Prove that $\binom{n}{r} + \binom{n}{r+1} = \binom{n+1}{r+1} $ [duplicate]

$$ \binom{n}{r} + \binom{n}{r+1} \\ \frac{n!}{(n-r)!r!} + \frac{n!}{(n-r-1)!(r+1)!} \\ \frac{n!}{(n-r)(n-r-1)!r!} + \frac{n!}{(n-r-1)!r!(r+1)} \\ \frac{n!}{(n-r-1)!r!}\left(\frac{1}{n-r} + \frac{1}{r+1}\right) \\ \frac{n!}{(n-r-1)!r!}\left(\frac{n+1}{(n-r)(r+1)}\right) \\ \frac{(n+1)!}{(n-r)!(r+1)!}\\ \binom{n+1}{r+1} $$

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 ${n+1\choose r+1}={n\choose r+1}+{n\choose r}$.