$\newcommand{\Fun}{F_\text{un}}$There was lots of "Fun with $\Fun$"  ([field with one element][1]) in recent years.
One of the points is that it provides bridge between geometrical and combinatorial questions, i.e. in the limit case $q=1$ geometry disappears and combinatorial structure distills, on the other hand looking through a $\Fun$ glass on combinatorics one may try to lift it to a geometric picture over $F_q$/any field.

   

My question is about giving examples of that kind. 

**Question:**
What are some combinatorial/probabilistic  identities/concepts which have a nice $\Fun$  interpretation and might be lifted to nice geometric identities/concepts  over $F_q$/any field? Something similar to the example below:

------------------------

**Example:** 

(See https://mathoverflow.net/questions/299581/is-there-a-lift-of-the-q-vandermonde-identity-to-some-geometric-motivic-identi?noredirect=1&lq=1)


***Combinatorial side:***

Consider the [Vandermonde identity][2]:
$$ \binom{m + n}{k} =\sum_{j} \binom{m}{k - j} \binom{n}{j}. $$



***Probability side:***
Consider $n+m$ balls, $m$ white, $n$ black,
what is the probability to choose $k-j$ white, $j$ black balls? 
The answer is:  
$$  \frac{  \binom{m}{k - j} \binom{n}{j} } { \binom{m + n}{k} }.  $$

And since the sum of probabilities over $j$ gives 1 we have the Vandermonde identity above.

***First $\Fun/F_q$ interpretation (projective geometry)***

**Interpretation:** 
$\DeclareMathOperator\Gr{Gr}$According to $\Fun$-wisdom we should think about the Grassmanian when we see a
binomial coefficient:  
$$\binom{n}{k} = \#( \Gr(k,n, \Fun) ). $$

Hence we might expect some geometric identity related to Grassmannians,
and indeed there is motivic identity which is true over any field:  
$$
[\Gr(k,m+n)] = \sum_j [\Gr(k-j,m)][\Gr(j,n)] [\mathbb{A}^{j(m-k-j)}],
$$
as Sasha [explained](https://mathoverflow.net/a/299582).

On the level of enumeration of $F_q$-points it gives the [$q$-Vandermonde identity][3].


***Second interpretation (linear algebra)***

I am not sure that this interpretation is fully correct, but let me give it.
$\DeclareMathOperator\GL{GL}$One of the curious things about $\Fun$ is that linear and projective geometry coincide over it. They should be different by $\GL(F_q)$ but over $\Fun$ it is just $1$  (at least I see it like that).
 
**Interpretation:** 
 
$$ \binom{n}{k} = \# ( {\bigwedge}^k V^n(F_1)) = \dim ( {\bigwedge}^k V^n). $$

I mean that number of elements in a vector space coincide with its dimension over $\Fun$. 

Now the Vandermonde identity can be interpreted like this:
consider  $V = V^{m} \oplus V^n$. Then

$$   {\bigwedge}^k V = \bigoplus_j {\bigwedge}^{k-j} V^m\otimes {\bigwedge}^{j}  V^{n}   $$

gives a lift from enumeration to isomorphism of objects (linear spaces)
which holds true not only over $\Fun$, but actually over any field.

----------------------

Some similar  **examples**  can be found here:
https://mathoverflow.net/questions/299748/can-one-divide-algebraic-manifolds-make-sense-gr2-n-gr2-nm-pn-1-p

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As a reincarnation of Gjergji Zaimi's question https://mathoverflow.net/questions/208560/q-catalan-numbers-from-grassmannians?noredirect=1&lq=1
it is natural to ask a particular case of question 1:

**Question 2:** Can one give any $\Fun/F_q$ interpretation/lift of any identity related to Catalan numbers? 

There are plenty of facts about the [Catalan numbers][4], but it seems not obvious to interpret them geometrically. See Will Sawin's [comment](https://mathoverflow.net/questions/208560/q-catalan-numbers-from-grassmannians#comment517485_208560) under the above mentioned  question
— the $q$-Catalan number itself is NOT the number of points of any smooth projective variety over $F_q$.


  [1]: https://en.wikipedia.org/wiki/Field_with_one_element
  [2]: https://en.wikipedia.org/wiki/Vandermonde%27s_identity
  [3]: https://en.wikipedia.org/wiki/Q-Vandermonde_identity
  [4]: https://en.wikipedia.org/wiki/Catalan_number