# Prove ${n - 1 \choose k - 1} + {n - 2 \choose k - 1} + {n - 3 \choose k - 1} + \dots + {k - 1 \choose k - 1} = {n \choose k}$

Can someone check my logic here.

Question: How many ways are there to choose a an $k$ person committee from a group of $n$ people?

Answer 1: there are ${n \choose k}$ ways.

Answer 2: condition on eligibility. Assume the creator of the committee is already in the committee. This leaves us with choosing $k - 1$ people from a group of $n - 1$ potentially eligible people. If all remaining people are eligible, there are ${n - 1 \choose k - 1}$ possible committees, if there are $n - 2$ eligible people, there are ${n - 2 \choose k - 1}$ committees, if there are $n - 3$ eligible people, there are ${n - 3 \choose k - 1}$ committees,..., if there are $k - 1$ eligible people there are ${k - 1 \choose k - 1}$ committees. Therefore,$${n - 1 \choose k - 1} + {n - 2 \choose k - 1} + {n - 3 \choose k - 1} + \dots + {k - 1 \choose k - 1} = {n \choose k}$$.

• All of your cases, there's always "the creator" in the committee? – Quang Hoang Sep 26 '15 at 3:05

It’s not clear whether your argument works or not, because you’ve not told us what you mean by eligible. Here is one way to make your argument valid.

Suppose that that the $n$ potential members all have different ages, and that the committee must be created by its oldest member. In other words, the creator may choose only from those potential members who are younger than he is. In other words, we’re defining eligible to mean younger than the creator.

If he’s the oldest person in the pool of potential members, he can pick any $k-1$ of the other $n-1$ people in the pool: all $n-1$ of them are younger than he. Of course this can be done in $\binom{n-1}{k-1}$ ways. If he’s only the second-oldest, then $n-2$ members of the pool are younger, and he can choose $k-1$ of them in $\binom{n-2}{k-1}$ ways. In general, if he’s the $i$-th oldest person in the pool, then $n-i$ are younger, and he can choose the rest of the committee in $\binom{n-i}{k-1}$ ways. With these added details you get a legitimate argument that the sum on the lefthand side counts the $k$-member committees that can be formed from the pool of $n$ people.

Did you define what you mean by "eligible"? I don't follow the argument.

Here is another approach: inductively apply Pascal's identity $$\binom{n}{k}=\binom{n-1}{k}+\binom{n-1}{k-1}.$$

• The answer is absolutely right for people who knows why it is right, but I'm afraid it does not address the OP's confusion of why the identity holds and why his reasoning does not hold. Maybe you want to add in more explanation? – dcstup Sep 26 '15 at 3:17
• His reasoning is incomplete, because he hasn't told us what he means by "eligible". Maybe his idea works, if he provides more information about what he means. – pre-kidney Sep 26 '15 at 3:19

Usually the logic is:

If $A$ is in the committee, the problem reduces to choosing $k-1$ people from $n-1$ people.

If $A$ is not in the committee, the problem reduces to choosing $k$ people from $n-1$ people, which we keep dividing:

$\,$ If $B$ is in the committee, the problem reduces to choosing $k-1$ people from $n-2$ people.

$\,$ If $B$ is not in the committee, the problem reduces to choosing $k$ people from $n-2$ people, which we keep dividing, and so on.

So summing them up gives you the left hand side, while directly solving the problem gives you the right hand side.

I don't think your logic applies because in the first place there is no such idea as "creator" in the question and assuming it with your real-world experience is not really mathematical.

Apart from the creator problem, the remaining part of the logic does not apply too because it does not follow MECE principle (i.e. you have to make sure you did not double count and you counted all cases), since in the case where you choose $k-1$ people from $n-1$ or $n-2$ eligible candidates, there is no guarantee that you will not pick the same combination.