In combinatorics, how can one verify that one has counted correctly?

Question

This is a soft question, but I've tried to be specific about my concerns. When studying basic combinatorics, I was struck by the fact that it seems hard to verify if one has counted correctly.

It's easiest to explain with an example, so I'll give one (it's fun!) and then pose my questions at the bottom. In this example there are two ways to count the number of ways $2n$ people can be placed into $n$ partnerships (from the book Introduction to Probability by Blitzstein and Hwang):

$$\frac{(2n)!}{2^n \cdot n!} = (2n - 1)(2n - 3) \cdots 3 \cdot 1$$

Story proof for the right side: For the first person you have $(2n - 1)$ choices for partner. Then for the next person you have $(2n - 3)$ choices, etc.

Story proof for the left side: Line all $2n$ people up, walk down the line, and select every 2 people as a pair. The ordering you chose for the line determines the pairs, and there are $2n!$ ways to order $2n$ people. But divide by $2^n$ because the order within each pairing doesn't matter. Also divide by $n!$ because the order of the pairings doesn't matter.

My question is:

What if I had been tasked with getting this number at work, and I chose the approach on the left side, but I neglected to divide by $2^n$ and only divided by $n!$? I'd be wrong of course, but how could I have known?

I suppose one answer to my question could just be "Try hard, try to think of ways you might be over/undercounting, look at other examples, and continue to study theory."

But the prospect of not knowing if I'm wrong makes me nervous. So I'm looking for more concrete things I can do. So, three questions of increasing rigor:

• What are specific "habits of mind" that people in combinatorics use to avoid error?
• What are specific validation techniques that such people use to check their work? (One does come to mind from my example: Calculate the same thing two ways and confirm equality)
• Is there any formal list of questions that, if answered, should verify that one's approach is correct?

Answer 1

Here are some methods I have used. Some of these have already been suggested in the comments, and none of them are fool-proof.

1. Work out a few small cases by hand, i.e. generate all the possibilities, and verify that your formula works in those cases. (Often working out the small cases will also suggest a method of solution.)
2. Solve the problem by two different methods. For example, often a problem which is solved by the principle of inclusion/exclusion can also be solved by generating functions. Or it may be that you can derive a recurrence that the solution must satisfy, and verify that your solution satisfies the recurrence; if it is too hard to do the general case, verify the recurrence is satisfied for a few small cases.
3. Write a computer program that solves a particular case of the problem by "brute force", i.e. enumerates all the possibilities, and check the count against your analytic answer. If you don't yet know how to program, this is one of many reasons why it's good to to learn how. For example, how many ways can you make change for a dollar? It's probably easier to write a program that generates all the ways than to solve this problem analytically.
4. Sometimes a combinatoric problem can be reinterpreted as a probability problem. For example, the problem of counting all the poker hands that form a full house is closely connected to the probability of drawing a full house. In this case you can check your answer by Monte Carlo simulation: use a random number generator to simulate many computer hands and count the number which are full houses. Once again, this requires computer programming skills. You won't get an exact match against your analytic answer, but the proportion of full houses should be close to your analytic result. Here it helps to know some basic statistics, e.g. the Student's t test, so you can check to see if your answer falls in the range of plausible results.

Answer 2

Here are some other methods:

• Estimate what the answer should be. For example, Let's say I want to count all 3-element subsets of $1,\dots,n$ where the sum of any two elements is not equal to the third. I think to myself "hmm, if $n$ is large then most three element subsets should be valid." So whatever the answer is, I know that it should be close to $\binom{n}{3}$ if $n \gg 0$.

• Also keep in mind that your intuition about how large the answer should be might be off. This is ok! It means you will put more thought into the problem.

• Have someone look over your work. This can work really well because sometimes you'll have something you are mistaken about in the back of your mind when you are writing. When this happens, what you write will (hopefully) look funny to other people whereas it might take a lot of thought on your part to catch that something is off.

