Binomial Coefficient Identities I am trying to simplify the following fraction, which I think is equal to 1 but I am not sure.
$$\frac{\frac{\left(\begin{array}{c}
b-1\\
k-1
\end{array}\right)\left(\begin{array}{c}
r\\
n-k
\end{array}\right)}{\left(\begin{array}{c}
r+b-1\\
n-1
\end{array}\right)}}{\frac{\left(\begin{array}{c}
b\\
k
\end{array}\right)\left(\begin{array}{c}
r\\
n-k
\end{array}\right)}{\left(\begin{array}{c}
r+b\\
n
\end{array}\right)}}$$
I tried to use the identity 
$$\left(\begin{array}{c}
n\\
r
\end{array}\right)=\left(\begin{array}{c}
n-1\\
r-1
\end{array}\right)+\left(\begin{array}{c}
n-1\\
r
\end{array}\right)
 $$
I have done the following
Step 1: 
$$\frac{\left(\begin{array}{c}
b-1\\
k-1
\end{array}\right)\left(\begin{array}{c}
r\\
n-k
\end{array}\right)}{\left(\begin{array}{c}
r+b-1\\
n-1
\end{array}\right)}\cdot\frac{\left(\begin{array}{c}
r+b\\
n
\end{array}\right)}{\left(\begin{array}{c}
b\\
k
\end{array}\right)\left(\begin{array}{c}
r\\
n-k
\end{array}\right)}$$
Step 2:
$$\frac{\left(\begin{array}{c}
b-1\\
k-1
\end{array}\right)}{\left(\begin{array}{c}
b\\
k
\end{array}\right)}\cdot\frac{\left(\begin{array}{c}
r+b\\
n
\end{array}\right)}{\left(\begin{array}{c}
r+b-1\\
n-1
\end{array}\right)}$$
From there I get stuck here
$$\frac{\left[\left(\begin{array}{c}
b\\
k
\end{array}\right)-\left(\begin{array}{c}
b-1\\
k
\end{array}\right)\right]}{\left(\begin{array}{c}
b\\
k
\end{array}\right)}\cdot\frac{\left(\begin{array}{c}
r+b\\
n
\end{array}\right)}{\left[\left(\begin{array}{c}
r+b\\
n
\end{array}\right)-\left(\begin{array}{c}
r+b-1\\
n
\end{array}\right)\right]}$$
Is there any other identity that would be more useful for this problem? If not does anyone have a useful hint for where to proceed from here?
 A: If we use the identity $\displaystyle\binom{m}{n}=\frac{m}{n}\binom{m-1}{n-1}$, we obtain
$$\displaystyle\frac{\binom{b-1}{k-1}\binom{r+b}{n}}{\binom{b}{k}\binom{r+b-1}{n-1}}=\frac{\binom{b-1}{k-1}\frac{r+b}{n}\binom{r+b-1}{n-1}}{\frac{b}{k}\binom{b-1}{k-1}\binom{r+b-1}{n-1}}=\frac{k}{b}\cdot\frac{r+b}{n}$$
A: The Fraction above does not simplify down to 1 as I had thought it would. It simplifies down to $\frac{k}{b}\cdot\frac{r+b}{n}$.
If we go back to Step 2 of the original question, we have 
$$\frac{\left(\begin{array}{c}
b-1\\
k-1
\end{array}\right)}{\left(\begin{array}{c}
b\\
k
\end{array}\right)}\cdot\frac{\left(\begin{array}{c}
r+b\\
n
\end{array}\right)}{\left(\begin{array}{c}
r+b-1\\
n-1
\end{array}\right)}$$
Now, we can express the binomial coefficients as factorials, as suggest by André Nicolas above, and then we get 
$$\frac{\left(\frac{(b-1)!}{[(b-1)-(k-1)]!(k-1)!}\right)}{\left(\frac{b!}{(b-k)!k!}\right)}\cdot\frac{\left(\frac{(r+b)!}{(r+b-n)!n!}\right)}{\left(\frac{(r+b-1)!}{[(r+b-1)-(n-1)](n-1)!}\right)}$$
Which simplifies to
$$\frac{\left(\frac{(b-1)!}{(b-k)!(k-1)!}\right)}{\left(\frac{b!}{(b-k)!k!}\right)}\cdot\frac{\left(\frac{(r+b)!}{(r+b-n)!n!}\right)}{\left(\frac{(r+b-1)!}{(r+b-n)!(n-1)!}\right)}$$
Which in turn simplifies to 
$$\frac{\left(\frac{(b-1)!}{(k-1)!}\right)}{\left(\frac{b!}{k!}\right)}\cdot\frac{\left(\frac{(r+b)!}{n!}\right)}{\left(\frac{(r+b-1)!}{(n-1)!}\right)}$$
Which is equal to 
$$\left(\frac{(b-1)!k!}{(k-1)!b!}\right)\left(\frac{(r+b)!(n-1)!}{n!(r+b-1)!}\right)$$
Which at last reduces down to 
$$\frac{k}{b}\cdot\frac{r+b}{n}$$
If you know of any other clever ways to simplify this, please post you answer. Thanks.
