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In this lesson, the concepts of permutation and combination will be introduced and connected to the computation of probabilities of compound events.
### Catch-Up and Review

**Here are a few recommended readings before getting started with this lesson.**

Challenge

Vincenzo is playing with the following letters.

He wants to create as many different arrangements as possible using $7$ of the letters without repeating any letters in each arrangement. He also decides that the arrangements must consist of $3$ vowels and $4$ consonants. How many different arrangements can Vincenzo create?Discussion

Many situations involve the rearrangement of a specific set of objects. These are called *permutation* problems. Below, the definition of permutation and its corresponding formula are discussed.

Concept

A permutation is an arrangement of objects in which the order is important. For example, consider constructing a number using only the digits $4,$ $5,$ and $6$ without repetitions. Any of the three digits can be picked for the first position, leaving two choices for the second position, then only one choice for the third position.

In this case, there are six possible permutations.$456465546564645654 $

Although all these numbers are formed with the same three digits, the order in which the digits appear affects the number produced. Each different order of the digits creates a different number. The number of permutations can be calculated by using the Fundamental Counting Principle.
$Number of permutations=3⋅2⋅1⇕Number of permutations=6 $

Listing all the permutations may be a difficult task when many objects are being arranged. In these cases, the Permutation Formula can be used instead.

Rule

The number of permutations of $n$ different objects arranged $r$ at a time — denoted as $_{n}P_{r}$ — is given by the following formula.

$_{n}P_{r}=(n−r)!n! ,r≤n$

The exclamation sign in the formula indicates that the factorial of a value must be calculated. As a direct consequence, since $0!=1,$ when $n=r$ the number of permutations is given by the factorial of $n.$

$_{n}P_{n}=n!$

An alternative notation for $_{n}P_{r}$ is $P(n,r).$

The formula can be proven by using the Fundamental Counting Principle. In an arrangement with $r$ elements, there are $n$ choices for the first element, $n−1$ choices for the second element, $n−2$ choices for the third element, and so on.

Position | Number of Choices |
---|---|

$1$ | $n$ |

$2$ | $n−1$ |

$3$ | $n−2$ |

$⋮$ | $⋮$ |

$r$ | $(n−r+1)$ |

$_{n}P_{r}=n(n−1)…(n−r+1) $

The right-hand side of this equation consists of the first $r$ factors of the factorial of $n.$ The Multiplication Property of Equality can be used to multiply both sides by the last $n−r$ factors of the factorial of $n.$ The product of these factors can be written as $(n−r)!$
It is important to remember how to write $(n−r)!$ as a product.
$(n−r)!=(n−r)⋅(n−r−1)⋅…2⋅1 $

This expression will be substituted for $(n−r)!$ on the right-hand side of the equation.
$_{n}P_{r}(n−r)!=n(n−1)…(n−r+1)(n−r)!$

Write as a product

$_{n}P_{r}(n−r)!=n(n−1)…(n−r+1)(n−r)(n−r−1)…(2)(1)$

Write as a factorial

$_{n}P_{r}(n−r)!=n!$

DivEqn

$LHS/(n−r)!=RHS/(n−r)!$

$_{n}P_{r}=(n−r)!n! $

Example

The following cities are the ten most visited cities in Europe.

Rank | City |
---|---|

$1$ | London, UK |

$2$ | Paris, France |

$3$ | Istanbul, Turkey |

$4$ | Antalya, Turkey |

$5$ | Rome, Italy |

$6$ | Prague, Czech Republic |

$7$ | Amsterdam, Netherlands |

$8$ | Barcelona, Spain |

$9$ | Vienna, Austria |

$10$ | Milan, Italy |

a Dominika and her friend Heichi are planning to go to Europe next summer. How many different ways can they arrange their trip to see all ten cities?

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b Suppose that they can only visit $3$ of the ten places. In how many ways can they do it?

