Transforming Binomialcoefficients

Question

I don't understand the following equation which is presented in our lecture:

$${p+q-1 \choose p-1} - { p+q-1 \choose p} = \frac{p-q}{p+q}{p+q \choose p},$$ where $p>q$ and $p,q \in \mathbb{Z}$.

I have tried several manipulations of the Binomialcoefficients but I couldn't show the equality.

Can someone explain me the steps of how to show the equality?

Answer 1

$${p+q-1 \choose p-1} - { p+q-1 \choose p} \\ = \frac{(p+q-1)!}{(p-1)!q!} - \frac{(p+q-1)!}{(p)!(q-1)!} \\ = (p+q-1)! \frac{p-q}{(p)!(q)!} \\ = \frac{p-q}{p+q}{p+q \choose p}$$

where I have used ... $$ \frac 1{(p-1)!} = \frac{p}{p!} \\ \frac 1{(q-1)!} = \frac{q}{q!} \\ (p+q-1)! = \frac{(p+q)!}{p+q} $$

Answer 2

I think you must see this $${n\choose r}=\frac{n}{r}{n-1\choose r-1}$$
I suggest below relation $${n\choose r}=\frac{n!}{r!(n-r)!}\\now-check-this-out\\ \frac{n+1}{r+1}{n\choose r}=\frac{(n+1)n!}{(r+1)r!(n-r)!}={n+1\choose r+1}$$so

$$\displaystyle{{p+q-1 \choose p-1} - { p+q-1 \choose p} = \\ \frac{p+q-1+1}{p-1+1}{p+q-1 \choose p-1}\frac{p}{p+q} - { p+q-1 \choose (p+q-1)-p} = \\ \frac{p+q-1+1}{p-1+1}{p+q-1 \choose p-1}\frac{p}{p+q} - { p+q-1 \choose q-1} = \\ {p+q \choose p}\frac{p}{p+q} - \frac{p+q-1+1}{q-1+1}{ p+q-1 \choose q-1}\frac{q}{p+q} = \\ {p+q \choose p}\frac{p}{p+q} - { p+q \choose q}\frac{q}{p+q} = \\ {p+q \choose p}(\frac{p}{p+q} - \frac{q}{p+q}) = \\what-u-want}$$