why is this not $n-r-1$? why did we not multiply though by $n-r-1$?

Question

My question is why is this not $n-r-1$? Why did we not multiply through by $n-r-1$

I'm confused in my statistics class. Please give links if you have them. Thanks for the help.

Answer 1

Because $(n-r)!(n-r+1) = (n-r+1)!,$ hence

$$\frac{n-r+1}{\color{red}{n-r+1}} \frac{n!}{r(r-1)!\color{red}{(n-r)!}}=\frac{n-r+1}{r}\frac{n!}{(r-1)!\color{red}{(n-r+1)!}}$$

You can multiply $n-r-1$ to the numerator and the denominator but it might not simply things. Multiplying $n-r+1$ is a good choice because we can simplify

$$\frac{n!}{(r-1)!(n-r+1)!}=\binom{n}{r-1}$$

Answer 2

You are merely rearranging the original expression to show that it is equivalent to another expression. Here is a slightly different way to do the same thing.

$\begin{eqnarray} {n\choose r}&=&\frac{n!}{r!\,(n-r)!}\\ &=&\frac{(n-r+1)\,n!}{r\,(r-1)!\,(n-r+1)\,(n-r)!}\\ &=&\frac{n-r+1}{r}\cdot\frac{n!}{(r-1)!\,(n-r+1)!}\\ &=&\frac{n-r+1}{r}\cdot{n\choose r-1} \end{eqnarray}$

In the first step, we rewrote $r!$ as $r\,(r-1)!$ and multiplied numerator and denominator by $(n-r+1)$.

In the second step we factored out the fraction in front leaving a second fraction to be simplified in the third step.

In the third step we recoginzed that the second fraction in the previous step could be rewritten in the form ${n\choose r-1}$.