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a The number of permutations of $n$ out of $n$ is given by the factorial of $n.$

b Consider the permutation formula for $r$ objects out of $n.$

a Because the order in which the cities will be visited is important, the problem can be solved by using permutations. The number of permutations when taking $n$ items out of $n$ is given by the factorial of $n.$
Therefore, Dominika and Heichi have $3628800$ different ways to visit all ten most visited cities of Europe.

$_{n}P_{n}=n! $

In this case, since there are ten cities to be visited, the factorial of $10$ needs to be calculated.
$_{n}P_{n}=n!$

Substitute

$n=10$

$_{10}P_{10}=10!$

Write as a product

$_{10}P_{10}=10⋅9⋅8⋅8⋅7⋅6⋅5⋅4⋅3⋅2⋅1$

Multiply

Multiply

$_{10}P_{10}=3628800$

b Now suppose that Heichi and Dominika will visit only three places. Since the order of the cities they visit is important, permutations can be used again. The number of permutations when taking $r$ items out of $n$ is given by the following formula.
There are $720$ ways of visit $3$ of the ten cities.

$_{n}P_{r}=(n−r)!n! $

Of the $10$ places, only $3$ can be visited. Therefore, the number of permutations of $3$ out of $10$ needs to be calculated.
$_{n}P_{r}=(n−r)!n! $

SubstituteII

$n=10$, $r=3$

$_{10}P_{3}=(10−3)!10! $

▼

Evaluate right-hand side

SubTerm

Subtract term

$_{10}P_{3}=7!10! $

Write as a product

$_{10}P_{3}=7!10⋅9⋅8⋅7! $

CancelCommonFac

Cancel out common factors

$_{10}P_{3}=7!10⋅9⋅8⋅7! $

SimpQuot

Simplify quotient

$_{10}P_{3}=110⋅9⋅8 $

DivByOne

$1a =a$

$_{10}P_{3}=10⋅9⋅8$

Multiply

Multiply

$_{10}P_{3}=720$

Example

In the $2020$ Olympic Games, the competitors of the men's $100$ meter freestyle swimming finals came from the following countries.

Men’s $100$ Meter Freestyle Swimming Finals | |
---|---|

United States | Australia |

Russia | France |

South Korea | Italy |

Hungary | Romania |

a If there were no ties, in how many different ways could the gold, silver, and bronze medals have been awarded?

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b If all athletes have the same athletic ability, what is the probability that the Italian swimmer wins the gold medal, the French swimmer the silver medal, and the Australian swimmer the bronze medal? Approximate the answer to three decimals.

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a Because the order in which the medals are awarded is essential, the problem can be solved by using permutations. The number of permutations when taking $r$ items out of $n$ is given by the following formula.

$_{n}P_{r}=(n−r)!n! $

Of the $8$ swimmers, only $3$ can be on the podium and be awarded medals. Therefore, the number of permutations of $3$ out of $8$ needs to be calculated.
$_{n}P_{r}=(n−r)!n! $

SubstituteII

$n=8$, $r=3$

$_{8}P_{3}=(8−3)!8! $

▼

Evaluate right-hand side

SubTerm

Subtract term

$_{8}P_{3}=5!8! $

Write as a product

$_{8}P_{3}=5!8⋅7⋅6⋅5! $

CancelCommonFac

Cancel out common factors

$_{8}P_{3}=5!8⋅7⋅6⋅5! $

SimpQuot

Simplify quotient

$_{8}P_{3}=18⋅7⋅6 $

DivByOne

$1a =a$

$_{8}P_{3}=8⋅7⋅6$

Multiply

Multiply

$_{8}P_{3}=336$

b Recall that the probability of an event is the ratio of the number of favorable outcomes to the number of possible outcomes.

The number of favorable outcomes is the number of ways the Italian, French, and Australian athletes can win the gold, silver, and bronze medals, respectively. Although there is only one way for the order of the first three positions, there are several ways for the order of the remaining five positions. All these are favorable outcomes.

