How to find k here $\displaystyle\frac{n^{\underline k}}{n^k}<a$ ?
With, $n^{\underline k}=n\cdot(n-1)\cdot...(n-k+1)$ and $(n>k,\ a>0)$
Of course if $k\approx n/2$ the inequality holds, but it is an horrible estimation. The main exercise was the birthday problem, the idea is clear, but not how to compute.
Can you help please ? Thanks in advance.