## Comparison of binomial coefficients

Compare and sort the following binomial coefficients:

$$\binom{80}{20}, \binom{90}{10}, \binom{90}{70}, \binom{90}{30}, \binom{80}{70}, \binom{80}{60}.$$

• #### Solution

From the 80th line of Pascal’s triangle we have: $$\binom{80}{10}=\binom{80}{70}<\binom{80}{20}=\binom{80}{60}$$

and analogously from the 90th line: $$\binom{90}{10}<\binom{90}{20}=\binom{90}{70}<\binom{90}{30}$$

Also $$\binom{80}{10}$$ is slanted above $$\binom{90}{10}$$, thus $$\binom{80}{10}<\binom{90}{10}$$ and similarly $$\binom{80}{20}<\binom{90}{20}$$ – can also be proved by factoring out the binomial coefficients.

It remains to compare $$\binom{80}{20}$$ and $$\binom{90}{10}$$. We first decompose $$\binom{80}{20}=\frac{80{\cdot}79\cdots61}{20{\cdot}19\cdots11{\cdot} 10!}$$ and $$\binom{90}{10}=\frac{90{\cdot}79\cdots81}{10!}$$. Since for every $$i=1,…,10$$ it holds that $$\frac{(70+i)(60+i)}{10+i}>80+i$$, we get that $$\binom{80}{20}>\binom{90}{10}$$.

The desired ordering is: $${80 \choose 70} < {90 \choose 10} < {80 \choose 20} = {80 \choose 60} < {90 \choose 70} < {90 \choose 30}$$.