How to reduce large combinations? The result of a hypergeometric distribution question that I posted about earlier this evening is what follows:
$$\frac{{30 \choose 10}{20 \choose 5}}{{50 \choose 15}}$$
This becomes:
$$\frac{\frac{30!}{10!20!}\frac{20!}{5!15!}}{ \frac{50!}{15!35!}}$$
$$\frac{30! \cdot 35!}{10!\cdot5!\cdot50!}$$
$$\frac{30! \cdot 35!}{10!\cdot5!\cdot50!}$$
I see that I can, for example, cancel out 35! on the numerator and then that makes the 50! become $50\cdot49\cdots36$, so you get:
$$\frac{30!}{10!\cdot5!\cdot(50\cdot49\cdots36)}$$
and subsequently you can do:
$$\frac{(30\cdot29\cdots11)}{5!\cdot(50\cdot49\cdots36)}$$, but is there some way to compute these if I didn't have a computer or a long time to perform the arithmetic? I know there are a few expansions but they seemed to be not particularly helpful either...
Thanks
 A: I actually performed the calculation(all by hand,none with calculator).
$$\zeta=\frac{30.29.28.27.26.25.24.23.22.21.20.19.18.17.16.15.14.13.12.11}{5.4.3.2.1.50.49.48.47.46.45.44.43.42.41.40.39.38.37.36}
$$
Cancelling(Writing explicitly took me 1 min $[9:10\to9:11]$ and cancelling took me 3 mins$[9:12\to9:15]$):
$$\zeta=\frac{9.11.13.17.29}{37.41.43.47}$$
Multiplying(Took me 2 mins $[9:17\to9:19]$)
$$\zeta=\frac{634491}{3065857}$$
Dividing(Took me 2 mins$[9:21\to9:23]$)
$$\zeta=0.206+\frac{2924458\times10^{-\text{something}}}{3065857}$$
So it's good to approx:
$$\zeta\approx0.206$$
Total $8$ mins. I hope you'll do better.
A: It depends on the accuracy you are looking for the approximation. A standard tool to approximate large binomial coefficients is the Central Limit Theorem. We have that:
$$\frac{1}{2^{30}}\binom{30}{10}\approx \sqrt{\frac{2}{30\pi}}\cdot e^{-\frac{2(15-10)^2}{30}},$$
$$\frac{1}{2^{20}}\binom{20}{5}\approx \sqrt{\frac{2}{20\pi}}\cdot e^{-\frac{2(10-5)^2}{20}},$$
$$\frac{1}{2^{50}}\binom{50}{15}\approx \sqrt{\frac{2}{50\pi}}\cdot e^{-\frac{2(25-15)^2}{50}},$$
hence your ratio is around:
$$\sqrt{\frac{1}{6\pi}}\cdot\exp\left(-\frac{5}{3}-\frac{5}{2}+4\right)=\frac{e^{-1/6}}{\sqrt{6\pi}}\approx 0.195.$$
This is not much different from approximating the original hypergeometric distribution with a normal distribution. Using the Stirling's approximation 
$$ n! \approx n^n e^{-n} \sqrt{2\pi n}\,e^{\frac{1}{12n}}$$
is another option.
