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 \(\binom{80}{20}>\binom{90}{10}\).
Answer
The desired ordering is: \({80 \choose 70} < {90 \choose 10} < {80 \choose 20} = {80 \choose 60} < {90 \choose 70} < {90 \choose 30}\).