*This is a continuation of Part 1 of this series of posts on $q$-analogs.*

## Counting by $q$’s

Another important area in which $q$-analogs come up is in combinatorics. In this context, $q$ is a formal variable, and the $q$-analog is a generating function in $q$, but viewed in a different light than usual generating functions. We think of the $q$-analog of as “$q$-counting” a set of weighted objects, where the weights are given by powers of $q$.

Say you’re trying to count permutations of $1,\ldots,n$, that is, ways of rearranging the numbers $1,\ldots,n$ in a row. There are $n$ ways to choose the first number, and once we choose that there are $n-1$ remaining choices for the second, then $n-2$ for the third, and so on. So there are $n!=n\cdot (n-1)\cdot \cdots \cdot 2\cdot 1$ ways to rearrange the entries. For instance, the $3!=6$ permutations of $1,2,3$ are $123$, $132$, $213$, $231$, $312$, $321$.

Now, say we weight the permutations according to how “mixed up” they are, in the sense of how many pairs of numbers are out of order. An **inversion** is a pair of entries in the permutation in which the bigger number is to the left of the smaller, and $\mathrm{inv}(\pi)$ denotes the number of inversions of the permutation $\pi$. The table below shows the permutations of 3 along with the number of inversions they contain.

$$\begin{array}{ccc}

p & \mathrm{inv}(p) & q^{\mathrm{inv}(p)}\\\hline

123 & 0 & 1 \\

132 & 1 & q\\

213 & 1 & q\\

231 & 2 & q^2\\

312 & 2 & q^2 \\

321 & 3 & q^3

\end{array}

$$

We weight each permutation $p$ by $q^{\mathrm{inv}(p)}$, and **$q$-count** by summing these $q$-powers, to form the sum $$\sum_{p\in S_n}q^{\mathrm{inv}(p)}$$ where $S_n$ is the set of all permutations of $1,\ldots,n$. So for $n=3$, the sum is $1+2q+2q^2+q^3$ by our table above.

We now come to an important philosophical distinction between $q$-analogs and generating functions. As a generating function, the sum $1+2q+2q^2+q^3$ is thought of in terms of the sequence of *coefficients*, $1,2,2,1$. Generatingfunctionologically, we might instead write the sum as $\sum_{i=0}^\infty c_i q^i$ where $c_i$ is the number of permutations of length $n$ with $i$ inversions. But in $q$-analog notation, $\sum_{p\in S_n}q^{\mathrm{inv}(p)}$, we understand that it is not the coefficients but rather the *exponents* of our summation that we are interested in..

In general, a *combinatorial $q$-analog* can be defined as a summation of $q$-powers $q^{\mathrm{stat}(p)}$ where $p$ ranges over a certain set of combinatorial objects and $\mathrm{stat}$ is a statistic on these objects. Recall that we defined an “interesting $q$-analog” of an expression $P$ to be an expression $P_q$ such that

- Setting $q=1$ or taking the limit as $q\to 1$ results in $P$,
- $P_q$ can be expressed in terms of (possibly infinite) sums or products of rational functions of $q$ over some field,
- $P_q$ gives us more refined information about something that $P$ describes, and
- $P_q$ has $P$-like properties.

Certainly setting $q=1$ in a combinatorial $q$-analog results in the total number of objects, and the $q$-analog gives us more information about the objects than just their total number. It’s also a polynomial in $q$, so it satisfies properties 1, 2, and 3 above. Let’s now see how our $q$-analog $\sum_{p\in S_n}q^{\mathrm{inv}(p)}$, which is a $q$-analog of $n!$, also satisfies property 4.

Notice that $1+2q+2q^2+q^3$ factors as $(1)(1+q)(1+q+q^2)$. Indeed, in general this turns out to be the same $q$-factorial we saw in the last post! That is, $$\sum_{p\in S_n}q^{\mathrm{inv}(p)}=(1)(1+q)(1+q+q^2)\cdots(1+q+\cdots+q^n)=(n)_q!.$$ So it satisfies property 4 by exhibiting a product formula like $n!$ itself. I posted a proof of this fact in this post, but let’s instead prove it by building up the theory of $q$-counting from the ground up.

The *multiplication principle* in combinatorics is the basic fact that the number of ways of choosing one thing from a set of $m$ things and another from a set of $n$ things is the product $m\cdot n$. But what if the things are weighted?

$q$-Multiplication Principle: Given two weighted sets $A$ and $B$ with $q$-counts $M(q)$ and $N(q)$, the $q$-count of the ways of choosing one element from $A$ and another from $B$ is the product $M(q)N(q)$, where the weight of a pair is the sum of the weights of the elements.

Let’s see how this plays out in the case of $(n)_q!$. If each entry in the permutation is weighted by the number of inversions it forms with smaller entries (to its right), then the first entry can be any of $1,2,\ldots,n$, which contributes a factor of $1+q+q^2+\cdots+q^{n-1}$ to the total. The next entry then can be any of the $n-1$ remaining entries, and since the first entry cannot be the smaller entry of an inversion with the second, this choice contributes a factor of $1+q+q^2+\cdots+q^{n-2}$ to the total by the same argument. Continuing in this fashion we get the $q$-factorial as our $q$-count.

Notice that even the *proof* was a $q$-analog of the proof that $n!$ is the number of permutations of $1,\ldots,n$, now that we have the $q$-Multiplication Principle.

That’s all for now! In the next post we’ll talk about how to use the $q$-Multiplication Principle to derive a combinatorial interpretation of the $q$-binomial coefficient, and discuss $q$-Catalan numbers and other fun $q$-analogs. Stay tuned!