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Permutations with Repetitions 1Permutations with Repetitions 1
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Question 1 of 5
1. Question
Find the number of ways the letters in `DAD` can be arranged if each letter is:`(i)` distinguishable `(D``A``D``)``(ii)` indistinguishable `(DAD)`-
`(i)` (6)`(ii)` (3)
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
$$ _\color{purple}{n}P_{\color{green}{r}}=\frac{\color{purple}{n}!}{(\color{purple}{n}-\color{green}{r})!} $$Permutation with Repetitions
$$\frac{\color{#9a00c7}{n}!}{\color{#004ec4}{a}!\color{#004ec4}{b}!\color{#004ec4}{c}!…}$$`n``=`total number of items
`a,b,c``=`count of each repeated items`(i)` Distinguishable letters `(D``A``D``)`We need to arrange `3` letters `(r)` into `3` positions `(n)``r=3``n=3`$$_\color{purple}{n}P_{\color{green}{r}}$$ `=` $$\frac{\color{purple}{n}!}{(\color{purple}{n}-\color{green}{r})!}$$ Permutation Formula $$_\color{purple}{3}P_{\color{green}{3}}$$ `=` $$\frac{\color{purple}{3}!}{(\color{purple}{3}-\color{green}{3})!}$$ Substitute the values of `n` and `r` `=` $$\frac{3!}{0!}$$ `=` $$3\cdot2\cdot1$$ `0! =1` `=` $$6$$ There are `6` ways to arrange `D``A``D``D``A``D` `D``D``A` `A``D``D` `D``A``D` `D``D``A` `A``D``D` `(ii)` Indistinguishable letters `(DAD)`First, list down the letters that are repeated.`DAD``D` is repeated `2` times`a=2`The total number of letters in `DAD` is `3`, which means `n=3`Finally, solve for the permutationPermutation with Repetitions `=` $$\frac{\color{#9a00c7}{n}!}{\color{#004ec4}{a}!\color{#004ec4}{b}!\color{#004ec4}{c}!…}$$ `=` $$\frac{\color{#9a00c7}{3}!}{\color{#004ec4}{2}!}$$ Substitute `n` and `a` `=` $$\frac{3\cdot\color{#CC0000}{2\cdot1}}{\color{#CC0000}{2\cdot1}}$$ `=` $$3$$ Cancel like terms There are `3` ways to arrange `DAD``DAD` `DDA` `ADD` `(i) 6``(ii) 3` -
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Question 2 of 5
2. Question
Find the number of ways the letters in the following words can be arranged:`(i) W E A R``(i) W E R E`-
`(i)` (24)`(ii)` (12)
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
$$ _\color{purple}{n}P_{\color{green}{r}}=\frac{\color{purple}{n}!}{(\color{purple}{n}-\color{green}{r})!} $$Permutation with Repetitions
$$\frac{\color{#9a00c7}{n}!}{\color{#004ec4}{a}!\color{#004ec4}{b}!\color{#004ec4}{c}!…}$$`n``=`total number of items
`a,b,c``=`count of each repeated items`(i) W E A R`We need to arrange `4` letters `(r)` into `4` positions `(n)``r=4``n=4`$$_\color{purple}{n}P_{\color{green}{r}}$$ `=` $$\frac{\color{purple}{n}!}{(\color{purple}{n}-\color{green}{r})!}$$ Permutation Formula $$_\color{purple}{4}P_{\color{green}{4}}$$ `=` $$\frac{\color{purple}{4}!}{(\color{purple}{4}-\color{green}{4})!}$$ Substitute the values of `n` and `r` `=` $$\frac{4!}{0!}$$ `=` $$4\cdot3\cdot2\cdot1$$ `0! =1` `=` $$24$$ There are `24` ways to arrange `W E A R``(ii) WERE`First, list down the letters that are repeated.`E` is repeated `2` times`a=2`The total number of letters in `W E R E` is `4`, which means `n=4`Finally, solve for the permutationPermutation with Repetitions `=` $$\frac{\color{#9a00c7}{n}!}{\color{#004ec4}{a}!\color{#004ec4}{b}!\color{#004ec4}{c}!…}$$ `=` $$\frac{\color{#9a00c7}{4}!}{\color{#004ec4}{2}!}$$ Substitute `n` and `a` `=` $$\frac{4\cdot3\cdot\color{#CC0000}{2\cdot1}}{\color{#CC0000}{2\cdot1}}$$ `=` $$12$$ Cancel like terms and evaluate There are `12` ways to arrange `W E R E``(i) 24``(ii) 12` -
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Question 3 of 5
3. Question
How many arrangements can be made with the letters in `O B S E S S E D`?- (3360)
Hint
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Permutation with Repetitions
