I need to prove the following property of the binomial coefficient...please Help!

Question

$$\binom rk = \frac rk \binom{r-1}{k-1}$$

Any help is immensely appreciated! I am absolutely confused by this problem and have no real idea of how to solve it, my professor mentioned that the answer involved factorials and we are learning about combinatorics and binomial coefficients.

The way that I was trying to solve it:

Knowing that $$\binom rk = \frac {r!}{k!(r-k)!}$$

means that

$$\binom {r-1}{k-1} = \frac {(r-1)!}{(k-1)!((r-1)-(k-1))!}$$ and then I got stuck.

After reading through your responses, I am pretty sure that I was not exactly on the right track but getting close.

Thank you all for your responses and I do believe that FrodCube was the response I needed. I have not been able to confirm this yet but will hopefully be able to clarify tomorrow and then update.

Answer 1

$${r \choose k} = \frac {r!}{k!(r-k)!}=\frac{r}{k}\frac {(r-1)!}{(k-1)!(r-k)!}=$$ $$=\frac{r}{k}\frac {(r-1)!}{(k-1)!((r-1)-(k-1))!}=\frac{r}{k}{r-1 \choose k-1}$$

Answer 2

Equivalently we want to show that $$k\binom{r}{k}=r\binom{r-1}{k-1}.$$ There are $r$ different doughnuts in front of us. We want to choose one to eat immediately, and $k-1$ to eat a little later. We will count, in two different ways, the number of ways to carry out this task.

Way (i): There are $r$ ways to choose the doughnut to be eaten immediately, For each of these ways, there are $\binom{r-1}{k-1}$ ways to choose the doughnuts to be eaten later, for a total of $r\binom{r-1}{k-1}$.

Way (ii): We choose $k$ doughnuts, and eat one of the chosen ones, saving the rest for later. The $k$ doughnuts can be chosen in $\binom{r}{k}$ ways. For each such choice there are $k$ ways to choose from them the doughnut to be eaten immediately, for a total of $k\binom{r}{k}$.