\(\newcommand{\set}[1]{\{\, #1 \,\}}\newcommand{\card}[1]{|#1|}\newcommand{\pset}[1]{𝒫(#1)}\)

Recall that the cardinality or size of a set \(S\), written \(\card{S}\), is the number of elements contained in \(S\). In some cases, computing the size of a set is straightforward. For example, if \(S = \set{a, b, c, d, e}\) then \(\card{S} = 5\) by inspection of the definition of \(S\). However, suppose we have:
\[\begin{gather} S_1 = \set{a, b, c, d} \\ S_2 = \set{a, c, e, f}. \end{gather}\]

What is \(\card{\pset{S_1 \cup S_2}}\)? We can compute the contents of \(\pset{S_1 \cup S_2}\) and then count the number of elements. However, if there are many elements in the set, it might be impractical to compute-and-count. Furthermore, what if don’t know the contents of \(S_1\) and \(S_2\)? We would like to express the cardinality of this quantity in terms of \(\card{S_1}\) and \(\card{S_2}\).

In this chapter, we focus on techniques for calculating the cardinality of finite sets, a branch of mathematics called enumerative combinatorics. As computer scientists, we are interested in not just modeling data but also performing operations over this data. Thus, we care greatly about techniques for calculating the sizes of sets as their sizes ultimately influence the expected runtime of the algorithms we development.

For example, consider the problem of determining the optimal route for a delivery truck to visit a number of businesses in a city and return back to the delivery center. This problem, a variant of the traveling salesman problem, is a fundamental problem with applications to a wide range of domains. A simple algorithm is as follows:

Enumerate every possible path through the city that originates from the delivery center and pick the shortest among them that (a) visits every business and (b) returns to the center.

How long would this program take to run? This is equivalent to asking the following question: what is the size of the set of all possible paths through the city? It turns out for a sufficiently well-connected city, there are an exponential number of paths to consider, far too many to simply enumerate in a reasonable amount of time. To see why this is the case, we will develop principles for counting sets of increasing complexity, using our set operations as a guide.

The Sum and Product Rules

Let’s first explore how we might calculate the cardinality of the union of two sets. Suppose that we have the following sets:

\[\begin{gather} S_1 = \set{a, b, c} \\ S_2 = \set{d, e} \end{gather}\]

\(S_1 \cup S_2 = \set{a, b, c, d, e}\) so \(\card{S_1 \cup S_2} = 5\). \(\card{S_1} = 3\) and \(\card{S_2} = 2\) so it is tempting to infer that the cardinality of the union of two sets is the sum of their cardinalities. However, consider the following alternative sets:

\[\begin{gather} S_1 = \set{a, b, c} \\ S_2 = \set{a, c} \end{gather}\]

\(\card{S_1} = 3\) and \(\card{S_2} = 2\) but \(S_1 \cup S_2 = \set{a, b, c}\) and thus \(\card{S_1 \cup S_2} = 3\). Therefore, in order to assert that the cardinality of the union of sets is the sum of the cardinalities of the sets, we must also require that the sets do not possess elements in common! This gives rise to the sum rule for set cardinalities:

Note that we capture the notion of “sets do not possess elements in common” with the condition \(S_1 \cap S_2 = ∅\).

We can generalize the sum rule to any number of sets as long as they are pairwise disjoint.

Then we can say that for a collection of sets \(S_1, \ldots, S_k\), if they are pairwise disjoint, then \(\card{S_1 \cup \cdots \cup S_k} = \card{S_1} + \cdots + \card{S_k}\).

Next, let’s consider the Cartesian product, \(S_1 × S_2\). Suppose that we again have:

\[\begin{gather} S_1 = \set{a, b, c} \\ S_2 = \set{d, e} \end{gather}\]

Then:

\[\begin{align} S_1 \times S_2 =&\; \{\, (a, d), (a, e) \\ &\; , (b, d), (b, e) \\ &\; , (c, d), (c, e) \,\} \end{align}\]

So \(\card{S_1 \times S_2} = 6\). Because \(\card{S_1} = 3\) and \(\card{S_2} = 2\), we can infer that the cardinality of the Cartesian product is the product of the input sets.

Indeed, this is the case, which gives us the product rule for sets.

Counting as Choices

Set operations give us a formal, definition-based view of counting elements in sets. A useful, higher-level view of our counting principles are phrasing them as the number of possible choices we can make from a given set. This view of counting as choice is particularly useful for applying counting principles to real-world examples.

For example, let’s consider the sum rule. Suppose that we have on a field trip:

• 10 first grade students,
• 15 second grade students, and
• 8 third grade students.

There are \(10 + 15 + 8 = 33\) different students we can choose from, overall. If we label the sets \(S_1\), \(S_2\), and \(S_3\), respectively, then the sum rule tells us that:

\[ \card{S_1 \cup S_2 \cup S_3} = \card{S_1} + \card{S_2} + \card{S_3} = 10 + 15 + 8 = 33. \]

Set union and the sum rule allow us to consider choices when we combine sets into a single set. In contrast, Cartesian product allows us to make independent choices from a collection of sets. Each of the \(k\) sets of a Cartesian product represents a different pool of elements to choose from. The Cartesian product enumerates all the different ways we can generate a tuple of \(k\) elements by choosing one element from each pool.

