Suppose you have a list of distinct natural or integer numbers
$$\mathcal{L}=\{1,3,2,5,\dots\},$$
with length $N$.
Is there a formal proof of following relation? $$ N = \biggl\lceil \sqrt{\frac{N!}{(N-2)!}}\biggr\rceil = \bigl\lceil \sqrt{2\cdot\#}\bigr\rceil $$ Here $\#$ is the number of unique pairwise (different) combinations of the list $\mathcal{L}$ (= length of list $\mathcal{P}$), $\mathcal{P}$ is the list of unique pairwise (different) combinations of list $\mathcal{L}$, $\lceil \bullet \rceil$ is the ceil function (Gauss bracket).
Example 1: $$\mathcal{L}=\{2,5,7\},$$
$$\mathcal{P}=\left\{ \begin{matrix} \{2 , 5\}\\ \{2 , 7\}\\ \{5 , 7\}\\ \end{matrix} \right\},$$
$$ N=3, \quad \# = 3, $$
$$ 3 = \biggl\lceil \sqrt{\frac{3!}{(3-2)!}}\biggr\rceil = \bigl\lceil \sqrt{2\cdot3}\bigr\rceil $$
Example 2: $$\mathcal{L}=\{1,2,3,4,5\},$$
$$\mathcal{P}=\left\{ \begin{matrix} \{1 , 2\}\\ \{1 , 3\}\\ \{1 , 4\}\\ \{1 , 5\}\\ \{2 , 3\}\\ \{2 , 4\}\\ \{2 , 5\}\\ \{3 , 4\}\\ \{3 , 5\}\\ \{4 , 5\}\end{matrix} \right\},$$
$$ N=5, \quad \# = 10, $$
$$ 5 = \biggl\lceil \sqrt{\frac{5!}{(5-2)!}}\biggr\rceil = \bigl\lceil \sqrt{2\cdot10}\bigr\rceil $$
Thanks in advance.
You already know the right equality from $$\frac{N!}{(N-2)!}=N(N-1)=2\#$$ It remains to show that $N=\lceil\sqrt{N(N-1)}\rceil$. To do this, simply note that $N(N-1)=N^2-N$ is between $(N-1)^2=N^2-2N+1$ and $N^2$ if $N\ge2$.