10. Permutations and Combinations in Probability
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Chapter 6
10.
Permutations and Combinations in Probability
Understanding permutations and combinations is pivotal for calculating probabilities, especially in compound events. The lesson delves into the distinction between permutations, where the order of arrangement matters, and combinations, where the order is inconsequential. For instance, when considering the arrangement of letters in a word or the order of runners in a race, permutations come into play. On the other hand, combinations are used when selecting a subset, like choosing books from a list. The lesson also touches upon real-world scenarios, such as determining the likelihood of students sharing the same birthday. Grasping these mathematical concepts can significantly aid in solving complex problems in various fields.
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| | 10 Theory slides |
| | 10 Exercises - Grade E - A |
| | Each lesson is meant to take 1-2 classroom sessions |
Permutations and Combinations in Probability
Slide of 10
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
Creating Arrangements
Vincenzo is playing with the following letters.
Discussion
Defining Permutations and the Permutation Formula
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
Permutation
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.
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
Permutation Formula
The number of permutations of n different objects arranged r at a time — denoted as nPr — is given by the following formula.
nPr=(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.
nPn=n!
An alternative notation for nPr is P(n,r).
Proof
Permutation FormulaThe 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) |
nPr=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.
nPr(n−r)!=n(n−1)…(n−r+1)(n−r)!
Write as a product
nPr(n−r)!=n(n−1)…(n−r+1)(n−r)(n−r−1)…(2)(1)
Write as a factorial
nPr(n−r)!=n!
DivEqn
LHS/(n−r)!=RHS/(n−r)!
nPr=(n−r)!n!
Example
Investigating Permutations in Real-Life Situations
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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Hint
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.
Solution
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.
nPn=n!
In this case, since there are ten cities to be visited, the factorial of 10 needs to be calculated.
nPn=n!
Substitute
n=10
10P10=10!
Write as a product
10P10=10⋅9⋅8⋅8⋅7⋅6⋅5⋅4⋅3⋅2⋅1
Multiply
Multiply
10P10=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.
nPr=(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.
nPr=(n−r)!n!
SubstituteII
n=10, r=3
10P3=(10−3)!10!
▼
Evaluate right-hand side
SubTerm
Subtract term
10P3=7!10!
Write as a product
10P3=7!10⋅9⋅8⋅7!
CancelCommonFac
Cancel out common factors
10P3=7!10⋅9⋅8⋅7!
SimpQuot
Simplify quotient
10P3=110⋅9⋅8
DivByOne
1a=a
10P3=10⋅9⋅8
Multiply
Multiply
10P3=720
Example
Finding Probabilities Using Permutations
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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Solution
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.
nPr=(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.
nPr=(n−r)!n!
SubstituteII
n=8, r=3
8P3=(8−3)!8!
▼
Evaluate right-hand side
SubTerm
Subtract term
8P3=5!8!
Write as a product
8P3=5!8⋅7⋅6⋅5!
CancelCommonFac
Cancel out common factors
8P3=5!8⋅7⋅6⋅5!
SimpQuot
Simplify quotient
8P3=18⋅7⋅6
DivByOne
1a=a
8P3=8⋅7⋅6
Multiply
Multiply
8P3=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 Outcomes5P5=5!⇔5P5=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 Outcomes8P8=8!⇔8P8=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
Defining Combinations and the Combination Formula
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
Combination
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
Combination Formula
The number of combinations of n different objects taken r at a time — denoted as nCr — is given by the following formula.
nCr=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.
nCn=1
An alternative notation for cCr is C(n,r).
Proof
The formula can be proven by using the Permutation Formulas.
nPr=(n−r)!n!andrPr=r!
Let nCr be the number of combinations of n objects chosen r at a time. By the Fundamental Counting Principle, the product of nCr by rPr equals the number of permutations of r objects out of n.
nCr⋅rPr=nPr⇓nCr⋅r!=(n−r)!n!
Finally, by applying the Division Property of Equality, the Combination Formula is obtained. nCr⋅r!=(n−r)!n!⇕nCr=r!(n−r)!n!
