Unexpected use of polynomials in combinatorics

Question

Can someone please post some (relatively easy, say high school level) combinatorial problems which can be solved with polynomials but NOT generating functions.

Related to this post in ME.SE.

Answer 1

Problem: Is it possible to weight two dice (possibly unequally) in such a way that every sum from $2$ to $12$ is equally probable?

The Answer is No.

Comment: I set this problem for high school students a few years back. These were students who did well at competition style problems. They found this one difficult, but were able to make very good progress on it, and arrived at the first proof entirely on their own. They got the second after the basic idea was proposed to them. It's not intended to be an easy problem and it may well be more difficult than you were looking for.

Motivating Example: to see that you can not do this with equally weighted dice, suppose you could and let $p_i$ be the probability a given die comes up $i$. Then $p_1^2=p_6^2=\frac 1{11}\implies p_1=p_6=\sqrt {\frac 1{11}}$ But then the probability of getting a $7$ is at least $2p_1p_6=\frac 2{11}$, a contradiction.

Direct proof (no polynomials): Say the probabilities for the first die are $p_i$ and for the second $q_i$. Then $p_1q_1=\frac 1{11}=p_6q_6$. But then we can get a lower bound on the probability of throwing a $7$ as $$\text {Prob}(Sum = 7)≥p_1q_6+p_6q_1=p_1\times \frac {1}{11p_6}+p_6\times \frac {1}{11p_1}=\frac 1{11}\times \left(\frac {p_1}{p_6}+\frac {p_6}{p_1}\right)≥\frac 2{11}$$ so we reach the same contradiction as in the equally weighted case. Here, of course, we have used the fact that $x+\frac 1x≥2$.

Proof using polynomials: Let the $p's$ and $q's$ be as above. Define two polynomials $$P(x)=\sum_{i=1}^6p_ix^i\quad \&\quad Q(x)=\sum_{i=1}^6q_ix^i$$

Note that both $P(x), Q(x)$ are divisible by $x$ and that $P(1)=Q(1)=1$.

If the weightings were to work we'd have the product $$P(x)Q(x)=\frac 1{11}\sum_{i=2}^{12}x^i=\frac {x^2}{11}\frac {x^{11}-1}{x-1}$$

However, this is impossible. Indeed, factoring out the factors of $x$ from $P(x),Q(x)$ we get two polynomials of degree $5$ each (noting that neither $p_6$ nor $q_6$ can be $0$). As $5$ is odd, these must have at least one real root each...but $x^{11}-1$ has no real root other than $1$ (which is not a root of $P(x)$ nor $Q(x)$).

Answer 2

Here is an example I was talking about:

We have $2n$ different numbers $a_1,...a_n, b_1,...b_n$. A table $n\times n$ is divided on $n^2$ unit cells and in cell $(i,j)$ we write a number $a_i+b_j$. Suppose that all products of numbers written in cells in each column are the same. Prove that then all products of numbers written in cells in each row are the same.

Idea for a solution: Observe a polynomial $$P(x) = (x+a_1)...(x+a_n)-(x-b_1)...(x-b_n)$$

Answer 3

There is a classical trick for deriving combinatorial identities that proceeds as follows:

1. Find some combinatorial identity that you can prove for all $n \in \mathbb{N}$ and that has the form $P\left(n\right) = Q\left(n\right)$, where $P$ and $Q$ are two fixed polynomials. For example, the identity $\dbinom{n}{a}\dbinom{a}{b} = \dbinom{n}{b}\dbinom{n-b}{a-b}$ (for fixed nonnegative integers $a$ and $b$) fits this description (with the polynomials $P$ and $Q$ given by $P\left(x\right) = \dbinom{x}{a}\dbinom{a}{b}$ and $Q\left(x\right) = \dbinom{x}{b}\dbinom{x-b}{a-b}$); it can be proved for all $n \in \mathbb{N}$ by double-counting the number of ways to choose two nested subsets $B \subseteq A \subseteq \left\{1,2,\ldots,n\right\}$ with $\left|A\right| = a$ and $\left|B\right| = b$.

