Pascal's rule

$$ \binom{n}{k}=\binom{n-1}{k-1}+\binom{n-1}{k} \tag{1} $$

may be used recursively to obtain the hockey stick identity

$$ \binom{n+1}{k+1}=\binom{n}{k}+\binom{n-1}{k}+\cdots+\binom{k}{k}. \tag{2} $$

The reason for the name is that if all these binomials are highlighted in Pascal's triangle, they form what looks like a hockey stick. This is a special case of a more general identity,

$$ \binom{n+1}{r+s+1}=\sum_{a+b=n}\binom{a}{r}\binom{b}{s}, \tag{3} $$

where $\binom{a}{r}\binom{b}{s}$ counts the subsets of $\{1,\cdots,n+1\}$ where the element which separates $r$ terms on the left and $s$ terms on the right within the subset happens to separate $a$ terms on the left and $b$ terms on the right in the whole set (where we interpret left/right relative to the usual linear ordering of the numbers $1,\cdots,n+1$).

For $q$-binomials, Pascal's rule states

$$ \left[\begin{array}{c} n \\ k \end{array}\right] = \left[\begin{array}{c} n-1 \\ k-1 \end{array}\right]+q^k\left[\begin{array}{c} n-1 \\ k \end{array}\right] \tag{4} $$

which can be used recursively to obtain

$$ \left[\begin{array}{c} n+1 \\ k+1 \end{array}\right]= \left[\begin{array}{c} n \\ k \end{array}\right] + q^{k+1} \left[\begin{array}{c} n-1 \\ k \end{array}\right]+q^{2(k+1)} \left[\begin{array}{c} n-2 \\ k \end{array}\right]+\cdots \tag{5} $$

By hand, I drew out a hockey stick in $q$-Pascal's triangle, and used Pascal's rule to move one line up in the triangle. Guessing the pattern, I conjecture a generalized $q$-hockey stick identity:

$$ \left[\begin{array}{c} n+1 \\ r+s+1 \end{array}\right] = \sum_{a+b=n} q^{(b-1)(r+1)}\left[\begin{array}{c} a \\ r \end{array}\right] \left[\begin{array}{c} b \\ s \end{array}\right] \tag{6} $$

Presumably one could write the generating function proof for $(2)$ (multiply both sides by $x^n$ and sum over $n$) and then translate that into one for $(6)$, which would be interesting to see in an answer. But my major interest is:

Question. Does $(6)$ have an enumerative interpretation with objects associated to $\mathbb{F}_q^n$?

(Compare with this enumerative interpretation of the $q$-analog of the Chu-Vandermonde convolution identity.)


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