$\newcommand{\Fun}{F_\text{un}}$There was lots of "Fun with $\Fun$" (field with one element) 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 Is there a lift of the q-Vandermonde identity to some geometric (motivic) identity for Grassmannians over $F_q$?)

Combinatorial side:

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

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.

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

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: Can one divide algebraic manifolds ? Make sense: $Gr(2,n)/ Gr(2,n+m) = P^{n-1}/P^{n+m-1} P^{n-2}/P^{n+m-2}$

As a reincarnation of Gjergji Zaimi's question q-Catalan numbers from Grassmannians 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, but it seems not obvious to interpret them geometrically. See Will Sawin's comment under the above mentioned question — the $q$-Catalan number itself is NOT the number of points of any smooth projective variety over $F_q$.