Topics
>
Precalculus>
Combinations and Permutations>
Permutations with Restrictions>
Permutations with Restrictions 3Permutations with Restrictions 3
Try VividMath Premium to unlock full access
Time limit: 0
Quiz summary
0 of 7 questions completed
Questions:
- 1
- 2
- 3
- 4
- 5
- 6
- 7
Information
–
You have already completed the quiz before. Hence you can not start it again.
Quiz is loading...
You must sign in or sign up to start the quiz.
You have to finish following quiz, to start this quiz:
Loading...
- 1
- 2
- 3
- 4
- 5
- 6
- 7
- Answered
- Review
-
Question 1 of 7
1. Question
How many arrangements can be made with the letters in `INDEPENDENCE` if the vowels must always be together?Hint
Help VideoCorrect
Well Done!
Incorrect
Permutation with Repetition
$$\frac{\color{#9a00c7}{n}!}{\color{#004ec4}{a}!\color{#004ec4}{b}!\color{#004ec4}{c}!…}$$`n``=`total number of items
`a,b,c``=`count of each repeated itemsSolve the permutation for the letters if the vowels are treated as a single letter and the permutation for the vowels, then multiply them.First, treat the vowels as a single letter and list down the letters that are repeated.`N` `D` `P` `N` `D` `N` `C` `I E E E E``n=8``N` is repeated `3` times`a=3``D` is repeated `2` times`b=2`Apply the formulaPermutation with Repetition `=` $$\frac{\color{#9a00c7}{n}!}{\color{#004ec4}{a}!\color{#004ec4}{b}!\color{#004ec4}{c}!…}$$ `=` $$\frac{\color{#9a00c7}{8}!}{\color{#004ec4}{3}!\color{#004ec4}{2}!}$$ Substitute `n,a` and `b` `=` $$\frac{8\cdot7\cdot\color{#CC0000}{6}\cdot5\cdot4\cdot3\cdot\color{#CC0000}{2}\cdot\color{#CC0000}{1}}{\color{#CC0000}{3\cdot2\cdot1\cdot2\cdot1}}$$ `=` `8*7*5*4*3` Cancel like terms `=` $$3360$$ Hence, there are `3360` ways to arrange `NDPNDNC` and `I E E E E`Next, solve for the permutation of the vowels.`I``E E E E``n=5``E` is repeated `4` times`a=4`Apply the formulaPermutation with Repetition `=` $$\frac{\color{#9a00c7}{n}!}{\color{#004ec4}{a}!\color{#004ec4}{b}!\color{#004ec4}{c}!…}$$ `=` $$\frac{\color{#9a00c7}{5}!}{\color{#004ec4}{4}!}$$ Substitute `n` and `a` `=` $$\frac{5\cdot\color{#CC0000}{4\cdot3\cdot2\cdot1}}{\color{#CC0000}{4\cdot3\cdot2\cdot1}}$$ `=` `5` Cancel like terms Hence, there are `5` ways to arrange `I E E E E`Finally, multiply the two solved permutationsfirst permutation`=3360`second permutation`=5`$$3360\times5$$ `=` `16 800` Therefore, there are `16 800` ways of arranging the letters `INDEPENDENCE` if the vowels must always be together`16 800` -
Question 2 of 7
2. Question
How many arrangements can be made with the letters in `INDEPENDENCE` if the `E`’s and the `N`’s must always be grouped together?Hint
Help VideoCorrect
Great Work!