Example Favorable Outcomes | ||
---|---|---|

Italy | Italy | Italy |

France | France | France |

Australia | Australia | Australia |

United States | Hungary | South Korea |

Russia | Romania | Russia |

South Korea | Russia | United States |

Hungary | United States | Hungary |

Romania | South Korea | Romania |

$Favorable Outcomes_{5}P_{5}=5!⇔_{5}P_{5}=120 $

The number of possible outcomes is the number of all possible ways in which the medals can be awarded. This is found by calculating the permutations of $8$ swimmers taken from a group of $8.$
$Possible Outcomes_{8}P_{8}=8!⇔_{8}P_{8}=40320 $

With this information, the probability that the Italian swimmer wins the golden medal, the French swimmer the silver medal, and the Australian swimmer the bronze medal can be calculated.
$P(A)=Number of possible outcomesNumber of favorable outcomes $

SubstituteValues

Substitute values

$P(A)=40320120 $

ReduceFrac

$ba =b/120a/120 $

$P(A)=3361 $

FracToDiv

$ba =a÷b$

$P(A)=0.002976…$

RoundDec

Round to $3$ decimal place(s)

$P(A)≈0.003$

Discussion

In other situations, only the selected objects are important, not the order in which they come. These problems are called *combination* problems. Below, the definition of combination and its corresponding formula are developed.

Concept

A combination is a selection of objects in which the order is *not* important. Combinations focus on the selected objects. For example, consider choosing two different ingredients for a salad from five unique options in a salad bar.

Rule

The number of combinations of $n$ different objects taken $r$ at a time — denoted as $_{n}C_{r}$ — is given by the following formula.

$_{n}C_{r}=r!(n−r)!n! ,r≤n $

The exclamation mark in the formula indicates that the factorial of the value should be calculated. As a direct consequence of the above formula, since $0!=1,$ when $n=r$ the number of combinations is $1.$

$_{n}C_{n}=1 $

An alternative notation for $_{c}C_{r}$ is $C(n,r).$

The formula can be proven by using the Permutation Formulas.

$_{n}P_{r}=(n−r)!n! and_{r}P_{r}=r! $

Let $_{n}C_{r}$ be the number of combinations of $n$ objects chosen $r$ at a time. By the Fundamental Counting Principle, the product of $_{n}C_{r}$ by $_{r}P_{r}$ equals the number of permutations of $r$ objects out of $n.$
$_{n}C_{r}⋅_{r}P_{r}=_{n}P_{r}⇓_{n}C_{r}⋅r!=(n−r)!n! $

Finally, by applying the Division Property of Equality, the Combination Formula is obtained. $_{n}C_{r}⋅r!=(n−r)!n! ⇕_{n}C_{r}=r!(n−r)!n! $

Example

Kriz is going on vacation next month and wants to pack $4$ books from their must-read list. Each of the books belongs to one of the following genres.

Kriz’s List of Books By Genres | |
---|---|

Fantasy | Romance |

Mystery | Fiction |

Biography | Graphic Novel |

Drama | History |

Western | Poetry |

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The order in which the books are selected is not crucial.

As long as $4$ books are selected, the order is not important. Therefore, the different ways in which Kriz can select $4$ books can be found by using combinations. The number of combinations when selecting $r$ items out of $n$ is given by the following formula.
There are $210$ ways in which Kriz can select $4$ books to pack from their must-read list.

$_{n}C_{r}=r!(n−r)!n! $

By substituting $10$ for $n$ and $4$ for $r,$ the number of combinations can be calculated. $_{n}C_{r}=r!(n−r)!n! $

