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

Question

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)!} $

Answer 1

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

Answer 2

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

Answer 3

Hint: Recall that $(d+1)!=(d+1)*d!$ From here, try combining the fractions in the sum by giving them a common denominator.

Answer 4

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}