Suppose you have enrolled in the courses that are required to complete your diploma, but there is room in your schedule for three more courses. You can choose the courses you want, but you must choose three. Suppose you have narrowed down your choices and organized them by category, and you plan to choose one from each category:

Math course: Trigonometry or Precalculus
Science course: Chemistry or Anatomy & Physiology
Other: Choir, Weight Lifting I, Woodworking, Drama, French I

How many different ways could you choose three courses? Explain your process for finding the answer.

Let's consider another scenario where we are building ice cream sundaes. We are given the following choices and can choose only one from each category.

Ice cream types: vanilla, chocolate, or strawberry
Toppings: cherries, sprinkles, cookie crumbles, or peanuts
Sauces: hot fudge or caramel

We can draw a tree diagram to represent all of the possible choices.

A tree diagram about building icecream sundaes. Starts at Vanilla, Chocolate and Strawberry. The next set of branches show hot fudge and caramel. The last set of branches show peanuts, sprinkles, cherries and cookie crumbles. Ask your teacher for more information.

From the last set of branches in the tree diagram, we can count a total of 24 different ways the ice cream sundaes can be created. We can also see from the tree diagram that we could have calculated the number of possible sundaes by multiplying 3\cdot 2 \cdot 4 = 24. This is an example of the fundamental counting principle.

Fundamental counting principle

If one decision can be made x ways and another can be made y ways, then the two decisions can be made x\cdot y ways.

This essentially means, if we know the number of ways each decision can be made, we can multiply those numbers together to find the total number of ways all of the decisions can be made.

Examples

Example 1

Tyson's mom is buying a brand new car, and she asked him to help her make a final decision. She has narrowed down her options for the manufacturer, type of car, color, and type of seats to the following:

Ford or GM
Sedan, minivan, pickup truck, SUV
Black, silver, red
Leather seats or fabric seats

Determine the number of unique car choices from the options above.

Worked Solution

Create a strategy

From the options that Tyson's mom has narrowed down, there are 2 manufacturers, 4 types of cars, 3 colors, and 2 types of seats.

Apply the idea

By the fundamental counting principle, we need to multiply to number of choices in each category together to find the total number of possible cars.2\cdot 4\cdot 3\cdot 2=48 There are 48 unique cars that could be created.

Reflect and check

Note that we could have changed the list of options slightly, but still ended up with the same number of possibilities. If the options were:

Ford, Chrysler-Dodge, GM
Sedan, mini-van, truck, SUV
White, black, silver, red

There would still be 3\cdot 4\cdot 4=48 unique cars that could be created.

Example 2

Suppose you have the digits 1,\,3,\,4,\,7,\,8,\,9. How many 3-digit odd numbers which are greater than 200 can be made without repeating digits?

Worked Solution

Create a strategy

To have an odd number, we know the last digit will have to be odd. To be greater than 200, we know the first digit cannot be 1. We have two cases here: one where the first number is odd in which case we lose one possibility for the last digit, and one where the first number is even.

Apply the idea

Case 1: The first number is odd.

For the first digit, it could be 3,\,7 or 9. Now that one of those has been used, one of them cannot be included in the choices for the last digit. That means there are 2 of those could be used for the last digit, and we also have the choice of 1. In other words, there are 3 possibilities for the first digit and 3 possibilities for the second digit.

\dfrac{3}{1\text{st digit}}\text{ }\cdot\text{ }\dfrac{}{2\text{nd digit}}\text{ }\cdot \text{ }\dfrac{3}{3\text{rd digit}}

Since we have already used 2 of the 6 options for the first and last digits, there are 4 possibilities left for the 2nd digit.

\dfrac{3}{1\text{st digit}}\text{ }\cdot\text{ }\dfrac{4}{2\text{nd digit}}\text{ }\cdot \text{ }\dfrac{3}{3\text{rd digit}}

Therefore, there are =3\cdot4\cdot3=36 possibilities for a 3-digit code that begins with an odd number greater than 2.

Case 2: The first number is even.

The first digit could be 4 or 8. The last digit could be any of the odd numbers 1,\,3,\,7,\,9. Since 2 of the 6 options have already used for the first and last digits, there are 4 possibilities left for the 2nd digit.

