CSCE 235

Handout 18: Counting

Assigned March 3, 2004

Principle of Inclusion-Exclusion

Given a finite number of finite sets, , the number of elements in the union

is

Some readers may notice that the problems discussed so far could also have been solved by Venn diagrams, which are an acceptable approach wherever possible. When more than three sets are involved, however, Venn diagrams can no longer be used.

Example

Of 30 personal computers (PCs) owned by faculty members in a certain university department, 20 run Windows, 8 have 21 inch monitors, 25 have CD-ROM drives, 20 have at least two of these features, and 6 have all three.

(a) How many PCs have at least one of these features?

(b) How many have none of these features?

(c) How many have exactly one feature?

Let W = { the set of PCs running under Windows }, M = { set of PCs with 21 inch monitors }, and C = { the set with CD-ROM drives }. We are given that , , , (“20 have at least two of these features”), and (“6 have all three”). Now, we have set up the problem. Now, what do we do? How do we proceed? The first strategy is to locate the unions. Out of the five sets, only is a union. Why do we look for unions? This is because we can break a union down to a sum using the Principle of Inclusion-Exclusion. By the Principle of Inclusion-Exclusion,

.

Now, we can simplify the above equation. How? , , , and . So, the above equation can be simplified to:

Now, the above equation is a bit familiar! We know that . Thus, we have . So, . Since we do not have any more unions to break down, we are ready to attempt to answer the three questions.

(a) The number of PCs with at least one feature is

(b) How many have none of these features? Since there are 27 PCs with at least one feature, then there are 30 – 7 = 3 PCs that have none of these features.

(c) How many have exactly one feature? We know that the number of PCs with at least one feature is 27. We know that the number of PCs with at least two features is 20 (given). Thus, the number of PCs with exactly one feature is simply 27 – 20 = 7.

Principle of Inclusion-Exclusion (Disjoint Sets)

Given n pairwise disjoint finite sets, , then

This principle allows us to have the addition rule and the multiplication rule.

Addition Rule (Sum Rule)

The number of ways in which precisely one of a collection of mutually exclusive events can occur is the sum of the number of ways in which each event can occur (basically the above).

If a first task can be done in ways and a second task in ways, and if these tasks cannot be done at the same time, then there are ways to do one of these tasks

Example

In how many ways can you get a total of six when rolling two dice?

The event “get a six” is the union of the following mutually exclusive subevents: = { two 3s }, = { a 2 and a 4 }, and = { a 1 and a 5 }. Are these subevents mutually exclusive, i.e., disjoint? Yes! That means we can use the principle of inclusion-exclusion (disjoint sets) and the addition rule. So, . Now, we need to find out the size of each set: , , . Thus, there are 1+2+2 = 5 ways to get a total of six when rolling two dice.

Multiplication Rule (Product Rule)

The number of ways in which a sequence of events can occur is the product of the number of ways in which each individual event can occur.

(a) For finite sets , we have

(b) More generally, suppose that a given set can be viewed as a set of ordered k-tuples with the following structure. There are possible choices of ; given , there are possible choices of ; and so on. So, the set has elements. We can use this to count cards.

Suppose that a procedure can be broken down into a sequence of two tasks. If there are ways to do the first task and ways to do the second task after the first task has been done, then there are ways to do the procedure.

Example

A license plate of a Nebraska vehicle has 3 digits and 3 letters. For example, NYZ 123. However, the first letter is always the letter ‘N’. Suppose the other letters can range between ‘A’ and ‘Z’, and each of the last three digits can range between 0 and 9, how many unique license plates can the state of Nebraska issue?

In this case, we have six sets: = { first letter: ‘N’ }, = { second letter: ‘A’ through ‘Z’ }, = { third letter: ‘A’ through ‘Z’ }, = { 1st digit after letter: 0 to 9 }, = { 2nd digit after letter: 0 to 9 }, and = { 3^rd digit after letter: 0 to 9 }. Are these sets mutually exclusive, i.e., disjoint? Yes! Once again, that means we can use the principle of inclusion-exclusion (disjoint sets) and the multiplication rule. So, . We know that . We know the size of the alphabet is 26, . And we know that . Thus, there are (1)(26)(26)(10)(10)(10) = 676,000 unique license plates that can be issued by the state. By the way, does that mean that the state of Nebraska expect only 676,000 cars registered within the state? What about vanity plates?

The Pigeon-Hole Principle

If n objects are put into m boxes and n > m, then at least one box contains two or more of the objects.

The Pigeon-Hole Principle (Strong Form)

If n objects are put into m boxes and n > m, then some box must contain at least objects.

Example

In a classroom of 60 students, at least how many students will have birthdays in one of the months of a year?

There are twelve months in a year. In this case, we have 60 students that we want to put into these 12 boxes. So, n = 60 and m = 12. So, n > m. So, now we can invoke the Pigeon-Hole Principle. Thus, at least = 60/12 = 5 students will have birthdays in one of the months.

Based on (Rosen 2004), (Ross and Wright 1988) and (Goodaire and Parmenter 2002).