Suppose that we have two hats, five shirts, three pairs of pants, and two pairs of shoes. The total number of outfits we can put together consisting of a hat, shirt, pants, and shoes is:

\[ 2 × 5 × 3 × 2 = 60. \]

Alternatively, we can think of the problem as having four sets, one for hats, shirts, pants, and shoes. An outfit is, therefore, a 4-tuple with elements drawn from each of these sets. The Cartesian product of these four sets then gives us all possible outfits as 4-tuples.

Combinatorial Descriptions

Consider the total quantity of outfits we derived previous:

\[ 2 × 5 × 3 × 2. \]

This unsimplified formula actually tells us quite a bit about the quantity in question. Because of the product rule, we know that the quantity represents the number of ways we can form choices from pools of two, five, three, and two choices, respectively.

In contrast, consider in isolation the value that this formula evaluates to:

\[ 60. \]

While technically accurate, this value tells us very little about the structure of the quantity or object that we are counting!

When counting quantities, we will universally favor giving unsimplified formulae for our set cardinalities rather than simplifying the formulae to a final result. We call these formulae combinatorial descriptions because these unsimplified cardinality formulae communicate the various choices we made in constructing an object in terms of set operations and our counting principles. In effect, a combinatorial description serves as a terse proof that an object can be decomposed used our counting principles as long as we know how to interpret the arithmetic operations contained within!

The Power Set

We saw in the previous reading that the power set of a set \(S\) is the set of all subsets that you can make from \(S\). If \(S = \set{a, b, c}\), then:

\[\begin{align} S =&\; \{\, \emptyset \\ &\; , \set{a}, \set{b}, \set{c} \\ &\; , \set{a, b}, \set{b, c}, \set{a, c} \\ &\;, \set{a, b, c} \,\}. \end{align}\]

So \(\card{S} = 3\) and \(\card{\pset{S}} = 8\).

With additional data points, we can see that the seems to grow exponentially with the size of the input set! Indeed, the following property of the size of a power set is true:

It is difficult to validate this empirically. If this proposition is true, then trying to calculate the power set of a 10-element set ought to result in \(2^10 = 1024\) elements which is certainly not reasonable to do by-hand! Instead, we would like to prove that this formula holds through a counting argument that establishes the cardinality of a set without needing to enumerate elements. We instead use our counting principles in a systematic fashion to describe how to build or choose elements from that set.

Let’s see how we can do this to justify our claim about power sets.

In other words, we can choose a subset of a set by forming a \(k\)-tuple. Each element of the \(k\)-tuple corresponds to one of the elements of the set. We can then assign a boolean value to each position indicating whether that element is in or out of the subset.

As a concrete example, suppose we have \(S = \set{a, b, c}\). Then the tuple \((t, f, t)\) corresponds to the subset \(\set{a, c}\).

This particular counting argument, the in-out argument, is particularly useful in computer science because we frequently work with binary (0–1 data) or boolean choices (yes-or-no). As an example, suppose we have a piece of datum that is \(k\) bits wide, e.g., 32 bits for an integer. Recall that a bit can either be set to 0 or 1. We can think of each integer as the collection of bits in the 32 bit sequence that are set 1. Since there are 32 bits in the collection, there must be \(2^{32}\) such possible sets and thus \(2^{32}\) possible integers. (Note that other bits of the datum might be devoted to other tasks, so the number of effective datum possible from \(k\)-bits might be different than this raw quantity!)

The Inclusion-Exclusion Principle

The sum rule gives us a cardinality of the union of sets provided they are pairwise disjoint. However, can we use intersection to precisely characterize the size of the union without the need for a restriction?

To explore this idea, consider the following statistics about college majors:

• There are 45 computer science majors.
• There are 20 math majors.
• There are 30 economics majors.

And suppose that we want to compute the total number of CS, math, and econ majors at the college. It would be tempting to say that the total is \(45 + 20 + 30 = 95\) but what about a double major? If a person majors in both computer science and math, they would be counted twice, once for the count of CS majors and once for the count of math majors. If we knew how many double majors there were of all possible combinations, we can subtract them out once to account for this overcounting.

Suppose that we know that:

• There are 10 computer science and math double majors.
• There are 5 computer science and economics double majors.
• There are 5 math and economics double majors.

Then, we might say that the total number of majors is:

\[ 45 + 20 + 30 - 10 - 5 - 5 = 75. \]

However, what about those rare triple majors? Consider such a single triple major:

• The triple major appears once in each of the individual major counts, so we triple counted them in our addition.
• The triple major appears once in each of the double major counts, so we triple counted them again in our subtraction.

That means we need to add them back in one last time! If we know that:

• There are 2 computer science, math, and economics triple majors.

Then the total number of majors is:

\[ 45 + 20 + 30 - 10 - 5 - 5 + 2 = 77. \]

Intuitively, adding up all the singleton sets of majors overcounts the double major overlap, so we subtract them out. But by subtracting out the double major overlap, we undercount the triple majors, so we add them back in.

We generalize this alternation of addition and subtraction to account for overlap in the Inclusion-exclusion Principle or Generalized sum rule:

In other words, the cardinality of the union of a set of sets \(S_1 ∪ … ∪ S_k\), written \(\card{S_1 ∪ … ∪ S_k}\) is the sum of all the cardinalities of the singleton sets, the difference of all the intersection of pairs of sets, the sum of all the intersection of triples of sets, and so forth.