• Be more rigorous. This is easier said than done, obviously. For example if you are applying a theorem, look to make sure that you satisfy all the necessary hypotheses to be able to apply it.

• Slow down and think about "why am I multiplying here?" or dividing, or differentiating. Make sure the algebra makes sense combinatorially. If you are dividing, make sure you end up with integers where there should be integers.

Answer 3

It is actually possible to be rigorous in questions like this – though it can be a lot of work.

The first step is to build a mathematical model that corresponds to the question. This part deserves some thought, and there is scope for error, but usually just expressing the question mathematically is easier to do without error than finding the actual answer.

For this example, we might say that a partnering of a group of $2n$ people (whom we'll refer to as $G=\left\{0, 1, \ldots, 2n-1\right\}$) is any function $p: G \rightarrow G$ such that:

• For all $m$: $p(p(m)) = m$ (i.e. a person's partner is paired with the person)
• For all $m$: $p(m) \neq m$ (i.e. nobody is partnered with themself)

Hopefully it's fairly clear that the above corresponds to the definition of a partnering from the question - if not, you can do it in a way that's clearer to you.

Having made that definition, the question becomes: how many such functions $p$ are there?

One way to establish the answer rigorously is to come up with a bijection $B$ between the set of partnerings $P = \{p: p \,\hbox{is a partnering of $2n$ people}\}$ and an initial subset of the natural numbers $\mathbb N$.

That can be done as follows – I'll omit a lot of details for brevity, but hopefully it should be clear that this can be made fully rigorous. Given a partnering $p$:

Start with a set $U_0=G$ of unpartnered people, and a value $v_0=0$. At each step $k$, starting from 0:

• Consider the person from $U_k$ with the smallest number – call them $g_k$.
• Find out who they're partnered with – that is, evaluate $p(g_k)$ - and find that person's "rank" in $U_k\setminus g_k$ starting from 0. That is, if (of all people other than $g_k$ who are so far unpartnered) $p(g_k)$ has the lowest number, call them rank 0, and if there are three with numbers lower than $p(g_k)$ that are unpartnered, call them rank 3, and so on. Call that rank $r_k$.
• Set $U_{k+1} = U_k \setminus \{g_k, p(g_k)\}$ and $v_{k+1} = v_k\left|U_k-1\right| + r_k$. That is, take out the two people we just partnered, and calculate a new value from the previous step's value and the rank of the person we just partnered with $g_k$.

Repeat for $k$ from 0 to $n-2$, and take $v_{n-1}$ as our result for $B(p)$.

It's clear that, given any partnering $p$, this procedure gives us back a number $B(p)$. What may not yet be clear is that, given $B(p)$, we can determine what $p$ was – that is, $B$ is a bijection.

I won't go into the details of that, but note that, when calculating the value, the person who gets partnered with person 0 has their rank multiplied by $(2n-3)(2n-5)\ldots1$, and we can get it back by dividing $B(p)$ by that number – the integer part of the answer gives us $p(0)$, and then we can use the remainder in a similar way to deduce the rest of the pairings in $p$.

So (again I'm leaving out some proof details) we have a function that is a bijection between $\left\{0,1,2,\ldots,m-1\right\}$ and the set of possible partnerings of $2n$ people. This is enough to show that there are $m$ such partnerings – and it's reasonably easy to see (by taking the maximum possible value at each step in the procedure above) that the value of $m$ is exactly $(2n-1)(2n-3)\cdots 1$ as you already knew.

Answer 4

One practical method, which I can illustrate with your example, is to get rid of fractions.

This gives you more concrete things to count and avoids 'story proof' phrases like "divide by ... because order doesn't matter''.

It means that if we want to check the claim that $$\mathrm{Answer} = \frac{(2n)!}{2^nn^!}, $$ then we should instead to consider proving that $$\mathrm{Answer} \times n! \times 2^n = (2n)!$$

This more readily suggests how to count: The idea is now clearly to count something of cardinality $(2n)!$ in two ways. Obviously, the problem suggests we should count the elements of $S_{2n}$ i.e. the number of ways of ordering the $2n$ people.

