Proving $r \binom{n}{r}=\left( n-(r-1) \right) \binom{n}{r-1}$ using counting arguements

Question

Show that : $\binom{n}{r}=\frac{\left[ n-(r-1)\right]}{r} \binom{n}{r-1}$ , using counting arguments

I have attempted the question but I can't identify where the factor of $r$ comes from.

My counting argument: Suppose you have a store with $n$ donuts and you need to choose $r$, you can first choose $r-1$ elements in $\binom{n}{r-1}$ ways then for the last element you have $\left[ n-(r-1)\right]$ choices, hence total choices are $\binom{n}{r-1} \left[ n-(r-1) \right]$

I can't understand why a division by $r$ is required in terms of counting..

Answer 1

You have a collection of $n$ memes and you want to read $r$ of them. You choose them and then see the first chosen meme again. Total number of ways are thus $\binom nr\times r$.

Instead, choose $r-1$ memes and select one from the remaining one to see twice. Total number of ways are $\binom n{r-1}\times(n-(r-1))$.

Why doesn't your method work? When you chose the last one separately and multiplied the number of ways, that takes into account the number of permutations (of the last chosen with others) as well. By your argument, we have $\binom nr = \binom nk\binom kr$ for any $k\ge r$.

Answer 2

From among $n$ people, you want to choose a team of $r$ people, along with a captain of the team (who will be one of the team members).

Method 1: Choose $r$ people from $n$. This can be done in ${n \choose r}$ ways.

Then choose one of the $r$ people on the team to be the captain. This can be done in $r$ ways.

Method 2: Choose $r-1$ people to be on the team but not the captain. This can be done in ${n\choose r-1}$ ways. Then choose a captain from the remaining (so far unchosen) people. This can be done in $n-(r-1)$ ways.