Combinatorial proof $\sum_{i=k}^n {i-1 \choose k-1} = {n \choose k}$

Question

Could someone help me as I am stuck with coming up with a proof for this?

Assume n is the total number of people in a town. Assume k is the number of possible ways to select a chief of the town. So the RHS is saying that there are k ways to choose a chief from n people.

on the LHS, From $i=k$, and $k=n$, it is referring to from k to n, which is the sum of the remaining people in the town who were not selected $(n-k)$, that there is $k-1$ ways to choose from $i-1$ objects. Since $i=k$, i could be the number of ways to possibly select a chief. If one person is chosen from i, who also belongs to $k, k-1$. But how does this lead to $${n \choose k}$$?

Possible duplicate of Prove ${n - 1 \choose k - 1} + {n - 2 \choose k - 1} + {n - 3 \choose k - 1} + \dots + {k - 1 \choose k - 1} = {n \choose k}$, and also Combinatorial Analysis: Fermat's Combinatorial Identity — Sil, Dec 16 '18 at 15:28
The linked question may help you to obtain an answer, but I think an important prerequisite is to learn how to interpret combinatorial expressions properly. In the quantity $\binom{n}{k}$, the number $k$ is the number being chosen, not the number of ways to choose. The very first thing you need to get straight is that $\binom{n}{k}$ is the number of ways to choose $k$ items out of a set of $n$ items. In the problem of choosing a single chief from a town of $n$ people, the answer would be $\binom{n}{1}$ or $n$. — Will Orrick, Dec 16 '18 at 16:18

score 0 · Answer 1 · answered Dec 16 '18 at 15:27

Hint: Partition the set of all $k$-element subsets of $[n] = \{ 1, 2, \dots, n \}$ according to their greatest elements.

If you must use a 'real-world' example, consider putting the $n$ townfolk in order—say, by height—and counting the number of ways to choose a committee of $k$ people which will be chaired by the tallest person on the committee. The left-hand side partitions the set of all possible committees according to who will be the committee chair.

Combinatorial proof $\sum_{i=k}^n {i-1 \choose k-1} = {n \choose k}$

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