\dfrac{2}{1\text{st digit}}\text{ }\cdot\text{ }\dfrac{4}{2\text{nd digit}}\text{ }\cdot \text{ }\dfrac{4}{3\text{rd digit}}

Therefore, there are =2\cdot 4\cdot 4=32 possibilities for a 3-digit code that begins with an even number.

We can find the total number of possibilities by adding the answers from both cases. There are 36+32=68 ways we can create a 3-digit odd number that is greater than 200 using the given numbers.

Idea summary

To find the number of ways multiple events can happen, we use the fundamental counting principle by multiplying together the number of ways each event can happen.

Permutations

When we're calculating the number of possibilities for events, sometimes the order in which we make our choices makes a difference in how we calculate the possibilities. For example, consider the letters A, B, C, D, and E. We want to see how many different ways we can select and arrange a set of 3 letters.

\dfrac{}{1\text{st letter}}\text{ }\dfrac{}{2\text{nd letter}}\text{ }\dfrac{}{3\text{rd letter}}

For each position, we write down how many possible choices we have. For the 1st position, we can choose any of the 5 letters: A, B, C, D, or E.

\dfrac{5}{1\text{st letter}}\text{ }\dfrac{}{2\text{nd letter}}\text{ }\dfrac{}{3\text{rd letter}}

We also write down how many possible choices we have for the 2nd position. Because we had 5 letters to start with, and we will have already selected one to go in the 1st position, there are 4 options left for the letter that goes in this next position.

\dfrac{5}{1\text{st letter}}\text{ }\dfrac{4}{2\text{nd letter}}\text{ }\dfrac{}{3\text{rd letter}}

For the last position, we will have already picked 2 of the 5 letters for the first and second positions, so there are only 3 letters left to choose from.

\dfrac{5}{1\text{st letter}}\text{ }\dfrac{4}{2\text{nd letter}}\text{ }\dfrac{3}{3\text{rd letter}}

By the fundamental counting principle, we multiply these together to determine there are {5\cdot 4\cdot 3=60} possible ways to arrange 3 letters. We call a situation like this, when we calculate the number of different ways a certain number of objects can be arranged from a larger set, a permutation.

Permutation

The number of ways in which a set of r objects can be ordered or arranged from a set of n objects. The notation is \text{}_{n} P_{r}.

Example:

\begin{aligned} \text{}_5 P_3 &= \dfrac{5!}{(5-3)!} \\&= 60 \end{aligned}

When calculating a permutation, the order the objects are chosen matters, so the number of ways the event can happen has a decreasing numerical pattern. There may be 4 ways the first event can happen, then 3 ways the second event can happen, 2 ways the third event can happen, and only 1 way for the final event.

Mathematicians have created a shorthand notation to account for this pattern, called a factorial. A factorial, denoted by n!, is the product of the first n positive integers. It is calculated by multiplying n by every positive integer less than n all the way down to 1. That is:

\displaystyle n!=n \cdot \left(n-1\right) \cdot \left(n-2\right) \cdot \ldots \cdot 2 \cdot 1

\bm{n}

a positive integer greater or equal to zero

An important property of factorial notation is 0!=1. This property makes it possible to perform more complex calculations. Calculators and computers can evaluate factorials to help us calculate them efficiently.

A permutation is an important application of the fundamental counting principle. We calculate a permutation using the formula:

\displaystyle \text{}_{n} P_{r} = \dfrac{n!}{(n-r)!}

\bm{n}

the total number of objects in the set

\bm{r}

the number of objects being ordered or arranged

Examples

Example 3

Suppose we need to choose a 4-digit passcode for our phones. We can use any number from 0–9, but we can only use a number once. Determine the number of passcodes possible.

Worked Solution

Create a strategy

Let's begin by imagining the place values in the passcode.

\dfrac{}{1\text{st digit}}\text{ }\dfrac{}{2\text{nd digit}}\text{ }\dfrac{}{3\text{rd digit}}\text{ }\dfrac{}{4\text{th digit}}\text{ }

We need to determine how many numbers are available to be chosen for each digit in the passcode.

For a permutation, the order in which the objects are arranged matters. In other words, a passcode of 1234 is very different from a passcode of 2143. Although the digits in both passcodes are the same, the way in which they are arranged is different. The order in which you type the numbers into your phone's lock screen matters.

Apply the idea

For the first digit of the code, we can use any of the 10 available numbers.

