How $(r-k){r \choose r-k}$ becomes $r{r-1 \choose r-k-1}$?

Question

I'm reading Knuth/Graham/Patashnik's Concrete Mathematics:

I don't understand how he goes from $(r-k){r \choose r-k}$ to $r{r-1 \choose r-k-1}$ using $(5.6)$. The mentioned property has a $k$ in one side and an $r$ in the other. I'm confused at what I should do. I'm using the RHS of $(5.6)$, and $r$ is $r$, then what would be the $k$ in the other side and why?

Answer 1

$(5.6)$ states: $$ \color{green}{k} \binom{\color{red}{r}}{\color{green}{k}} = \color{red}{r} \binom{\color{red}{r} - 1}{\color{green}{k} - 1} $$ So, in this case: $$ (\color{green}{r - k}) \binom{\color{red}{r}}{\color{green}{r - k}} = \color{red}{r} \binom{\color{red}{r} - 1}{\color{green}{r - k} - 1} $$

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To answer your question directly, the "$k$" in $(5.6)$ is simply $r - k$ now.

Answer 2

• Using the symmetry:

$\displaystyle(r-k){r\choose k}=(r-k){r\choose r-k}$

• Now using $(5.6)$:

$\begin{eqnarray*} {(r-k){r\choose r-k}}&=&{(r-k) \frac{r}{r-k}{n-1 \choose n-k-1}} \\ {}&&{}\\ {}&&{}\\ {}&=&{\frac{(r-k)r}{(r-k)} {n-1\choose n-k-1}} \\ {}&&{}\\ {}&&{}\\ {}&=&{r{n-1\choose n-k-1}} \end{eqnarray*}$