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,$${n – 1 \choose k – 1} + {n – 2 \choose k – 1} + {n – 3 \choose k – 1} + \dots + {k – 1 \choose k – 1} = {n \choose k}$$.
Best Answer
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.