\dfrac{10}{1\text{st digit}}\text{ }\dfrac{}{2\text{nd digit}}\text{ }\dfrac{}{3\text{rd digit}}\text{ }\dfrac{}{4\text{th digit}}\text{ }

But once the number for the first digit has been used, we only have 9 numbers left to choose for the second digit.

\dfrac{10}{1\text{st digit}}\text{ }\dfrac{9}{2\text{nd digit}}\text{ }\dfrac{}{3\text{rd digit}}\text{ }\dfrac{}{4\text{th digit}}\text{ }

Following this pattern, there would be 8 numbers left to choose for the third digit, and 7 numbers left to choose for the last digit of the passcode.

\dfrac{10}{1\text{st digit}}\text{ }\dfrac{9}{2\text{nd digit}}\text{ }\dfrac{8}{3\text{rd digit}}\text{ }\dfrac{7}{4\text{th digit}}\text{ }

By the fundamental counting principle, we multiply the options together to get a total of {10\cdot 9\cdot 8\cdot 7=5040} possibilities for a 4-digit passcode without repeated digits.

Reflect and check

Since the digits could not be repeated, we simply removed one of the options each time. This resulted in multiplying each integer less than 10 like we would for a factorial. However, we do not want to multiply each integer all the way down to 1. We can still calculate 10! in order to use that notation, but we can remove the unwanted numbers with division.

\dfrac{10\cdot9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}{6\cdot5\cdot4\cdot3\cdot2\cdot1}=\dfrac{10!}{6!}

Recall that we began with 10 digits to choose from, and we needed to choose 4 for the passcode. The denominator is just 10-4=6, so we can rewrite the expression as \dfrac{10!}{6!}=\dfrac{10!}{(10-4)!}. This process can be generalized into the formula we use to find permutations.

Example 4

For each of the following scenarios, determine if a permutation would be an appropriate method for finding the total number of ways to select the objects.

A local pizza place is offering a special on a large pizza with 3 toppings of your choice. They have a total of 9 pizza toppings to choose from.

Worked Solution

Create a strategy

If we use the permutation formula in this situation, we will find the number of different ways the three toppings could be arranged in order when selected from 9 options. We need to determine if the order that we pick the 3 toppings from the 9 options matters.

Would picking sausage then pepperoni then olives be different from picking olives, then pepperoni, then sausage? If the order does matter, then we would use a permutation.

Apply the idea

The order in which you pick the toppings does not matter because it is the same 3 toppings. The permutation formula would count 6 different ordered versions of olives, pepperoni, and sausage:

\begin{aligned}\text{Olives Pepperoni Sausage }\text{ }\text{ Olives Sausage Pepperoni}\\ \text{Pepperoni Olives Sausage }\text{ }\text{ Pepperoni Sausage Olives}\\ \text{Sausage Olives Pepperoni }\text{ }\text{ Sausage Pepperoni Olives}\end{aligned}

But, we would count all of these as one unique type of pizza with 3 toppings. This is not a situation where we would use a permutation.

Students need to elect a new president, vice president, and secretary for their school's student council. There are 14 students running for any position within the student council.

Worked Solution

Create a strategy

When we think of "order" in this case, we are not necessarily looking at who we choose 1st, 2nd, or 3rd. We are determining if the students we choose for each position on the council matters.

In other words, would choosing Jackie for president and Jillian for vice president be different from choosing Jillian for president and Jackie for vice president? If it is different, then the order matters.

Apply the idea

When counting the outcomes, each ordered spot represents a council position:\dfrac{}{\text{President}}\text{ }\text{ }\dfrac{}{\text{Vice president}}\text{ }\text{ }\dfrac{}{\text{Secretary}}

So, the outcomes \dfrac{\text{Jillian}}{\text{President}}\text{ }\text{ }\dfrac{\text{Jackie}}{\text{Vice president}}\text{ }\text{ }\dfrac{\text{Jerome}}{\text{Secretary}} and \dfrac{\text{Jerome}}{\text{President}}\text{ }\text{ }\dfrac{\text{Jackie}}{\text{Vice president}}\text{ }\text{ }\dfrac{\text{Jillian}}{\text{Secretary}}indicate the students were chosen for different positions, and should be counted separately.

The order that students choose the candidates or the positions the candidates receive on the council matters, so a permutation should be used.

