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{{ printedBook.courseTrack.name }} {{ printedBook.name }} 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.**

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?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.

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 $2$ choices for the second position, and $1$ choice for the third position.

In this case, there are $6$ possible permutations. $456465546564645654 $ Although all these numbers are formed with the same three digits, the order in which they appear will make 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 having many objects. In these cases, the Permutation Formula can be used instead.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 |
---|---|

$1$ | $n$ |

$2$ | $n−1$ |

$3$ | $n−2$ |

$⋮$ | $⋮$ |

$r$ | $(n−r+1)$ |

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 $n.$
$_{n}P_{n}=n! $
In this case, since there are ten cities to be visited, the factorial of $10$ needs to be calculated.
Therefore, Dominika and Heichi have $3628800$ different ways to visit all ten most visited cities of Europe.

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

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. $_{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.
There are $720$ ways of visit $3$ of the ten cities.

$_{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

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

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! $

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 |

$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 ${\textstyle 3 \, \ifnumequal{3}{1}{\text{decimal}}{\text{decimals}}}$

$P(A)≈0.003$

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.

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.

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! $

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

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

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

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! $

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

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

$_{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

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

$3$ 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 ${\textstyle 3 \, \ifnumequal{3}{1}{\text{decimal}}{\text{decimals}}}$

$P(A)≈0.083$

WritePercent

Convert to percent

$P(A)≈8.3%$

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.

a Note that the phrase

### General Expression For $P(A_{′})$ and $P(A)$

Suppose that the group consists of only $2$ students. The first student can have their birthday on *any* day. The probability of the second student not having their birthday on the same day is the ratio of $364$ to $365.$ Therefore, this expression is the probability of the two students not having their birthday on the same day.
$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.
Now, the numerator will be simplified.
The above expression represents the permutations of $365$ taking $n$ at a time. $_{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}_{365}P_{n} $ ### Calculating $P(A)$ if $n=10$

To apply the formula, $P(A_{′})$ will be calculated first.
Finally, by subtracting $P(A_{′})$ from $1,$ the probability of $A$ can be calculated.

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 |

Suppose that instead of $10,$ there are $n$ students. Let $A$ be the event that at least $2$ students out of $n$ have their birthday on the same day. If $n$ is greater than $365,$ then certainly at least $2$ students would share their birthday. For the sake of the example, $n$ is assumed to be less than or equal to $365.$ $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.

$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! $

In this case, $A$ is the event that at least $2$ students out of $10$ have their birthday on the same day.

$A$ | $A_{′}$ |
---|---|

At leat $2$ students out of $10$ have their birthday on the same day. | No one of the $10$ students shares a birthday. |

$P(A_{′})=365_{n}_{365}P_{n} $

Substitute

$n=10$

$P(A_{′})=365_{10}_{365}P_{10} $

Evaluate right-hand side

Substitute

$_{365}P_{10}=(365−10)!365! $

$P(A_{′})=365_{10}(365−10)!365! $

SubTerm

Subtract term

$P(A_{′})=365_{10}355!365! $

Write as a product

$P(A_{′})=365_{10}355!365⋅364⋅…⋅356⋅355! $

CancelCommonFac

Cancel out common factors

$P(A_{′})=365_{10}355!365⋅364⋅…⋅356⋅355! $

SimpQuot

Simplify quotient

$P(A_{′})=365_{10}1365⋅364⋅…⋅356 $

DivByOne

$1a =a$

$P(A_{′})=365_{10}365⋅364⋅…⋅356 $

UseCalc

Use a calculator

$P(A_{′})=0.883051…$

RoundDec

Round to ${\textstyle 2 \, \ifnumequal{2}{1}{\text{decimal}}{\text{decimals}}}$

$P(A_{′})≈0.88$

b It was found that if $A$ is the event that at least $2$ out of $n$ students have their birthday on the same day, $P(A)$ is given by the following equation.
$P(A)=1−365_{n}_{365}P_{n} $
By substituting $30$ for $n,$ $P(A)$ can be found for a group of $30$ students.
The probability that at least $2$ students out of $30$ have their birthday on the same day is about $0.71.$ What is more, and extremely curious, only $57$ people are required to bring the probability that at least two people have the same birthday up to $0.99.$