Looking at the factors on the left it is clear that a rigorous proof can be obtained if we can come up with a bijection $$ A_1 \times A_2 \times A_3 \longleftrightarrow S_{2n},$$ where $$ |A_1| = \mathrm{Answer}; \qquad |A_2| = n!; \qquad |A_3| = 2^n$$

There are of course very natural candidates for things of size $n!$ and $2^n$, leading us to see that an arrangement of $\{1,...,2n\}$ is obtained by

1. Picking a partition of $\{1,...,2n\}$ into $n$ pairs (in one of $|A_1|$ ways).
2. Picking an arrangement for those $n$ pairs (in one of $|A_2| = n!$ ways).
3. Choosing an ordering for each pair in the partition (in one of $|A_3| = 2^n$ ways).

If you want to make these steps more verifiable, then the notation is bound to get more complicated: Off the top of my head you could say:

1. Pick a partition $$X = \bigl\{\{x_{11},x_{12}\}, \{x_{21},x_{22}\},\dots,\{x_{n1},x_{n2}\}\bigr\}$$ of $\{1,...,2n\}$ into $n$ pairs.
2. Pick an arrangement $\alpha \in S_n$ for those pairs.
3. Choose an element from each pair to go 'first', by picking $\beta = (\beta_1,\dots,\beta_n) \in S_2^n$ (i.e. the cartesian product of $S_2$ with itself $n$ times).

Then we can actually define $F : A_1 \times S_n \times S_2^n \longrightarrow S_{2n}$ by saying $F(X,\alpha,\beta) = \sigma$, where, for odd $i$: $$ \sigma(i) = x_{\alpha(i)\beta_i(1)}; \qquad \text{and} \qquad \sigma(i+1) = x_{\alpha(i)\beta_i(2)} $$ Then one really can claim: "$F$ is surjective" and actually write down a proof. And then: "$F$ is injective" and write a proof.

I personally don't think these things should be glossed over and that in principle there's loads to be learned from completely thinking through a problem in the way I'm doing here, but typically this sort of thing isn't taught.

Answer 5

If you are working with specific numbers in the form $\binom{n}{k}$, there's a really neat thing you can do involving Pascal's Triangle. Although this works best for small numbers, it is a reliable way to check your method and ensure that you have gotten the correct answer. The table is as follows:

\begin{array}{|r|rrrrrr} \hline & 0 & 1 & 2 & 3 & 4 & \cdots \\ \hline 0 & 1 & & & \\ 1 & 1 & 1 & & \\ 2 & 1 & 2 & 1 & \\ 3 & 1 & 3 & 3 & 1 \\ 4 & 1 & 4 & 6 & 4 & 1 \\ \cdots & . & . & . & . & . & . \\ \end{array}

Using the vertical column for $n$ and the horizontal row for $k$, we can use this property to check an answer for a specific combinatorics question. For instance, if we calculate $\binom{4}{2}$, we get the answer of $6$.

This works because Pascal's Triangle gives us binomial coefficients: for example, expanding the binomial $(a + b)(a + b)$, we get $a^2 + 2ab + b^2$. The coefficients of each match the 3rd line of Pascal's Triangle, [1, 2, 1], as there is 1 way to write $aa$, 2 ways to write $ab$ ($ab$ or $ba$), and 1 way to write $bb$. That's the link to combinatorics, and that's why this is an effective way to check if your answer is correct, for small values of $n$ and $k$. That said, using this as verification is best to check if your method is correct, and extrapolate with larger $n$ from there. Unless you have a big enough Pascal's Triangle lying around! (:

Useful links:

Pascal's Triangle - Wikipedia

Combinatorics - Wikipedia, especially the subfields section

Combinatorial proof - Wikipedia, for proving the formula used to help understand it.