Ariana has 7 bottles of nail polish, and she wants to choose 2 different colors for her nails. She wants each nail to be half one color and half the other color.

Worked Solution

Create a strategy

The way in which Ariana paints her nails is not specified. For this question, we are not worried about which color is on which half of the nail. We are only looking at the way in which she chooses the colors. So, does the order in which Ariana chooses two colors matter?

Apply the idea

The order in which Ariana picks two nail polish colors does not matter because all her nails will have both colors on them. If she chooses pink first then yellow, her nails would still have the same colors if she had chosen yellow first then pink. A permutation would not be used in this scenario.

Reflect and check

Note that the way in which Ariana paints her nails is not specified. We do not know if she is painting the bottom half of her nails with the first color and the top of her nails with the second color. All we know is that she wants to paint her nails, and she is picking two colors.

If the question had specified that it was important to know which color was on top and which color was on the bottom of her nails, then it would be a different situation. In that case, pink below yellow would be different from yellow below pink, and a permutation would be used.

Example 5

Users of a website are required to create a unique PIN ID consisting of 5 characters, and they can be arranged in any order they choose.

Alicia wants to use her two favorite letters, \text{Z} and \text{S}, and her three favorite digits, 7, 8 and 2.

How many unique PIN IDs can she create if the letters and digits can appear in any particular order (and no character can be repeated)?

Worked Solution

Create a strategy

We can use the permutation formula _{n}P_{r}=\dfrac{n!}{\left(n-r \right)!}, where n is the number of characters Alicia can choose from and r is the number of characters in a PIN ID.

Apply the idea

Since Alicia only wants to use \text{Z, S, 7, 8,} and 2, there are 5 characters she can choose from. This will be the value of n in the permutation formula.

The total number of characters in a PIN ID is 5, so this will be the value of r.

Using the permutation formula:

\displaystyle _{n}P_{r}	\displaystyle =	\displaystyle \dfrac{n!}{\left(n-r \right)!}	Permutation formula
	\displaystyle =	\displaystyle \frac{5!}{\left(5-5\right)!}	Substitute n=5 and r=5
	\displaystyle =	\displaystyle \frac{5\cdot 4\cdot 3\cdot 2\cdot 1}{0!}	Expand the factorial
	\displaystyle =	\displaystyle 120	Evaluate

There are 120 unique PIN IDs that Alicia could create.

Reflect and check

We can verify our answer using technology. In the GeoGebra scientific calculator, we will use the built-in function \text{nPr}\left(\text{Number n, Number r}\right). Using n=5 and r=5. we can verify there are 120 combinations.

A screenshot of the GeoGebra scientific tool showing how to calculate the permutation of 5 taken 5. Speak to your teacher for more details.

How many PIN IDs can she create if the two letters are followed by the three digits (and no characters can be repeated)?

Worked Solution

Create a strategy

Since a specific arrangement of having the two letters first followed by the three digits is given, we should find the number of arrangements for the letters and digits separately using the permutation formula.

By the fundamental counting priciple, the product of these will be the number of unique PIN IDs that can be created.

Apply the idea

Let's find the number of arrangements for the letters first:

Since Alicia only wants to use the letters \text{Z} and \text{S}, the number of letters she can choose from \left(n\right) is 2.
The number of letters being arranged in the PIN ID \left(r\right) is 2.

Using the permutation formula:

\displaystyle _{n}P_{r}	\displaystyle =	\displaystyle \frac{2!}{\left(2-2\right)!}	Substitute r=2 and n=2
	\displaystyle =	\displaystyle \frac{2\cdot 1}{0!}	Expand the factorial
	\displaystyle =	\displaystyle 2	Evaluate

Next, we will find the number of arrangements for the digits:

Since Alicia only wants to use the digits 7, 8 and 2, the total number of digits she can choose from \left(n\right) is 3.
The number of digits being arranged in the PIN ID \left(r\right) is 3.

Using the permutation formula:

\displaystyle _{n}P_{r}	\displaystyle =	\displaystyle \frac{3!}{\left(3-3\right)!}	Substitute r=3 and n=3
	\displaystyle =	\displaystyle \frac{3\cdot 2\cdot 1}{0!}	Expand the factorial
	\displaystyle =	\displaystyle 6	Evaluate

Multiplying the total number of arrangements for the letters and digits \left(2 \cdot 6\right), there are only 12 unique PIN IDs that Alicia could create.