SubstituteII

$n=10$, $r=4$

$_{10}C_{4}=4!(10−4)!10! $

▼

Evaluate right-hand side

SubTerm

Subtract term

$_{10}C_{4}=4!⋅6!10! $

Write as a product

$_{10}C_{4}=4!⋅6!10⋅9⋅8⋅7⋅6! $

CancelCommonFac

Cancel out common factors

$_{10}C_{4}=4!⋅6!10⋅9⋅8⋅7⋅6! $

SimpQuot

Simplify quotient

$_{10}C_{4}=4!10⋅9⋅8⋅7 $

Write as a product

$_{10}C_{4}=4⋅3⋅2⋅110⋅9⋅8⋅7 $

Multiply

Multiply

$_{10}C_{4}=245040 $

CalcQuot

Calculate quotient

$_{10}C_{4}=210$

Example

Kriz has decided that they will select $5$ of their books at random instead of $4.$ However, they would prefer to bring at least one book from the fantasy, mystery, and drama genres. What is the probability of them choosing these three genres if the selection pool consists of $10$ books from $10$ different genres? Write the answer in percentage form rounded to $1$ decimal place.

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The probability of an event is the ratio of the number of favorable outcomes to the number of possible outcomes.

$_{n}C_{r}=r!(n−r)!n! $

By substituting $10$ for $n$ and $5$ for $r,$ the number of possible outcomes can be calculated.
$_{n}C_{r}=r!(n−r)!n! $

SubstituteII

$n=10$, $r=5$

$_{10}C_{5}=5!(10−5)!10! $

▼

Evaluate right-hand side

SubTerm

Subtract term

$_{10}C_{5}=5!⋅5!10! $

Write as a product

$_{10}C_{5}=5!⋅5!10⋅9⋅8⋅7⋅6⋅5! $

CancelCommonFac

Cancel out common factors

$_{10}C_{5}=5!⋅5!10⋅9⋅8⋅7⋅6⋅5! $

SimpQuot

Simplify quotient

$_{10}C_{5}=5!10⋅9⋅8⋅7⋅6 $

Write as a product

$_{10}C_{5}=5⋅4⋅3⋅2⋅110⋅9⋅8⋅7⋅6 $

Multiply

Multiply

$_{10}C_{5}=12030240 $

CalcQuot

Calculate quotient

$_{10}C_{5}=252$

$Possible Outcomes:252 $

Since the order does not matter, there is only one way of selecting a fantasy, a mystery, and a drama book. The other $2$ books need to be selected from the remaining $7.$ Therefore, the combinations of $2$ out of $7$ books will be calculated.
$_{n}C_{r}=r!(n−r)!n! $

SubstituteII

$n=7$, $r=2$

$_{7}C_{2}=2!(7−2)!7! $

▼

Evaluate right-hand side

SubTerm

Subtract term

$_{7}C_{2}=2!⋅5!7! $

Write as a product

$_{7}C_{2}=2!⋅5!7⋅6⋅5! $

CancelCommonFac

Cancel out common factors

$_{7}C_{2}=2!⋅5!7⋅6⋅5! $

SimpQuot

Simplify quotient

$_{7}C_{2}=2!7⋅6 $

$2!=2$

$_{7}C_{2}=27⋅6 $

Multiply

Multiply

$_{7}C_{2}=242 $

CalcQuot

Calculate quotient

$_{7}C_{2}=21$

$Favorable Outcomes:1⋅21=21 $

There are $21$ favorable outcomes and $252$ possible outcomes. Let $A$ be the event that $3$ of the $5$ books selected are a fantasy, a mystery, and a drama.By substituting the number of favorable outcomes and the number of possible outcomes into the Probability Formula, $P(A)$ can be found.

$P(A)=Number of possible outcomesNumber of favorable outcomes $

SubstituteII

$Number of favorable outcomes=21$, $Number of possible outcomes=252$

$P(A)=25221 $

ReduceFrac

$ba =b/21a/21 $

$P(A)=121 $

FracToDiv

$ba =a÷b$

$P(A)=0.083$

RoundDec

Round to $3$ decimal place(s)

$P(A)≈0.083$

WritePercent

Convert to percent

$P(A)≈8.3%$

Example

Magdalena teaches algebra to a group of $10$ students. While making a list to track their attendance, she wonders whether at least $2$ students have the same birthday. For simplicity, suppose that all years have exactly $365$ days.

a What is the probability that at least two students have the same birthday? Approximate the answer to two decimal places.