$P(A)=1−365_{n}_{365}P_{n} $

Substitute

$n=30$

$P(A)=1−365_{30}_{365}P_{30} $

Evaluate right-hand side

Substitute

$_{365}P_{30}=(365−30)!365! $

$P(A)=1−365_{30}(365−30)!365! $

SubTerm

Subtract term

$P(A)=1−365_{30}335!365! $

Write as a product

$P(A)=1−365_{30}335!365⋅364…⋅336⋅335! $

CancelCommonFac

Cancel out common factors

$P(A)=1−365_{30}335!365⋅364…⋅336⋅335! $

SimpQuot

Simplify quotient

$P(A)=1−365_{30}1365⋅364…⋅336 $

DivByOne

$1a =a$

$P(A)=1−365_{30}365⋅364…⋅336 $

UseCalc

Use a calculator

$P(A)=1−0.293683…$

SubTerm

Subtract term

$P(A)=0.706317…$

RoundDec

Round to ${\textstyle 2 \, \ifnumequal{2}{1}{\text{decimal}}{\text{decimals}}}$

$P(A)≈0.71$

Permutations and combinations can be used in many situations. Understanding these mathematical concepts can help solve many intricate problems. With this in mind, reconsider the problem in which Vincenzo wants to create arrangements by using the following letters.

How many different arrangements with $3$ vowels and $4$ consonants can be created?{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":[],"constants":[]}},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":null,"formTextAfter":"words","answer":{"text":["756000","756 000"]}}

Begin by calculating the number of ways of selecting $3$ vowels and $4$ consonants. The order of the arrangements is essential.

Because the arrangements consist of $3$ vowels and $4$ consonants, they have $7$ letters. An example arrangement is shown.

Note that $3$ vowels must be selected out of $5,$ which means that the number of possible combinations must be calculated. To do so, the combination formula can be used. The number of combinations of $n$ objects taken $r$ at a time is given by the following formula. $_{n}C_{r}=r!(n−r)!n! $ Therefore, the number of possible combinations can be calculated by substituting $5$ for $n$ and $3$ for $r$ into the formula.$_{n}C_{r}=r!(n−r)!n! $

SubstituteII

$n=5$, $r=3$

$_{5}C_{3}=3!(5−3)!5! $

Evaluate right-hand side

SubTerm

Subtract term

$_{5}C_{3}=3!⋅2!5! $

Write as a product

$_{5}C_{3}=3!⋅2!5⋅4⋅3! $

CancelCommonFac

Cancel out common factors

$_{5}C_{3}=3!⋅2!5⋅4⋅3! $

SimpQuot

Simplify quotient

$_{5}C_{3}=2!5⋅4 $

Multiply

Multiply

$_{5}C_{3}=2!20 $

$2!=2$

$_{5}C_{3}=220 $

CalcQuot

Calculate quotient

$_{5}C_{3}=10$

$_{n}C_{r}=r!(n−r)!n! $

SubstituteII

$n=6$, $r=4$

$_{6}C_{4}=4!(6−4)!6! $

Evaluate right-hand side

SubTerm

Subtract term

$_{6}C_{4}=4!⋅2!6! $

Write as a product

$_{6}C_{4}=4!⋅2!6⋅5⋅4! $

CancelCommonFac

Cancel out common factors

$_{6}C_{4}=4!⋅2!6⋅5⋅4! $

SimpQuot

Simplify quotient

$_{6}C_{4}=2!6⋅5 $

Multiply

Multiply

$_{6}C_{4}=2!30 $

$2!=2$

$_{6}C_{4}=230 $

CalcQuot

Calculate quotient

$_{6}C_{4}=15$

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