Example 6

Suppose that 8 people enter a room and randomly stand in a line along the back wall. Find the probability that they stand from tallest to shortest, left to right.

Worked Solution

Create a strategy

Recall probability is defined as the following: P\left(\text{Event}\right)=\dfrac{\text{The number of desired outcomes}}{\text{The number of total outcomes}}

To find the number of total outcomes, we need to find the total possible arrangements that 8 people can stand in a line. Because the order in which they stand matters, this is a permutation.

Apply the idea

We need all 8 of the people to line up along the wall, so we can find the total by calculating {}_8P_8.

\displaystyle {}_8P_{8}	\displaystyle =	\displaystyle \dfrac{8!}{\left(8-8\right)!}	Substitute n=8, r=8 in the permutation formula
	\displaystyle =	\displaystyle \dfrac{8!}{0!}	Evaluate the subtraction
	\displaystyle =	\displaystyle \dfrac{40\,320}{1}	Evaluate the factorials
	\displaystyle =	\displaystyle 40\,320	Evaluate the division

There is only 1 way the people can line up from tallest to shortest, left to right. The probability of this happening is

\dfrac{1}{40\,320}=0.0000248 \text{ or } 0.00248\%

Reflect and check

Whenever we need to select or order all of the objects in a permutation, whenever n=r, we can simply use n! instead. This is because the denominator will always be \left(n-n\right)!=0!=1, and anything divided by 1 is itself.

Idea summary

We use a permutation to find the number of ways in which a set of r objects can be ordered or arranged from a set of n objects.

\displaystyle \text{}_n P_r = \dfrac{n!}{(n-r)!}

\bm{n}

the total number of objects in the set

\bm{r}

the number of objects being ordered or arranged

Permutations can also be used to calculate probabilities.

Combinations

Exploration

A store has notebooks that come in 6 different colors. You need 3 new notebooks for various classes, and you want them to be different colors to easily distinguish between them.

How many different color combinations can you make with the 3 notebooks you choose? Explain your process.
Why is this answer different from the answer you would get if you had used the permutation formula?

In the scenario of choosing 3 different colored notebooks from 6 color options, we can determine the total number of ways to order 3 textbooks from 6 color choices by using the permutation formula.

\displaystyle {}_6P_3	\displaystyle =	\displaystyle \dfrac{6!}{(6-3)!}
	\displaystyle =	\displaystyle \dfrac{6!}{3!}
	\displaystyle =	\displaystyle \dfrac{6\cdot5\cdot4\cdot3\cdot2\cdot1}{3\cdot2\cdot1}
	\displaystyle =	\displaystyle 6\cdot5\cdot4
	\displaystyle =	\displaystyle 120

There are 120 different color combinations of notebooks we can create when the order matters. However, in the context of this situation, the order in which we choose the notebooks does not matter because a set of yellow, green, and blue notebooks is the same set of notebooks, regardless of order. Considering just 3 possible colors:

\begin{aligned}\text{Yellow Green Blue }\text{ }\text{ Yellow Blue Green}\\ \text{Green Blue Yellow }\text{ }\text{ Green Yellow Blue}\\ \text{Blue Green Yellow }\text{ }\text{ Blue Yellow Green}\end{aligned}

We can see that the 3 colors are repeated 3\cdot2\cdot1=6 times. We can remove these repeated possibilities by dividing it from our total. In other words, we need to divide the permutation formula by r! where r is the number of objects we are choosing. This process can be generalized into the formula we use to find combinations.

\displaystyle {}_n C_r = \dfrac{n!}{r!(n-r)!}

\bm{n}

the total number of objects in the set

\bm{r}

the number of objects that are being chosen

Combination

The number of ways in which a set of r objects can be chosen from a set of n objects. The notation is \text{}_n C_r.

Example:

\begin{aligned} \text{}_5 C_3 &= \dfrac{5!}{3!(5-3)!} \\&= 10 \end{aligned}

For combinations, the notation \binom{n}{r} may also be used, and we can read it as "n choose r".

Examples

Example 7

For each of the following scenarios, use either a permutation, a combination, or the counting principle to solve the problem.

There are 10 parts in a school play, and 10 students auditioned for the play. How may ways can the parts be assigned to the students?

