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?

  • 1
    $\begingroup$ Express the binomial coefficients in terms of factorials. There should be a lot of cancellation. Recall that $\binom{b}{a}=\frac{b!}{a!(b-a)!}$. $\endgroup$ – André Nicolas Sep 26 '13 at 1:37
  • $\begingroup$ @AndréNicolas So, there are no other tricks? I will try the factorials now. $\endgroup$ – JimmyJackson Sep 26 '13 at 1:39
  • $\begingroup$ There are endlessly many ideas one could bring to bear. But the one I mentioned will be useful. $\endgroup$ – André Nicolas Sep 26 '13 at 1:42
  • $\begingroup$ @AndréNicolas I think I will go back to step two, and then express it as factorial. I think that using the identity above will only make cancellation more difficult. $\endgroup$ – JimmyJackson Sep 26 '13 at 1:45
  • 2
    $\begingroup$ I maybe should have said, but didn't: don't use the identity you started to use. $\endgroup$ – André Nicolas Sep 26 '13 at 1:56

If we use the identity $\displaystyle\binom{m}{n}=\frac{m}{n}\binom{m-1}{n-1}$, we obtain


  • $\begingroup$ That is a much nicer way of dealing with this mess! $\endgroup$ – JimmyJackson Sep 28 '13 at 23:33

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.