$\frac{(k+(n-k))(n-1)!}{k!(n-k)!} = \frac{n(n-1)!}{k!(n-k)!}$$
At this point the last step is obvious, and with that last step, we arrive at the first expression, which is $\binom nk$.