2. Recall the fact that if two polynomials (over $\mathbb{Q}$ or $\mathbb{R}$) are equal on infinitely many points, then they are identical. In Step 1, you have shown that the polynomials $P$ and $Q$ are equal on infinitely many points (namely, on all $n \in \mathbb{N}$). Thus, they are identical. Hence, $P\left(n\right) = Q\left(n\right)$ holds not only for all $n \in \mathbb{N}$, but also for all $n \in \mathbb{R}$.

3. (Optional) This allows you to substitute $-n$ for $n$ in the identity. If you want, you can rewrite the resulting identity by getting rid of negative numbers on top of binomial coefficients (using the upper negation formula $\dbinom{-n}{k} = \left(-1\right)^k\dbinom{n+k-1}{k}$ for all $k \in \mathbb{Z}$). For example, if you substitute $-n$ for $n$ in the above-mentioned identity $\dbinom{n}{a}\dbinom{a}{b} = \dbinom{n}{b}\dbinom{n-b}{a-b}$, then you obtain $\dbinom{-n}{a}\dbinom{a}{b} = \dbinom{-n}{b}\dbinom{-n-b}{a-b}$; then, using upper negation, you can rewrite this as $\left(-1\right)^a \dbinom{n+a-1}{a} \dbinom{a}{b} = \left(-1\right)^b \dbinom{n+b-1}{b} \left(-1\right)^{a-b} \dbinom{n+b+a-b-1}{a-b}$. Okay, this time you haven't found anything new (you can easily derive this new identity directly from the original one), but often you do.

Here is another example. For any nonnegative integers $m$ and $i$, we let $\operatorname{sur}\left(m,i\right)$ denote the number of surjective maps from $\left\{1,2,\ldots,m\right\}$ to $\left\{1,2,\ldots,i\right\}$. (The cognoscienti will recognize this number as $i! {m \brace i}$, where ${m \brace i}$ is a Stirling number of the second kind. More important to us is the obvious fact that $\operatorname{sur}\left(m,i\right)$ counts not only the surjective maps from $\left\{1,2,\ldots,m\right\}$ to $\left\{1,2,\ldots,i\right\}$, but also the surjective maps from any given $m$-element set to any given $i$-element set.) It is easy to prove (by double-counting) that \begin{align} n^m = \sum_{i=0}^m \operatorname{sur}\left(m,i\right) \dbinom{n}{i} \label{darij1.eq.nmsur1} \tag{1} \end{align} for any $n \in \mathbb{N}$ and $m \in \mathbb{N}$. (Indeed, you can double-count the number of all maps from $\left\{1,2,\ldots,m\right\}$ to $\left\{1,2,\ldots,n\right\}$. The left hand side is the obvious count; the right hand side counts them by first choosing their image and then mapping $\left\{1,2,\ldots,m\right\}$ surjectively onto that image.)

Now, the identity \eqref{darij1.eq.nmsur1} has the form $P\left(n\right) = Q\left(n\right)$ for two polynomials $P$ and $Q$: namely, for $P\left(x\right) = x^m$ and for $Q\left(x\right) = \sum_{i=0}^m \operatorname{sur}\left(m,i\right) \dbinom{x}{i}$. Therefore, the above trick (specifically, its step 2) shows that it must hold not only for all $n \in \mathbb{N}$, but also for all $n \in \mathbb{R}$. In particular, we can substitute $-n$ for $n$ into it, and thus obtain \begin{align} \left(-n\right)^m = \sum_{i=0}^m \operatorname{sur}\left(m,i\right) \dbinom{-n}{i} . \end{align} In light of the upper negation formula $\dbinom{-n}{i} = \left(-1\right)^i\dbinom{n+i-1}{i}$, we can rewrite this as \begin{align} \left(-n\right)^m = \sum_{i=0}^m \operatorname{sur}\left(m,i\right) \left(-1\right)^i \dbinom{n+i-1}{i} . \end{align} Multiplying this by $\left(-1\right)^m$, we obtain \begin{align} n^m = \sum_{i=0}^m \operatorname{sur}\left(m,i\right) \left(-1\right)^{m-i} \dbinom{n+i-1}{i} . \end{align} Do you see how to prove this combinatorially? (Note that the particular case $n = 1$ of this identity takes the particularly simple form $\left(-1\right)^m = \sum_{i=0}^m \operatorname{sur}\left(m,i\right) \left(-1\right)^{m-i}$, since $\dbinom{1+i-1}{i}=\dbinom{i}{i}=1$. This identity is equivalent to Theorem 2.2.2 in Bruce Sagan, Combinatorics: The Art of Counting, version 2020-01-20, where it is proved using a sign-reversing involution. Perhaps the same method applies to the general case; but it is not as easy as ours!)

