General proof for combinatorics Prove that ${n \choose k} + {n \choose k+1} = {n+1 \choose k+1}$. 
I can prove it using numeric examples, but i need a broader proof. I think it might have to do with Pascal's Triangle. 


Argumentative Method:

We know that the number of ways to choose $k$ objects from $n$ objects is $\binom{n}{k}$. Let us add 1 more object into the group. Now we have to choose $k+1$ objects from $n+1$ objects. 
Lets say we include the newest item in the group. Then the number of ways to choose from the remaining $n$ objects is $\binom{n+1-1}{k+1-1} = \binom{n}{k}$.
Lets say we do not include the newest item in the group. Then the number of ways to choose from the remaining $n$ objects is $\binom{n+1-1}{k+1} = \binom{n}{k+1}$.
Since there are no overlaps between the 2 cases, $\binom{n+1}{k+1} = \binom{n}{k} + \binom{n}{k+1}$.

Mathematical Method:

$$
\begin{align*}
\binom{n}{k} + \binom{n}{k+1} & = \frac{n!}{k!(n-k)!} + \frac{n!}{(k+1)!(n-k-1)!} \\
& = \frac{n!(k+1)}{(k+1)!(n-k)!} + \frac{n!(n-k)}{(k+1)!(n-k)!}\\
& = \frac{n!(n-k+k+1)}{(k+1)!(n-k)!}\\
& = \frac{n!(n+1)}{(k+1)!(n-k)!}\\
& = \frac{(n+1)!}{(k+1)!(n-k)!}\\
& = \binom{n+1}{k+1}
\end{align*}
$$