Example
Investigating Combinations in Real-Life Situations
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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Hint
The order in which the books are selected is not crucial.
Solution
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.
nCr=r!(n−r)!n!
By substituting 10 for n and 4 for r, the number of combinations can be calculated. nCr=r!(n−r)!n!
SubstituteII
n=10, r=4
10C4=4!(10−4)!10!
▼
Evaluate right-hand side
SubTerm
Subtract term
10C4=4!⋅6!10!
Write as a product
10C4=4!⋅6!10⋅9⋅8⋅7⋅6!
CancelCommonFac
Cancel out common factors
10C4=4!⋅6!10⋅9⋅8⋅7⋅6!
SimpQuot
Simplify quotient
10C4=4!10⋅9⋅8⋅7
Write as a product
10C4=4⋅3⋅2⋅110⋅9⋅8⋅7
Multiply
Multiply
10C4=245040
CalcQuot
Calculate quotient
10C4=210
Example
Finding Probabilities Using Combinations
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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Hint
The probability of an event is the ratio of the number of favorable outcomes to the number of possible outcomes.
Solution
The probability of an event is the ratio of the number of favorable outcomes to the number of possible outcomes.
nCr=r!(n−r)!n!
By substituting 10 for n and 5 for r, the number of possible outcomes can be calculated.
nCr=r!(n−r)!n!
SubstituteII
n=10, r=5
10C5=5!(10−5)!10!
▼
Evaluate right-hand side
SubTerm
Subtract term
10C5=5!⋅5!10!
Write as a product
10C5=5!⋅5!10⋅9⋅8⋅7⋅6⋅5!
CancelCommonFac
Cancel out common factors
10C5=5!⋅5!10⋅9⋅8⋅7⋅6⋅5!
SimpQuot
Simplify quotient
10C5=5!10⋅9⋅8⋅7⋅6
Write as a product
10C5=5⋅4⋅3⋅2⋅110⋅9⋅8⋅7⋅6
Multiply
Multiply
10C5=12030240
CalcQuot
Calculate quotient
10C5=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.
nCr=r!(n−r)!n!
SubstituteII
n=7, r=2
7C2=2!(7−2)!7!
▼
Evaluate right-hand side
SubTerm
Subtract term
7C2=2!⋅5!7!
Write as a product
7C2=2!⋅5!7⋅6⋅5!
CancelCommonFac
Cancel out common factors
7C2=2!⋅5!7⋅6⋅5!
SimpQuot
Simplify quotient
7C2=2!7⋅6
2!=2
7C2=27⋅6
Multiply
Multiply
7C2=242
CalcQuot
Calculate quotient
7C2=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
The Birthday Problem
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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Hint
a Consider the Complement Rule of Probability.
b Substitute 30 for n and 2 for r in the expression found in Part A.
Solution
a Note that the phrase
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.
Now, the numerator will be simplified.
The above expression represents the permutations of 365 taking n at a time.
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 |
A= At least 2 students out of n have their 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 the n students shares 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. 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 of 2 Students 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 of 3 Students Not HavingTheir Birthday on the Same Day365364⋅365363 ⇔ 3652364⋅363
By following the same reasoning, the probability that n students do not share a birthday, P(A′), can be written.
P(A′)=n−1 times365364⋅365363⋅…365365−n+1⇕P(A′)=365n−1364⋅363⋅…⋅(365−n+1)
The obtained expression will now be simplified.
P(A′)=365n−1364⋅363⋅…⋅(365−n+1)
▼
Simplify right-hand side
IdPropMult
Identity Property of Multiplication
P(A′)=1⋅365n−1364⋅363⋅…⋅(365−n+1)
OneToFrac
Rewrite 1 as 365365
P(A′)=365365⋅365n−1364⋅363⋅…⋅(365−n+1)
MultFrac
Multiply fractions
P(A′)=365⋅365n−1365⋅364⋅363⋅…⋅(365−n+1)
MultBasePow
a⋅am=a1+m
P(A′)=365n365⋅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!
365Pn=(365−n)!365!
