### Permutations and Combinations Compared

Permutations and combinations are the next step in the learning of probability. It is by using permutations and combinations that we can find the probabilities of various events occurring at the same time, such as choosing 3 boys and 3 girls from a class of grade 12 math students.

In mathematics, we use more precise language:

If the order doesn't matter, it is a combination.

If the order does matter, it is a permutation.

Say, for example, you are making a salad. You throw in some lettuce, carrots, cucumbers, and green peppers. The order in which you throw in these vegetables doesn't really matter. Here we are talking about a combination. For combinations, you are merely selecting. Say, though, that Jack went to the ATM to get out some money and that he has to put in his PIN number. Here the order of the digits in the PIN number is quite important. In this case, we are talking about a permutation. For permutations, you are ordering objects in a specific manner.

The **Fundamental Counting Principle** states that if an event can be chosen in \begin{align*}p\end{align*}**Permutations** are the number of possible arrangements in an ordered set of objects.

#### Rearranging Letters

How many ways can you arrange the letters in the word MATH?

You have 4 letters in the word, and you are going to choose 1 letter at a time. When you choose the first letter, you have 4 possibilities ('M', 'A', 'T', or 'H'). Your second choice will have 3 possibilities, your third choice will have 2 possibilities, and your last choice will have only 1 possibility.

Therefore, the number of arrangements is: \begin{align*}4 \times 3 \times 2 \times 1 = 24\end{align*}

The notation for a permutation is: \begin{align*}{_n}P_r\end{align*}

where:

\begin{align*}n\end{align*}*total* number of objects.

\begin{align*}r\end{align*}

For simplifying calculations, when \begin{align*}n = r\end{align*}

The **factorial function (!)** requires us to multiply a series of descending natural numbers.

Examples:

\begin{align*}5! &= 5 \times 4 \times 3 \times 2 \times 1 = 120\\
4! &= 4 \times 3 \times 2 \times 1 = 24\\
1! &= 1\end{align*}

Note: It is a general rule that \begin{align*}0! = 1\end{align*}

#### Solving Permutations and Combinations

1. Solve for \begin{align*}{_4}P_4\end{align*}

\begin{align*}{_4}P_4 = 4 \cdot 3 \cdot 2 \cdot 1 = 24\end{align*}

This represents the number of ways to arrange 4 objects that are chosen from a set of 4 different objects.

2. Solve for \begin{align*}{_6}P_3\end{align*}

Starting with 6, multiply the first 3 numbers of the factorial:

\begin{align*}{_6}P_3 = 6 \cdot 5 \cdot 4 = 120\end{align*}

This represents the number of ways to arrange 3 objects that are chosen from a set of 6 different objects.

The formula to solve permutations like these is:

\begin{align*}{_n}P_r=\frac{n!}{(n-r)!}\end{align*}

Look at the example above. In this example, the total number of objects \begin{align*}(n)\end{align*}

\begin{align*}{_n}P_r &= \frac{n!}{(n-r)!}\\
{_6}P_3 &= \frac{6!}{(6-3)!}\\
{_6}P_3 &= \frac{6!}{3!}=\frac{6 \times 5 \times 4 \cancel{\times 3 \times 2 \times 1}}{\cancel{3 \times 2 \times 1}}\\
{_6}P_3 &= \frac{120}{1}\\
{_6}P_3 &= 120\end{align*}

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### Examples

**Example 1**

What is the total number of possible 4-letter arrangements of the letters 's', 'n', 'o', and 'w' if each letter is used only once in each arrangement?

In this problem, there are 4 letters to choose from, so \begin{align*}n = 4\end{align*}. We want 4-letter arrangements; therefore, we are choosing 4 objects at a time. In this example, \begin{align*}r = 4\end{align*}.

#### Example 2

A committee is to be formed with a president, a vice president, and a treasurer. If there are 10 people to select from, how many committees are possible?

In this problem, there are 10 committee members to choose from, so \begin{align*}n = 10\end{align*}. We want to choose 3 members to be president, vice-president, and treasurer; therefore, we are choosing 3 objects at a time. In this example, \begin{align*}r = 3\end{align*}.

\begin{align*}{_n}P_r &= \frac{n!}{(n-r)!}\\ {_{10}}P_3 &= \frac{10!}{(10-3)!}\\ {_{10}}P_3 &= \frac{10!}{7!}=\frac{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}\\ {_{10}}P_3 &= \frac{720}{1}\\ {_{10}}P_3 &= 720\end{align*}

### Review

- Solve for \begin{align*}{_7}P_5\end{align*}.
- Evaluate \begin{align*}{_4}P_2 \times {_5}P_3\end{align*}.
- How many different 4-digit numerals can be made from the digits of 56987 if a digit can appear just once in a numeral?
- In how many ways can 6 students be chosen from 9 students if the order in which the students are chosen matters?
- A TV station has 8 hour-long TV shows to choose from in order to fill 2 one-hour time slots. In how many ways can it fill the time slots?
- A basketball league consists of 12 teams. In how many ways can the teams finish in first, second, and third place?
- A secret code consist of 3 digits from 0 to 9 followed by 2 letters from 'A' to 'Z'. None of the digits or letters repeat. How many secret codes are possible?
- On a test, Robert has been presented with 15 vocabulary words and 15 definitions. He is being asked to match each vocabulary word to the appropriate definition. How many different ways are there for Robert to do the matching?
- A couple just had twin boys, but they can't decide between the names Mike, Mark, Peter, Paul, Sam, and Sonny. If the couple randomly chooses names for the 2 boys from the names listed, what is the probability that the first boy born will be named Sam and the second boy born will be named Mark? Assume that the boys will not have the same name.
- For question 9, what is the probability that one of the boys will be named Peter and the other boy will be named Paul?

### Review (Answers)

To view the Review answers, open this PDF file and look for section 2.2.