### Learning Outcomes

- Compute a conditional probability for an event
- Use Baye’s theorem to compute a conditional probability
- Calculate the expected value of an event

Counting? You already know how to count or you wouldn’t be taking a college-level math class, right? Well yes, but what we’ll really be investigating here are ways of counting *efficiently*. When we get to the probability situations a bit later in this chapter we will need to count some *very* large numbers, like the number of possible winning lottery tickets. One way to do this would be to write down every possible set of numbers that might show up on a lottery ticket, but believe me: you don’t want to do this.

### Basic Counting

We will start, however, with some more reasonable sorts of counting problems in order to develop the ideas that we will soon need.

### example

Suppose at a particular restaurant you have three choices for an appetizer (soup, salad or breadsticks) and five choices for a main course (hamburger, sandwich, quiche, fajita or pizza). If you are allowed to choose exactly one item from each category for your meal, how many different meal options do you have?

**Solution 1**: One way to solve this problem would be to systematically list each possible meal:

soup + hamburger soup + sandwich soup + quiche

soup + fajita soup + pizza salad + hamburger

salad + sandwich salad + quiche salad + fajita

salad + pizza breadsticks + hamburger breadsticks + sandwich

breadsticks + quiche breadsticks + fajita breadsticks + pizza

Assuming that we did this systematically and that we neither missed any possibilities nor listed any possibility more than once, the answer would be 15. Thus you could go to the restaurant 15 nights in a row and have a different meal each night.

**Solution 2**: Another way to solve this problem would be to list all the possibilities in a table:

hamburger | sandwich | quiche | fajita | pizza | |

soup | soup+burger | ||||

salad | salad+burger | ||||

bread | etc |

In each of the cells in the table we could list the corresponding meal: soup + hamburger in the upper left corner, salad + hamburger below it, etc. But if we didn’t really care *what* the possible meals are, only *how many* possible meals there are, we could just count the number of cells and arrive at an answer of 15, which matches our answer from the first solution. (It’s always good when you solve a problem two different ways and get the same answer!)

**Solution 3**: We already have two perfectly good solutions. Why do we need a third? The first method was not very systematic, and we might easily have made an omission. The second method was better, but suppose that in addition to the appetizer and the main course we further complicated the problem by adding desserts to the menu: we’ve used the rows of the table for the appetizers and the columns for the main courses—where will the desserts go? We would need a third dimension, and since drawing 3-D tables on a 2-D page or computer screen isn’t terribly easy, we need a better way in case we have three categories to choose form instead of just two.

So, back to the problem in the example. What else can we do? Let’s draw a **tree diagram**:

This is called a “tree” diagram because at each stage we branch out, like the branches on a tree. In this case, we first drew five branches (one for each main course) and then for each of those branches we drew three more branches (one for each appetizer). We count the number of branches at the final level and get (surprise, surprise!) 15.

If we wanted, we could instead draw three branches at the first stage for the three appetizers and then five branches (one for each main course) branching out of each of those three branches.

OK, so now we know how to count possibilities using tables and tree diagrams. These methods will continue to be useful in certain cases, but imagine a game where you have two decks of cards (with 52 cards in each deck) and you select one card from each deck. Would you really want to draw a table or tree diagram to determine the number of outcomes of this game?

Let’s go back to the previous example that involved selecting a meal from three appetizers and five main courses, and look at the second solution that used a table. Notice that one way to count the number of possible meals is simply to number each of the appropriate cells in the table, as we have done above. But another way to count the number of cells in the table would be multiply the number of rows (3) by the number of columns (5) to get 15. Notice that we could have arrived at the same result without making a table at all by simply multiplying the number of choices for the appetizer (3) by the number of choices for the main course (5). We generalize this technique as the *basic counting rule*:

### Basic Counting Rule

If we are asked to choose one item from each of two separate categories where there are *m* items in the first category and *n* items in the second category, then the total number of available choices is *m ***· ***n*.

This is sometimes called the multiplication rule for probabilities.

### example

There are 21 novels and 18 volumes of poetry on a reading list for a college English course. How many different ways can a student select one novel and one volume of poetry to read during the quarter?

The Basic Counting Rule can be extended when there are more than two categories by applying it repeatedly, as we see in the next example.

### example

Suppose at a particular restaurant you have three choices for an appetizer (soup, salad or breadsticks), five choices for a main course (hamburger, sandwich, quiche, fajita or pasta) and two choices for dessert (pie or ice cream). If you are allowed to choose exactly one item from each category for your meal, how many different meal options do you have?

### Try It

### Example

A quiz consists of 3 true-or-false questions. In how many ways can a student answer the quiz?

Basic counting examples from this section are described in the following video.

## Permutations

In this section we will develop an even faster way to solve some of the problems we have already learned to solve by other means. Let’s start with a couple examples.

### example

How many different ways can the letters of the word MATH be rearranged to form a four-letter code word?

In this example, we needed to calculate *n* · (*n* – 1) · (*n* – 2) ··· 3 · 2 · 1. This calculation shows up often in mathematics, and is called the **factorial**, and is notated *n*!

