Smallest $j$ for $\binom{n}{j} \geq (n-j)!$

Question

Obviously we have $\binom{n}{j} \geq (n-j)!$ for $j = n$ or $n-1$. What is the smallest $j$ such that this inequality occurs? (or largest $j$ such that $\binom{n}{j} \geq j!$ as pointed out below)

Thanks

Answer 1

Let's ask for an approximate solution of $\binom{n}k\ = k!$ where $j = n-k$.

If n and k are large, we may take the normal approximation to the binomial, $$ \binom{n}k \simeq \frac{2^n}{\sqrt{0.5 \pi n}}\exp \frac{2(k-n/2)^2}{n} $$ and Stirling's approximation for $k!$, $$ k! \simeq {\sqrt{2 \pi k}} \Big[\frac{k}{e}\Big]^k $$ The leading terms are $$ n \log 2 + \frac{2(k-n/2)^2}{n} \simeq k \log(k) $$ which solves for $n$: $$ n\simeq \frac{2 k + k \log(k) + k \sqrt{\log(k)^2 + 4 \log(k) - 8 \log 2}}{2\log 2 + 1} $$ and again, if all numbers are very large, $$ n \simeq \frac{2 k \log(k)}{2\log 2 + 1} $$ This should give the general trend.