Worked Solution

Create a strategy

The order that the parts are assigned matters. We can calculate this with the permutation formula.

Apply the idea

\displaystyle \text{}_{n} P_{r}	\displaystyle =	\displaystyle \dfrac{n!}{(n-r)!}	Permutation formula
	\displaystyle =	\displaystyle \dfrac{10!}{(10-10)!}	Substitute known values
	\displaystyle =	\displaystyle \dfrac{10!}{0!}	Evaluate the subtraction
	\displaystyle =	\displaystyle \dfrac{3\,628\,800}{1}	Evaluate the factorials
	\displaystyle =	\displaystyle 3\,628\,800	Evaluate the division

Reflect and check

Since there are 10 parts and 10 students to hand them out to, we can also calculate this with just a factorial.

10!=10\cdot 9\cdot 8\cdot 7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1=3\,628\,800

Suppose 7 people enter a marathon. Assuming there are no ties, determine the number of ways a gold, silver, and bronze medal could be awarded.

Worked Solution

Create a strategy

This problem requires finding the number of ways to order or arrange 3 objects from 7, so it is a permutation problem.

The order matters in this problem because a particular runner getting the gold medal is a different outcome than that same runner getting the silver medal. So the order in which they get the medals matters, making it a permutation problem rather than a combination problem.

Apply the idea

The formula {}_{7} P_{3} can be used to represent the permutation of 7 people taking 3 different medals (gold, silver, and bronze).

\displaystyle {}_{7} P_{3}	\displaystyle =	\displaystyle \dfrac{7!}{\left(7-3\right)!}	Substitution into formula
\displaystyle \text{ }	\displaystyle =	\displaystyle \dfrac{7\cdot 6\cdot 5\cdot4\cdot 3\cdot 2\cdot 1}{4 \cdot3 \cdot 2 \cdot 1}	Expand factorials
\displaystyle \text{ }	\displaystyle =	\displaystyle 7\cdot 6 \cdot 5	Simplify since \dfrac{4\cdot 3\cdot 2\cdot 1}{4\cdot 3\cdot 2\cdot 1}=1
\displaystyle \text{ }	\displaystyle =	\displaystyle 210	Evaluate the multiplication

There are 210 ways that the gold, silver and bronze medals can be awarded in a marathon with 7 people.

Reflect and check

Alternatively, this problem could have been solved using the counting principle. If there are three medals to win, 7 people could win gold, then there are 6 people left to win silver, then there are 5 people left to win bronze.

\displaystyle n(\text{gold}) \cdot n(\text{silver}) \cdot n(\text{bronze})	\displaystyle =	\displaystyle 7\cdot 6 \cdot 5
\displaystyle \text{ }	\displaystyle =	\displaystyle 210

A bakery has a selection of 10 different cupcakes, 8 different donuts, and 6 different muffins. If you want to buy one of each, how many different choices do you have?

Worked Solution

Create a strategy

This problem requires finding the number of ways multiple events can happen. The fundamental counting principle is appropriate for solving the problem.

To find the number of different choices we multiply the number of outcomes for each event.

Apply the idea

\begin{aligned} n(\text{cupcakes})\cdot n(\text{donuts})\cdot n(\text{muffins}) &= 10\cdot 8 \cdot 6 \\ &=480 \end{aligned}

Therefore, there are 480 different choices.

The manager of a company wants to create a group of 5 people from his 20 employees. How many different groups are possible?

Worked Solution

Create a strategy

This problem involves choosing 5 people from a group of 20. The order does not matter because there are no arrangements or positions within the group of 5. Therefore, this is a combination problem.

Apply the idea

There are n=20 employees and we want to choose r=5 members.

We can use the formula {}_{n} C_{r}.

\displaystyle {}_{20} C_{5}	\displaystyle =	\displaystyle \dfrac{20!}{5!\left(20-5\right)!}	Substitution into formula
\displaystyle \text{ }	\displaystyle =	\displaystyle \dfrac{20\cdot 19\cdot 18\cdot17\cdot 16\cdot 15!}{5\cdot 4\cdot 3 \cdot2 \cdot 1 \cdot 15!}	Expand factorials
\displaystyle \text{ }	\displaystyle =	\displaystyle \dfrac{20\cdot 19\cdot 18\cdot17\cdot 16}{5\cdot 4\cdot 3 \cdot2 \cdot 1 }	Simplify since \dfrac{15!}{15!}=1
\displaystyle \text{ }	\displaystyle =	\displaystyle 15\,504	Evaluate the multiplication and division

There are 15\, 504 possible groups.

