Prove that $\binom{n}{r} + \binom{n}{r+1} = \binom{n+1}{r+1} $ Prove that $\binom{n}{r} + \binom{n}{r+1} = \binom{n+1}{r+1}  $
Thanks in advance, my professor asked us to this a couple weeks ago, but I was enable to get to the right answer. 
Good luck!
Here is what I got up to;
$\frac{(n+1)!}{(n-r)!(r+1)!} = \frac{(n)!}{(r)!(n-r)!} + \frac{(n)!}{(r+1)!(n-r-1)!}    $
 A: $$
\binom{n}{r} + \binom{n}{r+1} \\
\frac{n!}{(n-r)!r!} + \frac{n!}{(n-r-1)!(r+1)!} \\ 
\frac{n!}{(n-r)(n-r-1)!r!} + \frac{n!}{(n-r-1)!r!(r+1)} \\ 
\frac{n!}{(n-r-1)!r!}\left(\frac{1}{n-r} + \frac{1}{r+1}\right) \\ 
\frac{n!}{(n-r-1)!r!}\left(\frac{n+1}{(n-r)(r+1)}\right) \\ 
\frac{(n+1)!}{(n-r)!(r+1)!}\\
\binom{n+1}{r+1} 
$$
A: Given $n+1$ people we can form a committee of size $r+1$ in ${n+1\choose r+1}$ ways. We can count the same thing by counting the number of ways in which person $x$ is in the committee and person $x$ is not in the committee. The number of ways person $x$ is not in the committee is ${n\choose r+1}$. We have $n$ people to work with because we are excluding the possibility of person $x$ being in the committee. The number of ways person $x$ is in the committee is ${n\choose r}$. We have $n$ people to work with since person $x$ is in the committee by default and we choose $r$ people because person $x$ is in the committee. Thus ${n+1\choose r+1}={n\choose r+1}+{n\choose r}$.
A: Hint: Recall that $(d+1)!=(d+1)*d!$ From here, try combining the fractions in the sum by giving them a common denominator.
A: Even though it seems a little far-fetched I will use the Binomial Theorem. The definition of number $\binom{n}{r}$ has a reason to come to exist in the development of $(x + y)^n$. So I think the natural and instructive. For all $x,y\in\mathbb{R}$ we have,
\begin{array}{rrl}
 \hspace{2cm}&( x+y)^{n+1}=     & (x+y)\cdot ( x+y)^n,
\\
\Longleftrightarrow & \sum_{k=0}^{n+1}\binom{n+1}{k}x^{(n+1)-k}y^{k}=  & (x+y)\cdot  \sum_{k=0}^{n}\binom{n}{k}x^{n-k}y^{k},
\\
\Longleftrightarrow & \sum_{k=0}^{n+1}\binom{n+1}{k}x^{(n+1)-k}y^{k}=  &  \sum_{k=0}^{n}\binom{n}{k}x^{n-k}y^{k+1}+ \sum_{k=0}^{n}\binom{n}{k}x^{n-k+1}y^{k}.
\end{array}
Then for all $k=1,2,\ldots n$ we have
\begin{array}{rl}
\binom{n+1}{k}x^{(n+1)-k}y^{k}=  &  \binom{n}{(k-1)}x^{n-(k-1)}y^{(k-1)+1}+ \binom{n}{k}x^{n-k+1}y^{k}.
\end{array}
For $x=1$ and $y=1$,
\begin{array}{rl}
\binom{n+1}{k}=  &  \binom{n}{(k-1)}+ \binom{n}{k},\qquad k=1,2,\ldots n.
\end{array}
Setting $k = r +1$ then
\begin{array}{rl}
\binom{n+1}{r+1}=  &  \binom{n}{r}+ \binom{n}{r+1},\qquad r=0,1,2,\ldots n-1.
\end{array}
