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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

permanganate
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  • That depends on $n$. First example, take $j = n-2$. Then the question is whether $n (n-1) - 4 \geq 0$ holds, which is true only for $n\ge 3$. Second example, take $j = n-3$. Then the question is whether $n (n-1) (n-2) - 36 \geq 0$ holds, which is true only for $n\ge 5$. – Andreas Oct 20 '16 at 12:29
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    Note that the question is equivalent to asking for the largest $k$ such that $\binom{n}k\ge k!$. – Brian M. Scott Oct 20 '16 at 12:30
  • Yes, the exact $j$ depends on $n$ (but I don't think there is an easy way to characterize it). I was mainly looking for an estimate of the solution. – permanganate Oct 20 '16 at 12:33

1 Answers1

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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.

Andreas
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