Therefore, P(A′) is the quotient of 365Pn and 365n.
P(A′)=365n365Pn
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)=1−365n365Pn
Calculating P(A) if n=10
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′)=365n365Pn
Substitute
n=10
P(A′)=36510365P10
▼
Evaluate right-hand side
Substitute
365P10=(365−10)!365!
P(A′)=36510(365−10)!365!
SubTerm
Subtract term
P(A′)=36510355!365!
Write as a product
P(A′)=36510355!365⋅364⋅…⋅356⋅355!
CancelCommonFac
Cancel out common factors
P(A′)=36510355!365⋅364⋅…⋅356⋅355!
SimpQuot
Simplify quotient
P(A′)=365101365⋅364⋅…⋅356
DivByOne
1a=a
P(A′)=36510365⋅364⋅…⋅356
UseCalc
Use a calculator
P(A′)=0.883051…
RoundDec
Round to 2 decimal place(s)
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.
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)=1−365n365Pn
By substituting 30 for n, P(A) can be found for a group of 30 students.
P(A)=1−365n365Pn
Substitute
n=30
P(A)=1−36530365P30
▼
Evaluate right-hand side
Substitute
365P30=(365−30)!365!
P(A)=1−36530(365−30)!365!
SubTerm
Subtract term
P(A)=1−36530335!365!
Write as a product
P(A)=1−36530335!365⋅364…⋅336⋅335!
CancelCommonFac
Cancel out common factors
P(A)=1−36530335!365⋅364…⋅336⋅335!
SimpQuot
Simplify quotient
P(A)=1−365301365⋅364…⋅336
DivByOne
1a=a
P(A)=1−36530365⋅364…⋅336
UseCalc
Use a calculator
P(A)=1−0.293683…
SubTerm
Subtract term
P(A)=0.706317…
RoundDec
Round to 2 decimal place(s)
P(A)≈0.71
Closure
Creating Arrangements
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.
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Hint
Begin by calculating the number of ways of selecting 3 vowels and 4 consonants. The order of the arrangements is essential.
Solution
Because the arrangements consist of 3 vowels and 4 consonants, they have 7 letters. An example arrangement is shown.
nCr=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.
nCr=r!(n−r)!n!
SubstituteII
n=5, r=3
5C3=3!(5−3)!5!
▼
Evaluate right-hand side
SubTerm
Subtract term
5C3=3!⋅2!5!
Write as a product
5C3=3!⋅2!5⋅4⋅3!
CancelCommonFac
Cancel out common factors
5C3=3!⋅2!5⋅4⋅3!
SimpQuot
Simplify quotient
5C3=2!5⋅4
Multiply
Multiply
5C3=2!20
2!=2
5C3=220
CalcQuot
Calculate quotient
5C3=10
nCr=r!(n−r)!n!
SubstituteII
n=6, r=4
6C4=4!(6−4)!6!
▼
Evaluate right-hand side
SubTerm
Subtract term
6C4=4!⋅2!6!
Write as a product
6C4=4!⋅2!6⋅5⋅4!
CancelCommonFac
Cancel out common factors
6C4=4!⋅2!6⋅5⋅4!
SimpQuot
Simplify quotient
6C4=2!6⋅5
Multiply
Multiply
6C4=2!30
2!=2
6C4=230
CalcQuot
Calculate quotient
6C4=15
10⋅15=150
Now, keep in mind that a different arrangement of each possible combination is different. This means that the order is important, and the number of possible permutations of the seven letters must to be calculated. Recall that the factorial of n gives the number of permutations of n out of n.
7P7=7!⇔7P7=5040
Finally the Fundamental Counting Principle will be used one more time. For each of the 150 ways of selecting three vowels and four consonants, there are 5040 permutations.
Possible Arrangements150⋅5040=756000
Permutations and Combinations in Probability
Exercises
Maya is playing a tile-based word game and has three tiles with the following letters.
