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# 2.2: Order and Probability

Difficulty Level: At Grade Created by: CK-12

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.

Permutations

The Fundamental Counting Principle states that if an event can be chosen in $p$ different ways and another independent event can be chosen in $q$ different ways, the number of different ways the 2 events can occur is $p \times q$. In other words, the Fundamental Counting Principle tells you how many ways you can arrange items. Permutations are the number of possible arrangements in an ordered set of objects.

Example 3

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

Solution:

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: $4 \times 3 \times 2 \times 1 = 24$ possible arrangements.

The notation for a permutation is: ${_n}P_r$,

where:

$n$ is the total number of objects.

$r$ is the number of objects chosen.

For simplifying calculations, when $n = r$, then ${_n}P_r = n!$.

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

Examples:

$5! &= 5 \times 4 \times 3 \times 2 \times 1 = 120\\4! &= 4 \times 3 \times 2 \times 1 = 24\\1! &= 1$

Note: It is a general rule that $0! = 1$.

With TI calculators, you can find the factorial function using:

$\boxed{\text{MATH}} \ \boxed{\blacktriangleright} \ \boxed{\blacktriangleright} \ \boxed{\blacktriangleright} \ (\text{PRB}) \boxed{\blacktriangledown} \ \boxed{\blacktriangledown} \ \boxed{\blacktriangledown} \ (\boxed{4})$

Example 4

Solve for ${_4}P_4$.

Solution:

${_4}P_4 = 4 \cdot 3 \cdot 2 \cdot 1 = 24$

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

Example 5

Solve for ${_6}P_3$.

Solution:

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

${_6}P_3 = 6 \cdot 5 \cdot 4 = 120$

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:

${_n}P_r=\frac{n!}{(n-r)!}$

Look at Example 5 above. In this example, the total number of objects $(n)$ is 6, while the number of objects chosen $(r)$ is 3. We can use these 2 numbers to calculate the number of possible permutations (or the number of arrangements) of 6 objects chosen 3 at a time.

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

Example 6

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?

Solution:

In this problem, there are 4 letters to choose from, so $n = 4$. We want 4-letter arrangements; therefore, we are choosing 4 objects at a time. In this example, $r = 4$.

Example 7

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?

Solution:

In this problem, there are 10 committee members to choose from, so $n = 10$. 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, $r = 3$.

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

Permutations with Repetition

There is a subset of permutations that takes into account that there are double objects or repetitions in a permutation problem. In general, repetitions are taken care of by dividing the permutation by the factorial of the number of objects that are identical.

If you look at the word TOOTH, there are 2 O’s in the word. Both O’s are identical, and it does not matter in which order we write these 2 O’s, since they are the same. In other words, if we exchange 'O' for 'O', we still spell TOOTH. The same is true for the T’s, since there are 2 T’s in the word TOOTH as well.

If we were to ask the question, "In how many ways can we arrange the letters in the word TOOTH?" we must account for the fact that these 2 O’s are identical and that the 2 T’s are identical. We do this using the formula:

$\frac{_nP_r}{x_1! x_2!}$, where $x$ is the number of times a letter is repeated.

$\frac{_nP_r}{x_1! x_2!} &= \frac{_5P_5}{2!2!}\\\frac{_5P_5}{2!2!} &= \frac{5 \times 4 \times 3 \times 2 \times 1}{2 \times 1 \times 2 \times 1}\\\frac{_5P_5}{2!2!} &= \frac{120}{4}\\\frac{_5P_5}{2!2!} &= 30$

Tech Tip: Calculating Permutations on the Calculator

Permutations $(nPr)$

Enter the $n$ value. Press $\boxed{\text{MATH}}$. You should see menus across the top of the screen. You want the fourth menu: PRB (arrow right 3 times). You will see several options. $nPr$ is the second. Press $\boxed{2}$. Enter the $r$ value. Press $\boxed{\text{ENTER}}$.

Example 8

Compute ${_9}P_5$.

Solution:

$\boxed{9} \ \boxed{\text{MATH}} \ \boxed{\blacktriangleright} \ \boxed{\blacktriangleright} \ \boxed{\blacktriangleright} \ (\text{PRB}) \ \boxed{2} \ (\text{nPr}) \ \boxed{5} \ \boxed{\text{ENTER}}$

${_9}P_5= 15,120$

Example 9

How many different 5-letter arrangements can be formed from the word APPLE?

Solution:

There are 5 letters in the word APPLE, so $n = 5$. We want 5-letter arrangements; therefore, we are choosing 5 objects at a time. In this example, $r = 5$, and we are using a word with letters that repeat. In the word APPLE, there are 2 P’s, so $x_1 = 2$.

Example 10

How many different 6-digit numerals can be written using the following 7 digits? Assume the repeated digits are all used.

3, 3, 4, 4, 4, 5, 6

Solution:

There are 7 digits, so $n = 7$. We want 6-digit arrangements; therefore, we are choosing 6 objects at a time. In this example, $r = 6$, and we are using a group of digits with numbers that repeat. In the group of 7 digits (3, 3, 4, 4, 4, 5, 6), there are two 3’s and three 4’s, so $x_1 = 2$ and $x_2 = 3$.

When there are no repetitions, remember that we use the standard permutation formula:

${_n}P_r=\frac{n!}{(n-r)!}$

Example 11

In how many ways can first and second place be awarded to 10 people?

Solution:

There are 10 people $(n = 10)$, and there are 2 prize winners $(r = 2)$.

Example 12

In how many ways can 3 favorite desserts be listed in order from a menu of 10 (i.e., permutations without repetition)?

Solution:

There are 10 menu items $(n = 10)$, and you are choosing 3 favorite desserts $(r = 3)$ in order.