Reflect and check

We can verify our answer using technology. In the GeoGebra scientific calculator, we will use the built-in function \text{nCr}\left(\text{Number n, Number r}\right). Using n=20 and r=5, we can verify there are 15\,504 combinations.

A screenshot of the GeoGebra scientific tool showing how to calculate the combination of 20 taken 5. Speak to your teacher for more details.

Example 8

Evaluate each expression:

_{12}P_{8}

Worked Solution

Create a strategy

Use the permutation formula _{n}P_{r}=\frac{n!}{\left(n-r\right)!}

Apply the idea

\displaystyle _{n} P_{r}	\displaystyle =	\displaystyle \frac{n!}{\left(n-r\right)!}	Permutation formula
	\displaystyle =	\displaystyle \dfrac{12!}{(12-8)!}	Substitute known values
	\displaystyle =	\displaystyle \dfrac{12!}{4!}	Evaluate the subtraction
	\displaystyle =	\displaystyle \dfrac{12\cdot 11\cdot 10\cdot 9\cdot 8\cdot 7\cdot 6\cdot 5\cdot \cancel{4!}}{\cancel{4!}}	Expand and divide out common factors
	\displaystyle =	\displaystyle 19\,958\,400	Evaluate the multiplication

_{14}C_{5}

Worked Solution

Create a strategy

Use the combination formula _{n}C_{r}=\frac{n!}{r!\left(n-r\right)!}

Apply the idea

\displaystyle _{14} C_{5}	\displaystyle =	\displaystyle \frac{14!}{5!\left(14-5\right)!}	Substitute known values
	\displaystyle =	\displaystyle \frac{14!}{5!\cdot 9!}	Evaluate the subtraction
	\displaystyle =	\displaystyle \dfrac{14\cdot 13\cdot 12\cdot 11\cdot 10\cdot \cancel{9!} }{5\cdot 4\cdot 3 \cdot2 \cdot 1 \cdot \cancel{9!}}	Expand and divide out common factors
	\displaystyle =	\displaystyle \dfrac{14\cdot 13\cdot \cancel{12}\cdot 11\cdot \cancel{10}}{\cancel{5}\cdot \cancel{4\cdot 3} \cdot \cancel{2} \cdot 1 }	Simplify since 4\cdot 3=11 and 5\cdot 2=10
	\displaystyle =	\displaystyle 2002	Evaluate the multiplication

Example 9

4 letters are chosen at random from the word TRAMPOLINE. Find the number of ways to choose the letters such that the selection includes exactly 2 vowels.

Worked Solution

Create a strategy

In order for the selection to contain exactly 2 vowels, 2 letters will be vowels and the remaining 2 will be consonants. To find the number of ways this can happen, we need to:

Find the number of vowels in TRAMPOLINE
Find the number of ways to choose 2 vowels from the total number of vowels
Find the number of consonants in TRAMPOLINE
Find the number of ways to choose 2 consonants from the total number of consonants

The order in which we choose the letters does not matter, so we are finding combinations. Finally, we will use the fundamental counting principle to find the total number of combinations possible.

Apply the idea

The word TRAMPOLINE contains 4 vowels and 6 consonants.

\displaystyle \text{}_4 C_2	\displaystyle =	\displaystyle 6	Number of ways to choose two letters from the vowels
\displaystyle \text{}_6 C_2	\displaystyle =	\displaystyle 15	Number of ways to choose two letters from the consonants

Now, we will use the fundmental counting principle to find the number of ways to choose 4 letters when 2 of them are vowels:

\displaystyle \text{}_4 C_2 \cdot \text{}_6 C_2	\displaystyle =	\displaystyle 6\cdot 15
	\displaystyle =	\displaystyle 90	Evaluate the multiplication

Reflect and check

We can extend this question by determining the probability of choosing 4 letters such that 2 letters are vowels. To do this, we need to find the total number of ways to choose 4 letters from the 10 letters in TRAMPOLINE:\text{ }_{10} C_4=210Now, we can divide the total number of favorable outcomes by the total number of possible outcomes:

\displaystyle P(2 \text{ vowels})	\displaystyle =	\displaystyle \dfrac{90}{210}	Probability formula
\displaystyle \text{ }	\displaystyle =	\displaystyle \dfrac{3}{7}	Simplify the fraction

Therefore, the probability that the selection includes 2 vowels is \dfrac{3}{7}.