$$\frac{\color{#9a00c7}{n}!}{\color{#004ec4}{a}!\color{#004ec4}{b}!\color{#004ec4}{c}!…}$$`n``=`total number of items
`a,b,c``=`count of each repeated itemsFirst, list down the letters that are repeated.`O B S E S S E D``S` is repeated `3` times`a=3``E` is repeated `2` times`b=2`The total number of letters in `O B S E S S E D` is `8`, which means `n=8`Finally, solve for the permutationPermutation with Repetitions `=` $$\frac{\color{#9a00c7}{n}!}{\color{#004ec4}{a}!\color{#004ec4}{b}!\color{#004ec4}{c}!…}$$ `=` $$\frac{\color{#9a00c7}{8}!}{\color{#004ec4}{3}!\color{#004ec4}{2}!}$$ Substitute `n,a` and `b` `=` $$\frac{8\cdot7\cdot\color{#CC0000}{6}\cdot5\cdot4\cdot3\cdot\color{#CC0000}{2\cdot1}}{\color{#CC0000}{3\cdot2\cdot1\cdot2\cdot1}}$$ `=` $$3360$$ Cancel like terms and evaluate Therefore, there are `3360` ways to arrange `O B S E S S E D``3360` -
Question 4 of 5
4. Question
How many arrangements can be made with the letters in `A C C E L E R A T E`?Hint
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Permutation with Repetitions
$$\frac{\color{#9a00c7}{n}!}{\color{#004ec4}{a}!\color{#004ec4}{b}!\color{#004ec4}{c}!…}$$`n``=`total number of items
`a,b,c``=`count of each repeated itemsFirst, list down the letters that are repeated.`A C C E L E R A T E``A` is repeated `2` times`a=2``C` is repeated `2` times`b=2``E` is repeated `3` times`c=3`The total number of letters in `A C C E L E R A T E` is `10`, which means `n=10`Finally, solve for the permutationPermutation with Repetitions `=` $$\frac{\color{#9a00c7}{n}!}{\color{#004ec4}{a}!\color{#004ec4}{b}!\color{#004ec4}{c}!…}$$ `=` $$\frac{\color{#9a00c7}{10}!}{\color{#004ec4}{2}!\color{#004ec4}{2}!\color{#004ec4}{3}!}$$ Substitute `n,a,b` and `c` `=` $$\frac{10\cdot9\cdot8\cdot7\cdot6\cdot5\cdot\color{#CC0000}{4\cdot3\cdot2\cdot1}}{\color{#CC0000}{2\cdot1\cdot2\cdot1\cdot3\cdot2\cdot1}}$$ `=` `151 200` Cancel like terms and evaluate Therefore, there are `151 200` ways to arrange `A C C E L E R A T E``151 200` -
Question 5 of 5
5. Question
How many arrangements can be made with the letters in `E N E R G E T I C` if `IC` must always be together?Hint
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Permutation with Repetitions
$$\frac{\color{#9a00c7}{n}!}{\color{#004ec4}{a}!\color{#004ec4}{b}!\color{#004ec4}{c}!…}$$`n``=`total number of items
`a,b,c``=`count of each repeated itemsFirst, list down the letters that are repeated.`E N E R G E T I C``E` is repeated `3` times`a=3`The total number of letters in `E N E R G E T``I C` is `8`, if `I C` is considered as one. This means `n=8`Solve for the permutationPermutation with Repetitions `=` $$\frac{\color{#9a00c7}{n}!}{\color{#004ec4}{a}!\color{#004ec4}{b}!\color{#004ec4}{c}!…}$$ `=` $$\frac{\color{#9a00c7}{8}!}{\color{#004ec4}{3}!}$$ Substitute `n,a,b` and `c` `=` $$\frac{8\cdot7\cdot6\cdot5\cdot4\color{#CC0000}{\cdot3\cdot2\cdot1}}{\color{#CC0000}{3\cdot2\cdot1}}$$ `=` `6720` Cancel like terms and evaluate Remember that `IC` must always be together. Find the number of ways these `2` letters can be arranged`n=2`$$_\color{purple}{n}P_{\color{purple}{n}}$$ `=` $$n!$$ Permutation Formula if `(n=r)` $$_\color{purple}{2}P_{\color{purple}{2}}$$ `=` $$ 2! $$ Substitute the value of `n` `=` $$2\cdot1$$ `=` $$2$$ There are `2` ways of arranging `I C`. Multiply this to the numerator of the Permutation with RepetitionsFinally, multiply the two solved permutationsPermutation for `E N E R G E T``I C``=6720`Permutation for `I C``=2``6720times2` `=` `13 440` There are `13 440` ways of arranging `E N E R G E T I C` if `IC` must always be together`13 440`
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