$\frac{b*{b-1 \choose k-1}{r \choose n-k}}{n*{b \choose k}}*\frac{{r+b \choose n}}{{r \choose n-k}{b\choose k}}$
How would I go about reducing this into $\frac k n$?
My attempt was $\frac{b!}{n*(k-1)!(b-k-2)!}*\frac {k!}{b!}=\frac{k!}{n*(k-1)!(b-k-2)!}=\frac{k}{n*(b-k-2)!}$
Don't know where the errors are / where to go from here
$$\frac{b{b-1 \choose k-1}}{n{b \choose k}}=\frac{b\cdot\frac{(b-1)!}{(k-1)!(b-k)!}}{n\cdot\frac{b!}{k!(b-k)!}}=\frac{k}{n}$$ because $b\cdot(b-1)!=b!$ and $k!=k(k-1)!$.