Incorrect
Permutation with Repetition
$$\frac{\color{#9a00c7}{n}!}{\color{#004ec4}{a}!\color{#004ec4}{b}!\color{#004ec4}{c}!…}$$`n``=`total number of items
`a,b,c``=`count of each repeated itemsSolve three permutations: arrangement of all letters if the `E`’s and `N`’s are each treated as a single letter, arrangement of `E`’s and arrangement of `N`’s, then multiply them.First, treat the `E`’s and `N`’s as a single letter each and list down the letters that are repeated.`D` `P` `D` `C` `I` `E E E E` `N N N``n=7``D` is repeated `2` times`a=2`Apply the formulaPermutation with Repetition `=` $$\frac{\color{#9a00c7}{n}!}{\color{#004ec4}{a}!\color{#004ec4}{b}!\color{#004ec4}{c}!…}$$ `=` $$\frac{\color{#9a00c7}{7}!}{\color{#004ec4}{2}!}$$ Substitute `n` and `a` `=` $$\frac{7\cdot6\cdot5\cdot4\cdot3\cdot\color{#CC0000}{2\cdot1}}{\color{#CC0000}{2\cdot1}}$$ `=` `7*6*5*4*3` Cancel like terms `=` $$2520$$ Hence, there are `2520` ways to arrange `DPDCI`, `E E E E` and `N N N`Next, solve for the permutation of the `E`’s`E E E E``n=4``E` is repeated `4` times`a=4`Apply the formulaPermutation with Repetition `=` $$\frac{\color{#9a00c7}{n}!}{\color{#004ec4}{a}!\color{#004ec4}{b}!\color{#004ec4}{c}!…}$$ `=` $$\frac{\color{#9a00c7}{4}!}{\color{#004ec4}{4}!}$$ Substitute `n` and `a` `=` `1` Cancel like terms Hence, there is only `1` way to arrange `E E E E`Then, solve for the permutation of the `N`’s`N N N``n=3``N` is repeated `3` times`a=3`Apply the formulaPermutation with Repetition `=` $$\frac{\color{#9a00c7}{n}!}{\color{#004ec4}{a}!\color{#004ec4}{b}!\color{#004ec4}{c}!…}$$ `=` $$\frac{\color{#9a00c7}{3}!}{\color{#004ec4}{3}!}$$ Substitute `n` and `a` `=` `1` Cancel like terms Hence, there is only `1` way to arrange `N N N`Finally, multiply the three solved permutationsfirst permutation`=2520`second permutation`=1`third permutation`=1`$$2520\times1\times1$$ `=` `2520` Therefore, there are `2520` ways of arranging the letters `INDEPENDENCE` if the `E`’s and `N`’s must always be together`2520` -
Question 3 of 7
3. Question
A license plate contains `4` letters. What is the probability that a license plate picked at random would have the letters `A,A,B` and `C` in any order?Hint
Help VideoCorrect
Nice Job!
Incorrect
Permutation with Repetition
$$\frac{\color{#9a00c7}{n}!}{\color{#004ec4}{a}!\color{#004ec4}{b}!\color{#004ec4}{c}!…}$$`n``=`total number of items
`a,b,c``=`count of each repeated itemsProbability
$$\frac{\color{#e65021}{\mathsf{favourable\:outcome}}}{\color{#007DDC}{\mathsf{total\:outcome}}}$$Solve the number of ways `A,A,B` and `C` can be arranged and divide it by the total number of ways `4` letters can be arranged using any letters.First, solve for the permutation of `A,A,B` and `C`.`A` `A` `B` `C``n=4``A` is repeated `2` times`a=2`Apply the formulaPermutation with Repetition `=` $$\frac{\color{#9a00c7}{n}!}{\color{#004ec4}{a}!\color{#004ec4}{b}!\color{#004ec4}{c}!…}$$ `=` $$\frac{\color{#9a00c7}{4}!}{\color{#004ec4}{2}!}$$ Substitute `n` and `a` `=` $$\frac{4\cdot3\cdot\color{#CC0000}{2\cdot1}}{\color{#CC0000}{2\cdot1}}$$ `=` `4*3` Cancel like terms `=` $$12$$ Hence, there are `12` ways to arrange `A A BC`. This is the favourable outcomeNext, count the number of letters that can be chosen for each letter in the license plateRemember that the letters can be repeatedFirst to Fourth letter:We can choose any letter from the alphabet and have it repeated. Hence, we have `26` choices for each letter`=``26`Use the Fundamental Counting Principle and multiply each of the options per place value.number of ways `=` `m``times``n` Fundamental Counting Principle `=` `26``times``26``times``26``times``26` `=` `26^4` `=` `456 976` There are `456 976` ways of choosing four letters from the alphabet with repetition.
