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5. Probability
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United States of America
Grades 9-12

5.06 Permutations and combinations

Lesson
Practice

Introduction

When finding the probability, we need to know the total number of outcomes. Sometimes, the total number of outcomes is not given and needs to be calculated. This lesson explores different methods of finding the total number of ways an event can happen or ways objects can be arranged.

The fundamental counting principle

To find the number of ways a set of r objects can be chosen or arranged from a set of n objects, we can use the fundamental counting principle, factorials, combinations, or permutations as appropriate.

Exploration

Suppose there is a food truck that sells ice cream sundaes. You are given the following choices, and you 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
  1. How many different types of sundaes could be created? Explain your process for finding the answer.

To understand the fundamental counting principle better, we can draw a tree diagram. Using the situation from the exploration:

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\times 2 \times 4 = 24. This is an example of the Fundamental counting principle.

Fundamental Counting Principle

To find the number of ways multiple events can happen together, find the product of the number of ways each individual event can happen.

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\times 4\times 3\times 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\times 4\times 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{ }\times\text{ }\dfrac{}{2\text{nd digit}}\text{ }\times \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{ }\times\text{ }\dfrac{4}{2\text{nd digit}}\text{ }\times \text{ }\dfrac{3}{3\text{rd digit}}

Therefore, there are =3\times4\times3=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{ }\times\text{ }\dfrac{4}{2\text{nd digit}}\text{ }\times \text{ }\dfrac{4}{3\text{rd digit}}

Therefore, there are =2\times4\times4=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\times 4\times 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.

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 \times (n-1) \times (n-2) \times \ldots \times 2 \times 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.

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 from 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 from for the third digit, and 7 numbers left to choose from 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\times 9\times 8\times 7=5\,040} 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

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.

b

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.

c

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

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!}{(8-8)!}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\,320Evaluate 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.

  1. How many different color combinations can you make with the 3 notebooks you choose? Explain your process.
  2. 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)!}=20
\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 6

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

a

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

Apply the idea

10!=10\cdot 9\cdot 8\cdot 7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1=4\,354\,560

Reflect and check

This could also have been calculated with the permutation formula:

\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{4\,354\,560}{1}Evaluate the factorials
\displaystyle =\displaystyle 4\,354\,560Evaluate the division
b

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 5Simplify since \dfrac{4\cdot 3\cdot 2\cdot 1}{4\cdot 3\cdot 2\cdot 1}=1
\displaystyle \text{ }\displaystyle =\displaystyle 210Evaluate 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
c

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.

d

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\,504Evaluate the multiplication and division

There are 15\, 504 possible groups.

Example 7

4 letters are chosen at random from the word TRAMPOLINE. Find the probability that the selection includes exactly 2 vowels.

Worked Solution
Create a strategy

The probability of an event is defined as: P\left(\text{event}\right)=\dfrac{\text{number of outcomes resulting in event}}{\text{total possible number of outcomes}}

So we need to determine:

  • The number of ways 4 letters that can be selected with 2 of them being vowels.

  • The total number of ways in which 4 letters can be selected from 10 the letters without restriction.

  • Then divide the first result by the second.

In order for the selection to contain 2 vowels, we need to choose 2 letters from the 4 vowels in TRAMPOLINE, and 2 letters from the 6 consonants in TRAMPOLINE. For both these situations, the order of the letters does not matter.

Apply the idea
\displaystyle \text{}_4 C_2\displaystyle =\displaystyle 6Number of ways to choose two letters from the vowels
\displaystyle \text{}_6 C_2\displaystyle =\displaystyle 15Number of ways to choose two letters from the consonants
\displaystyle \text{}_4 C_2 \cdot \text{}_6 C_2\displaystyle =\displaystyle 6\cdot 15Use the counting principle to find the number of ways to choose the 4 letters with 2 vowels
\displaystyle \text{}\displaystyle =\displaystyle 90Evaluate the multiplication
\displaystyle \text{ }_{10} C_4\displaystyle =\displaystyle 210Number of ways to choose any 4 letters from the 10 letters in TRAMPOLINE
\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 8

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

a

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 2 of them, so r=2.

\displaystyle {}_nC_{r}\displaystyle =\displaystyle \dfrac{n!}{(n-r)!r!}State the combination formula
\displaystyle {}_6C_{2}\displaystyle =\displaystyle \dfrac{6!}{(6-2)!2!}Substitute known values
\displaystyle =\displaystyle \dfrac{6\cdot5\cdot4!}{4!\cdot2\cdot1}Expand the factorials
\displaystyle =\displaystyle \dfrac{6\cdot5}{2\cdot1}Simplify since \dfrac{4!}{4!}=1
\displaystyle =\displaystyle 15Evaluate the multiplication and division

There are 15 possible ways to choose the pens.

b

Determine the probability of choosing the green and black pens.

Worked Solution
Create a strategy

Because the order in which we choose the pens does not matter, drawing green then black is the same as drawing black then green. So, drawing these 2 colors is one possible combination out of all the possible combinations found above.

Apply the idea

The probability of drawing the green and black pens together is \dfrac{1}{15}.

Reflect and check

Whenever we see "without replacement", we usually think of conditional probability. If we use that method to find the probability, we would get the same answer.

The probability of drawing green first and then drawing blue is \dfrac{1}{6}\cdot\dfrac{1}{5}=\dfrac{1}{30}

The probability of drawing blue first and then drawing green is \dfrac{1}{6}\cdot\dfrac{1}{5}=\dfrac{1}{30}

The total probability is \dfrac{1}{30}+\dfrac{1}{30}=\dfrac{2}{30}=\dfrac{1}{15}

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

S.CP.B.9 (+)

Use permutations and combinations to compute probabilities of compound events and solve problems.

S.MD.B.6 (+)

Use probabilities to make fair decisions.

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