# Verify the following identity by a committee-selection model

$${n \choose 0} + {n+1 \choose 1} + {n+2 \choose 2} +\cdots + {n + r \choose r} = {n+r+1 \choose r}$$

The "committee selection model" is a model used to view combinations as selecting a committee from a group of people. E.g., ${15 \choose 4}$ is the number of 4-person committees that can be formed from a group of 15 people.

What I'm tasked with doing is a combinatorial proof using this model--showing that the left counts the same thing as the right using the committee model.

• What exactly do you need to know? Feb 8, 2017 at 3:16
• I edited in another clause to hopefully be more clear. Feb 8, 2017 at 3:25

On the RHS, you count the number of committees of $r$ people that can be chosen from a group on $n+r+1$ people.

Now select one particular person of the group, say John. Then you have two mutually exclusive cases:

1. John is not chosen in the committee, so the number of committees that ca be formed without John is ${n+r\choose r}$
2. John is chosen in the committee, so then you have to choose $r-1$ people out of a group of $n+r$. Now consider another person from the group, say Susie

2.1 John is chosen in the comitee, but Susie is not, which can be done in ${n+r-1\choose r-1}$ ways

2.2 Both John and Susie are chosen, in which case...

... and so on and so forth...

So if you label the people as $1,2,\ldots,n+r+1$, then the number of ways to choose $r$ people equals the sum of the ways to form the committee by mandatory choosing the labels $0,1,\ldots k-1$ and not to choose label $k$, for $k=1,2\ldots,r+1$

• What happens if you choose both John and Susie? I'm having trouble seeing how this argument satisfies the LHS. Feb 9, 2017 at 20:55
• If you choose both John an Susie then again you split into two cases. If you choose John and Susie and you don't chose Alice, you can do it in $n+r-2\choose r-2$ ways, otherwise you continue until all $r$ members are chosen.
– Momo
Feb 10, 2017 at 1:35

\begin{align} \sum_{i=0}^r\binom {n+i}i &=\sum_{i=0}^r\binom {n+i}n\\ &=\sum_{i=0}^r \binom {n+i+1}{n+1}-\binom {n+i}{n+1}\\ &=\binom {n+r+1}{n+1}\\ &=\binom {n+r+1}r \end{align}

You have that $\binom x y$ counts the ways to select $y$ persons from a list of $x$ persons.

For instance $\binom 1 1$ counts the $1$ way to select one person from a list of one person.

Hint: Notice that $\tbinom x y = \tbinom x {x-y}$ for all integers $x,y$ where $0\leq y\leq x$, and so likewise $\tbinom {n+k} k=\tbinom {n+k}n$ for all $0\leq n\leq n+k$.

Thus the LHS, $\sum_{k=0}^r \binom {n+k}k$ equals $\sum_{k=0}^r \binom {n+k}n$ and counts ...

Likewise the RHS, $\binom {n+r+1}{r}$ is $\binom {n+r+1}{n+1}$, thus counts...

Show what they count can be the same task.