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Permutations with Restrictions 3Permutations with Restrictions 3
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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
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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
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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
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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
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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
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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
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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
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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