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b If there were $30$ students, what would be the probability that at least $2$ students the same birthday? Approximate the answer to two decimal places.

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a Consider the Complement Rule of Probability.

b Substitute $30$ for $n$ and $2$ for $r$ in the expression found in Part A.

at leastmeans that any outcome with $2$ or more students with the same birthday is a favorable outcome. Therefore, the following outcomes are all the possible favorable outcomes.

Favorable Outcomes | ||
---|---|---|

$2$ have the same birthday | $3$ have the same birthday | $4$ have the same birthday |

$5$ have the same birthday | $6$ have the same birthday | $7$ have the same birthday |

$8$ have the same birthday | $9$ have the same birthday | $10$ have the same birthday |

$A= At least2students out ofnhavetheir birthday on the same day $

Finding the probability of every possible outcome — all nine outcomes in the table above — means finding nine different probabilities. The opposite of at least $2$ students having their birthday on the same day is that there are no students with their birthday on the same day. This is the complement of $A$ written as $A_{′}.$
$A_{′}= No one of thenstudentsshares a birthday $

The Complement Rule of Probability can help to deal with this situation. To do so, a general expression will be found for no students from a group of $n$ having the same birthday. Then, $P(A)$ can be calculated. $Probability of2Students Not HavingTheir Birthday on the Same Day365364 $

Suppose that a third student is added to the group. The probability that this student has their birthday on a different day from the previous two is the ratio of $363$ to $365.$ By the Multiplication Rule of Probability, the probability that no students out of these three have their birthdays on the same day can be found.
$Probability of3Students Not HavingTheir Birthday on the Same Day365364 ⋅365363 ⇔365_{2}364⋅363 $

By following the same reasoning, the probability that $n$ students do not share a birthday, $P(A_{′}),$ can be written.
$P(A_{′})=n−1times365364 ⋅365363 ⋅…365365−n+1 ⇕P(A_{′})=365_{n−1}364⋅363⋅…⋅(365−n+1) $

The obtained expression will now be simplified.
$P(A_{′})=365_{n−1}364⋅363⋅…⋅(365−n+1) $

▼

Simplify right-hand side

IdPropMult

Identity Property of Multiplication

$P(A_{′})=1⋅365_{n−1}364⋅363⋅…⋅(365−n+1) $

OneToFrac

Rewrite $1$ as $365365 $

$P(A_{′})=365365 ⋅365_{n−1}364⋅363⋅…⋅(365−n+1) $

MultFrac

Multiply fractions

$P(A_{′})=365⋅365_{n−1}365⋅364⋅363⋅…⋅(365−n+1) $

MultBasePow

$a⋅a_{m}=a_{1+m}$

$P(A_{′})=365_{n}365⋅364⋅363⋅…⋅(365−n+1) $

$365⋅364⋅363⋅…⋅(365−n+1)$

▼

Simplify

IdPropMult

Identity Property of Multiplication

$365⋅364⋅363⋅…⋅(365−n+1)⋅1$

OneToFrac

Rewrite $1$ as $(365−n)!(365−n)! $

$365⋅364⋅363⋅…⋅(365−n+1)⋅(365−n)!(365−n)! $

MoveLeftFacToNum

$a⋅cb =ca⋅b $

$(365−n)!365⋅364⋅363⋅…⋅(365−n+1)⋅(365−n)! $

Substitute

$365⋅364⋅363⋅…⋅(365−n+1)⋅(365−n)!=365!$

$(365−n)!365! $

$_{365}P_{n}=(365−n)!365! $

Therefore, $P(A_{′})$ is the quotient of $_{365}P_{n}$ and $365_{n}.$
$P(A_{′})=365_{n}_{365}P_{n} $

The Complement Rule of Probability states that $P(A)$ is the difference between $1$ and $P(A_{′}).$ By using the expression found for $P(A_{′}),$ an expression for $P(A)$ can be found.
$P(A)=1−P(A_{′})⇕P(A)=1−365_{n}$