Note that, while we have generalized \eqref{darij1.eq.nmsur1} to all $n \in \mathbb{R}$, we cannot generalize \eqref{darij1.eq.nmsur1} to all $m \in \mathbb{R}$, since $\operatorname{sur}\left(m,i\right)$ is not a polynomial in $m$ (and also because $m$ appears as a summation bound and as an exponent in \eqref{darij1.eq.nmsur1}).

More examples of how this trick can (and cannot) be used are found in §2.6 (particularly §2.6.4 and §2.6.5) of my Enumerative Combinatorics notes.

Answer 4

Have You heard about rook polynomials?

Link to wikipedia: https://en.wikipedia.org/wiki/Rook_polynomial

Link to reasonable question on mathstack:

Use of rook polynomials

One more link with nice explanation:

https://www.d.umn.edu/~jgreene/Combinatorics/Fall_2015/Rook_polynomials.pdf

Answer 5

Here is another one:

Given two sequences of natural numbers $\{a_k\}$ and $\{b_k\}$, $k=1,\ldots,n$ (with non-identical sets of elements) such that the sets of their pairwise sums $$\{a_1+a_2,a_1 + a_3,\ldots, a_{n-1}+a_n\}$$ and $$\{b_1+b_2,b_1 + b_3,\ldots, b_{n-1}+b_n\}$$ coincide, show that $n=2^m,\ m\in\mathbb{N}.$

Answer 6

I would consider the chromatic polynomial of a graph a combinatorial use of a polynomial, since it counts the number of colourings in a graph.

Given a graph, $G$, the chromatic polynomial, $P(G, k)$ counts the number of ways you can $k$-colour $G$. This is not a generating function because the values of $P(G, k)$ encode the counts not the coefficients.

Answer 7

See Lucas's Theorem. There is one proof with groups and another with polynomials. Both are very nice. (There is a third by induction but it is pretty bad.)

Answer 8

I'm very surprised no one has named Vandermonde's Identity.

$$\sum_{j=0}^m \binom{a}{j}\binom{b}{m-j} = \binom{a+b}{m}$$

It is quite a beautiful exercise to convince a high-school student that polynomials can be a powerful tool in combinatorics.

Of course, proving that the above identity is true can be done combinatorially interpreting that you have to count the number of ways of picking $m$ balls, of which exactly $a$ are red and $b$ are blue (and they are all labeled, say).

The polynomials may arise naturally if the students already know the Binomial Theorem. If you try to compute $(1+x)^{a+b}$ using that, then the coefficient of $x^m$ is exactly $\binom{a+b}{m}$. Also, if you do the same with $(1+x)^a (1+x)^b$ and use the convolution formula for the product of polynomials, this gives you the "polynomial proof".

Answer 9

Here is another example:

We start with any finite list of distinct positive integers. We may replace any pair $n$, $n + 1$ (not necessarily adjacent in the list) by the single integer $n−2$, now allowing negatives and repeats in the list. We may also replace any pair $n$, $n + 4$ by $n − 1$. We may repeat these operations as many times as we wish. Either determine the most negative integer which can appear in a list, or prove that there is no such minimum.