This is the total outcomeFinally, solve for the probabilityProbability `=` $$\frac{\color{#e65021}{\mathsf{favourable\:outcome}}}{\color{#007DDC}{\mathsf{total\:outcome}}}$$ `=` $$\frac{\color{#e65021}{12}}{\color{#007DDC}{456\:976}}$$ The probability that the license plate will have the letters `A A BC` is `12/(456 976)``12/(456 976)` -
Question 4 of 7
4. Question
Find the probability of having `I` and `E` together in arranging the letters in the word `SHIELD`Hint
Help VideoCorrect
Keep Going!
Incorrect
Permutation Formula
if `(n=r)`$$ _\color{purple}{n}P_{\color{purple}{n}}=\color{purple}{n}! $$Probability
$$\frac{\color{#e65021}{\mathsf{favourable\:outcome}}}{\color{#007DDC}{\mathsf{total\:outcome}}}$$Solve the number of ways `IE` can be together in the word `SHIELD` and divide it by the total arrangements for the letters in `SHIELD`.First, treat `IE` as a single letter. This leaves us with only `5` letters `(r)` to be arranged in `5` positions `(n)``n=r=5`$$_\color{purple}{n}P_{\color{purple}{n}}$$ `=` $$\color{purple}{n}!$$ Permutation Formula (if `n=r`) $$_\color{purple}{5}P_{\color{purple}{5}}$$ `=` $$\color{purple}{5}!$$ Substitute the value of `n` `=` $$5\cdot4\cdot3\cdot2\cdot1$$ `=` $$120$$ There are `120` ways to arrange `SHLD` and `IE`Next, count all possible arrangements for `IE`. This means `2` letters `(r)` are to be arranged in `2` positions `(n)``n=r=2`$$_\color{purple}{n}P_{\color{purple}{n}}$$ `=` $$\color{purple}{n}!$$ Permutation Formula (if `n=r`) $$_\color{purple}{2}P_{\color{purple}{2}}$$ `=` $$\color{purple}{2}!$$ Substitute the value of `n` `=` $$2\cdot1$$ `=` $$2$$ There are `2` ways to arrange `I` and `E`Multiply the two solved permutations`5``P``5``=120``2``P``2``=2`$$120\cdot2$$ `=` $$240$$ Hence, there are `240` ways of arranging `SHIELD` if `IE` must always be together.
This is the favourable outcomeNext, count all arrangements for the letters in `SHIELD`. This means we are arranging `6` letters `(r)` in `6` positions `(n)``n=r=6`$$_\color{purple}{n}P_{\color{purple}{n}}$$ `=` $$\color{purple}{n}!$$ Permutation Formula (if `n=r`) $$_\color{purple}{6}P_{\color{purple}{6}}$$ `=` $$\color{purple}{6}!$$ Substitute the value of `n` `=` $$6\cdot5\cdot4\cdot3\cdot2\cdot1$$ `=` $$720$$ There are `720` ways to arrange the letters in `SHIELD`. This is the total outcomeFinally, solve for the probabilityProbability `=` $$\frac{\color{#e65021}{\mathsf{favourable\:outcome}}}{\color{#007DDC}{\mathsf{total\:outcome}}}$$ `=` $$\frac{\color{#e65021}{240}}{\color{#007DDC}{720}}$$ `=` $$\frac{1}{3}$$ The probability that the letters `IE` will be together in arranging the letters in `SHIELD` is `1/3``1/3` -
Question 5 of 7
5. Question
How many ways can six books be arranged on a shelf if one of them must always be kept at the end?- (120)
Hint
Help VideoCorrect
Excellent!
Incorrect
Use the permutations formula to find the number of ways an item can be arranged `(r)` from the total number of items `(n)`.Remember that order is important in Permutations.Permutation Formula if `(n=r)`
$$ _\color{purple}{n}P_{\color{purple}{n}}=\color{purple}{n}! $$Since one book should be left at the end, we end up just arranging `5` books `(r)` in `5` positions `(n)``n=r=5`$$_\color{purple}{n}P_{\color{purple}{n}}$$ `=` $$\color{purple}{n}!$$ Permutation Formula (if `n=r`) $$_\color{purple}{5}P_{\color{purple}{5}}$$ `=` $$\color{purple}{5}!$$ Substitute the value of `n` `=` $$5\cdot4\cdot3\cdot2\cdot1$$ `=` $$120$$ There are `120` ways to arrange six books on a shelf if one of them must be left at the end.`120` -
Question 6 of 7
6. Question
How many ways can six books be arranged on a shelf if two of them must be kept together?- (240)
Hint
Help VideoCorrect
Fantastic!
