Question

Show \[\binom{x+r-1}{r-1}p^r(1-p)^x=\frac{r(1-p)}{p}\binom{y+s-1}{s-1}p^s(1-p)^y\]
where
\[y=x-1, s=r+1\]

Answer 1

The given "identity" isn't true...
\[\begin{align*}
\binom{x+r-1}{r-1}p^r(1-p)^x&=\frac{(x+r-1)!}{(r-1)!(x+r-1-(r-1))!}p^r(1-p)^x\\\\

&=\frac{(x+r-1)!}{(r-1)!x!}p^r(1-p)^x\\\\

&=\frac{r}{x}\frac{((x-1)+r)!}{r!(x-1)!}p^r(1-p)^x\\\\

&=\frac{r}{x}\frac{((x-1)+r)!}{r!(x-1+r-r)!}p^r(1-p)^x\\\\

&=\frac{r(1-p)}{xp}\frac{(y+s-1)!}{(s-1)!(y+s-1-(s-1))!}p^{r+1}(1-p)^{x-1}\\\\

&=\frac{r(1-p)}{\color{red}xp}\binom{y+s-1}{s-1}p^s(1-p)^y
\end{align*}\]

Answer 2

I missed an x but i can now do it. thanks