Example 10

A box contains 6 pens of different colors: red, green, blue, yellow, black and white. Three pens are drawn at random without replacement.

Determine the total number of possible selections.

Worked Solution

Create a strategy

Because the order in which we choose the pens does not matter, we can use the combination formula.

Apply the idea

There are 6 total pens, so n=6. We need to choose 3 of them, so r=3.

\displaystyle {}_nC_{r}	\displaystyle =	\displaystyle \dfrac{n!}{(n-r)!r!}	State the combination formula
\displaystyle {}_6C_{2}	\displaystyle =	\displaystyle \dfrac{6!}{(6-3)!3!}	Substitute known values
	\displaystyle =	\displaystyle \dfrac{6\cdot5\cdot4\cdot3!}{3!3\cdot2\cdot1}	Expand the factorials
	\displaystyle =	\displaystyle \dfrac{6\cdot5\cdot4}{3\cdot2\cdot1}	Simplify since \dfrac{3!}{3!}=1
	\displaystyle =	\displaystyle 20	Evaluate the multiplication and division

There are 20 possible ways to choose the pens.

Determine the probability of choosing the green, black, and red pens.

Worked Solution

Create a strategy

Because the order in which we choose the pens does not matter, drawing green then black then red is the same as drawing red then black then green, and similiar with the other combinations. So, drawing these 3 colors is one possible combination out of all the possible combinations found above.

Apply the idea

The probability of drawing a green, black, and red pen together is \dfrac{1}{20}.

Suppose the three pens are being placed in a specific order after they are chosen. How many different arrangements of 3 pens can be made?

Worked Solution

Create a strategy

Since the pens are being arranged in a specific order, we now use the permutation formula. The number of ways to arrange 3 pens is calculated by the permutation formula.

Apply the idea

\displaystyle {}_nP_{r}	\displaystyle =	\displaystyle \dfrac{n!}{(n-r)!}	State the permutation formula
\displaystyle {}_6P_{3}	\displaystyle =	\displaystyle \dfrac{6!}{(6-3)!}	Substitute the values
	\displaystyle =	\displaystyle 6\cdot5\cdot4 = 120

There are 120 possible arrangements.

Compare the answers to parts (a) and (c).

Worked Solution

Create a strategy

In part (a), we calculated the total number of ways to select 3 pens from 6, without considering their order, using the combination formula. In part (c), we calculated the total number of ways to arrange those 3 pens in a specific order, using the permutation formula.

Apply the idea

The answer from part (a) is 20 combinations, since order does not matter. The answer from part (c) is 120 arrangements, since order matters. The number of permutations (arrangements) is greater than the number of combinations because permutations consider different orders as distinct, whereas combinations do not.

Reflect and check

In general, the number of permutations is always greater than or equal to the number of combinations, as permutations account for more possibilities by considering different orders of the same items.

Idea summary

We use combinations to find the number of ways r objects can be chosen from n objects when the order in which we choose them does not matter.

\displaystyle \binom{n}{r}=\text{}_n C_r = \dfrac{n!}{r!(n-r)!}

\bm{n}

the total number of objects in the set

\bm{r}

the number of objects that are being chosen

Outcomes

A2.ST.3

The student will compute and distinguish between permutations and combinations.

A2.ST.3a

Compare and contrast permutations and combinations to count the number of ways that events can occur.

A2.ST.3b

Calculate the number of permutations of n objects taken r at a time.

A2.ST.3c

Calculate the number of combinations of n objects taken r at a time.

A2.ST.3d

Use permutations and combinations as counting techniques to solve contextual problems.

A2.ST.3e

Calculate and verify permutations and combinations using technology.

7.08 Permutations and combinations

Ideas

Fundamental counting principle

Exploration

Examples

Example 1

Example 2

Permutations

Examples

Example 3

Example 4

Example 5

Example 6

Combinations

Exploration

Examples

Example 7

Example 8

Example 9

Example 10

Outcomes

A2.ST.3

A2.ST.3a

A2.ST.3b

A2.ST.3c

A2.ST.3d

A2.ST.3e

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