### Factorial

*n*! = *n* · (*n* – 1) · (*n* – 2) ··· 3 · 2 · 1

### Try It

### example

How many ways can five different door prizes be distributed among five people?

### Try It

Now we will consider some slightly different examples.

### examples

A charity benefit is attended by 25 people and three gift certificates are given away as door prizes: one gift certificate is in the amount of $100, the second is worth $25 and the third is worth $10. Assuming that no person receives more than one prize, how many different ways can the three gift certificates be awarded?

### Example

Eight sprinters have made it to the Olympic finals in the 100-meter race. In how many different ways can the gold, silver and bronze medals be awarded?

Factorial examples are worked in this video.

We can generalize the situation in the two examples above to any problem *without replacement* where the *order of selection is important*. If we are arranging in order *r* items out of *n* possibilities (instead of 3 out of 25 or 3 out of 8 as in the previous examples), the number of possible arrangements will be given by

*n* · (*n* – 1) · (*n* – 2) ··· (*n* – *r* + 1)

If you don’t see why (*n *— *r *+ 1) is the right number to use for the last factor, just think back to the first example in this section, where we calculated 25 **·** 24 **·** 23 to get 13,800. In this case *n* = 25 and *r* = 3, so *n *— *r *+ 1 = 25 — 3 + 1 = 23, which is exactly the right number for the final factor.

Now, why would we want to use this complicated formula when it’s actually easier to use the Basic Counting Rule, as we did in the first two examples? Well, we won’t actually use this formula all that often; we only developed it so that we could attach a special notation and a special definition to this situation where we are choosing *r* items out of *n* possibilities *without replacement* and where the *order of selection is important*. In this situation we write:

### Permutations

*nPr* = *n* · (*n* – 1) · (*n* – 2) ··· (*n* – *r* + 1)

We say that there are *nPr* **permutations** of size *r* that may be selected from among *n* choices *without replacement* when *order matters*.

It turns out that we can express this result more simply using factorials.

[latex]{}_{n}{{P}_{r}}=\frac{n!}{(n-r)!}[/latex]

In practicality, we usually use technology rather than factorials or repeated multiplication to compute permutations.

### example

I have nine paintings and have room to display only four of them at a time on my wall. How many different ways could I do this?

### Example

How many ways can a four-person executive committee (president, vice-president, secretary, treasurer) be selected from a 16-member board of directors of a non-profit organization?

View this video to see more about the permutations examples.

### Try It

How many 5 character passwords can be made using the letters A through Z

- if repeats are allowed
- if no repeats are allowed

## Combinations

In the previous section we considered the situation where we chose *r* items out of *n* possibilities *without replacement* and where the *order of selection was important*. We now consider a similar situation in which the order of selection is *not* important.

### Example

A charity benefit is attended by 25 people at which three $50 gift certificates are given away as door prizes. Assuming no person receives more than one prize, how many different ways can the gift certificates be awarded?

We can generalize the situation in this example above to any problem of choosing a collection of items *without replacement* where the *order of selection is not important*. If we are choosing

*r*items out of

*n*possibilities (instead of 3 out of 25 as in the previous examples), the number of possible choices will be given by [latex]\frac{{}_{n}{{P}_{r}}}{{}_{r}{{P}_{r}}}[/latex], and we could use this formula for computation. However this situation arises so frequently that we attach a special notation and a special definition to this situation where we are choosing

*r*items out of

*n*possibilities

*without replacement*where the

*order of selection is*.

**not**important### Combinations

[latex]{}_{n}{{C}_{r}}=\frac{{}_{n}{{P}_{r}}}{{}_{r}{{P}_{r}}}[/latex]

We say that there are *nCr* **combinations** of size *r* that may be selected from among *n* choices *without replacement* where *order doesn’t matter*.

We can also write the combinations formula in terms of factorials:

[latex]{}_{n}{{C}_{r}}=\frac{n!}{(n-r)!r!}[/latex]

### Example

A group of four students is to be chosen from a 35-member class to represent the class on the student council. How many ways can this be done?

View the following for more explanation of the combinations examples.

### Try It

The United States Senate Appropriations Committee consists of 29 members; the Defense Subcommittee of the Appropriations Committee consists of 19 members. Disregarding party affiliation or any special seats on the Subcommittee, how many different 19-member subcommittees may be chosen from among the 29 Senators on the Appropriations Committee?

In the preceding Try It problem we assumed that the 19 members of the Defense Subcommittee were chosen without regard to party affiliation. In reality this would never happen: if Republicans are in the majority they would never let a majority of Democrats sit on (and thus control) any subcommittee. (The same of course would be true if the Democrats were in control.) So let’s consider the problem again, in a slightly more complicated form:

### Example

The United States Senate Appropriations Committee consists of 29 members, 15 Republicans and 14 Democrats. The Defense Subcommittee consists of 19 members, 10 Republicans and 9 Democrats. How many different ways can the members of the Defense Subcommittee be chosen from among the 29 Senators on the Appropriations Committee?

This example is worked through below.