Incorrect
Use the permutations formula to find the number of ways an item can be arranged `(r)` from the total number of items `(n)`.Remember that order is important in Permutations.Permutation Formula if `(n=r)`
$$ _\color{purple}{n}P_{\color{purple}{n}}=\color{purple}{n}! $$Solve the number of arrangements if the two books are treated as a single book and the number of arrangements for those two books, then multiply them.First, treat the two books that must be together as a single book. This leaves us with only `5` books `(r)` to be arranged in `5` positions `(n)``n=r=5`$$_\color{purple}{n}P_{\color{purple}{n}}$$ `=` $$\color{purple}{n}!$$ Permutation Formula (if `n=r`) $$_\color{purple}{5}P_{\color{purple}{5}}$$ `=` $$\color{purple}{5}!$$ Substitute the value of `n` `=` $$5\cdot4\cdot3\cdot2\cdot1$$ `=` $$120$$ There are `120` ways to arrange `5` books.Next, count all possible arrangements for the two books. This means `2` books `(r)` are to be arranged in `2` positions `(n)``n=r=2`$$_\color{purple}{n}P_{\color{purple}{n}}$$ `=` $$\color{purple}{n}!$$ Permutation Formula (if `n=r`) $$_\color{purple}{2}P_{\color{purple}{2}}$$ `=` $$\color{purple}{2}!$$ Substitute the value of `n` `=` $$2\cdot1$$ `=` $$2$$ There are `2` ways to arrange two booksFinally, multiply the two solved permutations`5``P``5``=120``2``P``2``=2`$$120\cdot2$$ `=` $$240$$ Therefore, there are `240` ways of arranging six books on a shelf is two of them must always be together`240` -
Question 7 of 7
7. Question
Find the number of ways `4` women and `2` men can be arranged in a straight line, given that the `2` men must be on both ends- (48)
Hint
Help VideoCorrect
Great Work!
Incorrect
Use the permutations formula to find the number of ways an item can be arranged `(r)` from the total number of items `(n)`.Remember that order is important in Permutations.Permutation Formula if `(n=r)`
$$ _\color{purple}{n}P_{\color{purple}{n}}=\color{purple}{n}! $$Solve the number of arrangements for the `4` women and the number of arrangements for the `2` men on both ends, then multiply them.First, arrange `4` women `(r)` in `4` places in the straight line `(n)``n=r=4`$$_\color{purple}{n}P_{\color{purple}{n}}$$ `=` $$\color{purple}{n}!$$ Permutation Formula (if `n=r`) $$_\color{purple}{4}P_{\color{purple}{4}}$$ `=` $$\color{purple}{4}!$$ Substitute the value of `n` `=` $$4\cdot3\cdot2\cdot1$$ `=` $$24$$ There are `24` ways to arrange the `4` women.Next, arrange the `2` men `(r)` on the `2` ends of the line `(n)``n=r=2`$$_\color{purple}{n}P_{\color{purple}{n}}$$ `=` $$\color{purple}{n}!$$ Permutation Formula (if `n=r`) $$_\color{purple}{2}P_{\color{purple}{2}}$$ `=` $$\color{purple}{2}!$$ Substitute the value of `n` `=` $$2\cdot1$$ `=` $$2$$ There are `2` ways to arrange two menFinally, multiply the two solved permutations`4``P``4``=24``2``P``2``=2`$$24\cdot2$$ `=` $$48$$ Therefore, there are `48` ways of arranging `4` women and `2` men if the men are at both ends`48`
Quizzes
- Factorial Notation
- Fundamental Counting Principle 1
- Fundamental Counting Principle 2
- Fundamental Counting Principle 3
- Combinations 1
- Combinations 2
- Combinations with Restrictions 1
- Combinations with Restrictions 2
- Combinations with Probability
- Basic Permutations 1
- Basic Permutations 2
- Basic Permutations 3
- Permutation Problems 1
- Permutation Problems 2
- Permutations with Repetitions 1
- Permutations with Repetitions 2
- Permutations with Restrictions 1
- Permutations with Restrictions 2
- Permutations with Restrictions 3
- Permutations with Restrictions 4