Combinations

If you think about the lottery, you choose a group of lucky numbers in hopes of winning millions of dollars. When the numbers are drawn, the order in which they are drawn does not have to be the same order as on your lottery ticket. The numbers drawn simply have to be on your lottery ticket in order for you to win. You can imagine how many possible combinations of numbers exist, which is why your odds of winning are so small!

Combinations are arrangements of objects without regard to order and without repetition, selected from a distinct number of objects. A combination of $n$ objects taken $r$ at a time $({_n}C_r)$ can be calculated using the formula:

${_n}C_r = \frac{n!}{r!(n-r)!}$

Example 13

Evaluate: ${_7}C_2$.

Solution:

${_7}C_2 &= \frac{7!}{2!(7-2)!}\\{_7}C_2 &= \frac{7!}{2!(5)!}\\ {_7}C_2 &= \frac{7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{(2 \times 1)(5 \times 4 \times 3 \times 2 \times 1)}\\{_7}C_2 &= \frac{5,040}{(2)(120)}\\ {_7}C_2 &= \frac{5,040}{240}\\ {_7}C_2 &= 21$

Example 14

In how many ways can 3 desserts be chosen in any order from a menu of 10?

Solution:

There are 10 menu items $(n = 10)$, and you are choosing 3 desserts in any order $(r = 3)$.

${_{10}}C_3 &= \frac{10!}{3!(10-3)!}\\{_{10}}C_3 &= \frac{10!}{3!(7)!}\\{_{10}}C_3 &= \frac{10 \times 9 \times 8}{3 \times 2 \times 1}\\{_{10}}C_3 &= 120$

Example 15

There are 12 boys and 14 girls in Mrs. Cameron's math class. Find the number of ways Mrs. Cameron can select a team of 3 students from the class to work on a group project. The team must consist of 2 girls and 1 boy.

Solution:

There are groups of both boys and girls to consider. From the 14 girls $(n = 14)$ in the class, we are choosing 2 $(r = 2)$.

$\text{Girls:}\\{_{14}}C_2 &= \frac{14!}{2!(14 -2)!}\\{_{14}}C_2 &= \frac{14!}{2!(12)!}\\{_{14}}C_2 &= \frac{87,178,291,200}{2(479,001,600)}\\{_{14}}C_2 &= \frac{87,178,291,200}{958,003,200}\\{_{14}}C_2 &= 91$

From the 12 boys $(n = 12)$ in the class, we are choosing $1 (r = 1)$.

$\text{Boys:}\\{_{12}}C_1 &= \frac{12!}{1!(12-1)!}\\{_{12}}C_1 &= \frac{12!}{1!(11)!}\\{_{12}}C_1 &= \frac{479,001,600}{1(39,916,800)}\\{_{12}}C_1 &= \frac{479,001,600}{39,916,800}\\{_{12}}C_1 &= 12$

Therefore, the number of ways Mrs. Cameron can select a team of 3 students (2 girls and 1 boy) from the class of 26 students to work on a group project is:

$\text{Total combinations} = {_{14}}C_2 \times {_{12}}C_1=91 \times 12=1,092$

Example 16

If there are 20 rock songs and 20 rap songs to choose from, in how many different ways can you select 12 rock songs and 7 rap songs for a mix CD?

Solution:

As in Example 15, we have multiple groups from which we are required to select, so we have to calculate the possible combinations for each group (rock songs and rap songs in this example) separately and then multiply together.

Using TI technology: for ${_n}C_r$, type the $n$ value (the total number of items), and then press $\boxed{\text{MATH}} \ \boxed{\blacktriangleright} \ \boxed{\blacktriangleright} \ \boxed{\blacktriangleright}$ $\text{(PRB)} \ \boxed{\blacktriangledown} \ \boxed{\blacktriangledown}$ (to number 3) $\boxed{\text{ENTER}}$. Then type the $r$ value (the number of items your want to choose), and finally, press $\boxed{\text{ENTER}}$.

$\text{Rock:}\\{_{20}}C_{12} &= \frac{20!}{12!(20-12)!}\\{_{20}}C_{12} &= \frac{20!}{12!(8)!}\\ {_{20}}C_{12} &= 125,970\\\\\text{Rap:}\\{_{20}}C_7 &= \frac{20!}{7!(20-7)!}\\{_{20}}C_7 &= \frac{20!}{7!(13)!}\\{_{20}}C_7 &= 77,520$

Therefore, the total number of possible combinations is:

$&\text{Rock} \quad \ \text{Rap}\\&{_{20}}C_{12} \times _{20}C_7 = 125,970 \times 77,520 = 9.765 \times 10^9 \ \text{possible combinations}$

Tech Tip: Calculating Combinations on the Calculator

Combinations $(nCr)$

Enter the $n$ value. Press $\boxed{\text{MATH}}$. You should see modes across the top of the screen. You want the fourth menu: PRB (arrow right 3 times). You will see several options. $nCr$ is the third. Press $\boxed{3}$. Enter the $r$ value. Press $\boxed{\text{ENTER}}$.

Example 17

Compute ${_{10}}C_6$.

Solution:

${_{10}}C_6 & \qquad \quad \boxed{1} \ \boxed{0} \ \boxed{\text{MATH}} \ \boxed{\blacktriangleright} \ \boxed{\blacktriangleright} \ \boxed{\blacktriangleright} \ (\text{PRB}) \ \boxed{3} \ (\text{nCr}) \ \boxed{6} \ \boxed{\text{ENTER}}\\{_{10}}C_6 &= 210$

## Date Created:

Feb 23, 2012

Jan 27, 2015
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