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Because we want the number of different arrangements of the given letters, we have a permutation problem. The permutation of n items out of n is given by the factorial of n. In this case, since we have 3 tiles, we need to calculate the factorial of 3 to find the different arrangements. 3!=3* 2* 1 ⇔ 3!= 6 Therefore, there are 6 different ways of arranging the given letters. Additionally, since the number of letters is still small enough, we can visualize the possible arrangements using a tree diagram.
We can use a graphing calculator to determine how many arrangements there are. Start by entering a 3 on your graphing calculator.
Next, push MATH, scroll to the fourth menu, PRB. Finally, scroll to the fourth row and push ENTER twice for the answer.
The probability of an event A is given by the ratio of the favorable outcomes to the number of possible outcomes.
P(A)=Number of favorable outcomes/Number of possible outcomes
The number of possible outcomes is given by all the different arrangements of the given letters. We have found that there are 6 possible arrangements. Therefore, the number of possible outcomes is 6.
Number of possible outcomes= 6
Note that with the three letters, only one word is possible.
Then, the number of favorable outcomes is 1. Therefore, the probability of randomly making a word using all the three letters is given by the ratio of 1 to 6. P(Form a word)=1/6
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Six students are running for president of the student council. Before the voting begins, they are each allowed to give a speech to the rest of the school. In how many different orders can the students be asked to give their speeches?
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We need to determine the number of possible orders for the speech. Because of this, we have a permutation problem. Recall that the number of permutations of n items out of n is given by the factorial of n. Since there are 6 candidates, we need to calculate the factorial of 6. 6!=6* 5* 4* 3* 2* 1 ⇔ 6!= 720 There are 720 different combinations to arrange the student's speeches. This cannot be easily visualized by hand with a tree diagram.
The problem can also be solved by using a graphing calculator. Begin by entering 6 on the calculator.
Next, push MATH, scroll to the fourth menu, PRB. Finally, scroll to the fourth row and push ENTER twice.
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Paulina and a group of friends are going to the high school football game this weekend. To cheer them on, each of them has printed out a letter in the team's name, MOHAWKS,
on large cards that they will hold up during the game.
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Since we want the number of the different arrangements of letters, we have a permutation problem. Note that there are seven unique letters in the word MOHAWKS.
From the permutation formula we know that the number of permutations of n items out of n is given by the factorial of n.
_nP_n=n!
Therefore, we need to calculate the factorial of 7.
7!=7* 6* 5* 4* 3* 2* 1 ⇕ 7!= 5040
There are 5040 ways of arranging the letters.
The probability of an event A is given by the ratio of the number of favorable outcomes to the number of possible outcomes. P(A)=Number of favorable outcomes/Number of possible outcomes The number of possible outcomes is given by the different possible arrangements of the seven letters. Previously, we found that there are 5040 different arrangements. Number of possible outcomes= 5040 Additionally, because all the letters are unique, there is only one possible arrangement that form the team's name. Therefore, we have only one favorable outcome. Number of favorable outcomes= 1 The probability that the cards still spell out the correct name given that they were mixed is given by the ratio of 1 to 5040. P(correct name)=1/5040
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Ten runners are about to race. How many ways could the runners finish in the top three?
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In a race, the order in which the runners finish the race is important. This means that we have a permutation problem. Because we are looking for the top three, we need to calculate the permutations when taking 3 out of 10. To do so, we can use the permutation formula. _nP_r=n!/(n-r)! Therefore, we need to substitute 3 for r and 10 for n into the formula.
There are 720 ways to arrange the top three spots.
We can use the calculator to determine the number of permutations. Start by entering the number of people in the race on your graphing calculator, which is 10.
Next, push MATH, scroll to the fourth menu, PRB. Then, scroll to the second row and push ENTER.
Finally, by entering the number of people to be selected, we can determine the number of permutations when n=10 and r=3.
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In how many ways can the letters in the following word be arranged?
RANGE
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In the word RANGE,
there are 5 unique letters. Because we want the different arrangements that can be formed with this letter set, we are looking for permutations. The number of permutations of n objects out of n is given by the factorial of n.
_nP_n=n!
Therefore, we need to calculate the factorial of 5.
There are 120 possible arrangements.
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