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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?

Philipp
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    It is trivial if you use the formula ${n\choose r}=\frac{n!}{r!(n-r)!}$ – almagest Jan 01 '20 at 19:43
  • @almagest, I already did this and tried to simplify the difference but I only got some complicated fractions where nothing cancels out. – Philipp Jan 01 '20 at 19:47
  • Then you made an error. Write each of the two binomials on the lhs as something x $(p+q-1)!/(p!q!)$ and the result is immediate. – almagest Jan 01 '20 at 19:51
  • $${p+q-1 \choose p-1} - { p+q-1 \choose p} = \frac{(p+q-1)!}{(p-1)!q!} - \frac{(p+q-1)!}{p! (q-1)!}...$$ This is where I get stuck. Every further manipulation makes it more complicated, I still don't see where my mistake is. – Philipp Jan 01 '20 at 19:57
  • Now multiply the top and bottom of the first binomial on the rhs by $p$ and the top and bottom of the second by $q$ and the result is immediate. – almagest Jan 01 '20 at 19:59
  • @almagest, Thank you :D, I don't know why but I really didn't see this. – Philipp Jan 01 '20 at 20:05

2 Answers2

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

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

WW1
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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

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

Khosrotash
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