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Permutations with Repetitions>
Permutations with Repetitions 2Permutations with Repetitions 2
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Question 1 of 5
1. Question
How many ways can the number `7 372 737` be arranged? (105)
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 digits that are repeated.`7 372 737``7` is repeated `4` times`a=4``3` is repeated `2` times`b=2`The total number of digits in `7 372 737` is `7`, which means `n=7`Finally, solve for the permutationPermutation with Repetitions `=` $$\frac{\color{#9a00c7}{n}!}{\color{#004ec4}{a}!\color{#004ec4}{b}!\color{#004ec4}{c}!…}$$ `=` $$\frac{\color{#9a00c7}{7}!}{\color{#004ec4}{4}!\color{#004ec4}{2}!}$$ Substitute `n,a` and `b` `=` $$\frac{7\cdot6\cdot5\cdot\color{#CC0000}{4}\cdot\color{#CC0000}{3}\cdot\color{#CC0000}{2}\cdot\color{#CC0000}{1}}{\color{#CC0000}{4}\cdot\color{#CC0000}{3}\cdot\color{#CC0000}{2}\cdot\color{#CC0000}{1}\cdot2\cdot1}$$ `=` $$\frac{210}{2}$$ Cancel like terms `=` $$105$$ Therefore, there are `105` ways to arrange `7 372 737``105` 
Question 2 of 5
2. Question
How many ways can the number `9 041 513 194` be arranged?Hint
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Fantastic!
Incorrect
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 digits that are repeated.`9 041 513 194``9` is repeated `2` times`a=2``4` is repeated `2` times`b=2``1` is repeated `3` times`c=3`The total number of digits in `9 041 513 194` 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` `=` $$\frac{10\cdot9\cdot8\cdot7\cdot6\cdot5\cdot\color{#CC0000}{4}\cdot\color{#CC0000}{3}\cdot\color{#CC0000}{2}\cdot\color{#CC0000}{1}}{\color{#CC0000}{2}\cdot1\cdot\color{#CC0000}{2}\cdot1\cdot\color{#CC0000}{3}\cdot\color{#CC0000}{2}\cdot\color{#CC0000}{1}}$$ `=` $$\frac{151200}{1}$$ Cancel like terms `=` $$151200$$ Therefore, there are `151 200` ways to arrange `9 041 513 194``151 200` 
Question 3 of 5
3. Question
Eight books are on a shelf. `4` are identical novels, `3` are identical selfhelp books and `1` is a cookbook. How many ways can these books be arranged? (280)
Hint
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Incorrect
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 books that are identical.`4` novels`a=4``3` selfhelp books`b=3`The total number of books on the shelf is `8`, which means `n=8`Finally, solve for the permutationPermutation with Repetitions `=` $$\frac{\color{#9a00c7}{n}!}{\color{#004ec4}{a}!\color{#004ec4}{b}!\color{#004ec4}{c}!…}$$ `=` $$\frac{\color{#9a00c7}{8}!}{\color{#004ec4}{4}!\color{#004ec4}{3}!}$$ Substitute `n,a` and `b` `=` $$\frac{8\cdot7\cdot\color{#CC0000}{6}\cdot5\cdot\color{#CC0000}{4\cdot3\cdot2\cdot1}}{\color{#CC0000}{3\cdot2\cdot1\cdot4\cdot3\cdot2\cdot1}}$$ `=` $$280$$ Cancel like terms and evaluate Therefore, there are `280` ways to arrange eight books if `4` are identical novels, `3` are identical selfhelp books and `1` is a cookbook`280` 
Question 4 of 5
4. Question
Ten marbles are in a jar. `5` of them are blue, `3` are black and `2` are red. How many ways can these marbles be arranged if they are lined up straight on a table? (2520)
Hint
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Nice Job!
Incorrect
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 marbles that are identical.`5` blue marbles`a=5``3` black marbles`b=3``2` red marbles`c=2`The total number of marbles is `10`, which means `n=10`Finally, solve for the permutationPermutation with Repetitions `=` $$\frac{\color{#9a00c7}{n}!}{\color{#004ec4}{a}!\color{#004ec4}{b}!\color{#004ec4}{c}!…}$$ `=` $$\frac{\color{#9a00c7}{10}!}{\color{#004ec4}{5}!\color{#004ec4}{3}!\color{#004ec4}{2}!}$$ Substitute `n,a,b` and `c` `=` $$\frac{10\cdot9\cdot8\cdot7\cdot\color{#CC0000}{6\cdot5\cdot4\cdot3\cdot2\cdot1}}{\color{#CC0000}{5\cdot4\cdot3\cdot2\cdot1\cdot3\cdot2\cdot1}\cdot2\cdot1}$$ `=` $$\frac{5040}{2}$$ Cancel like terms and evaluate `=` $$2520$$ Therefore, there are `2520` ways to arrange ten marbles if `5` are blue, `3` are black and `2` are red`2520` 
Question 5 of 5
5. Question
An electronics shop has `4` laptops, `3` monitors and `3` printers. The owner wants to know how many arrangements can be made with these gadgets. (4200)
Hint
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Well Done!
Incorrect
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 gadgets that are identical.`4` laptops`a=4``3` monitors`b=3``3` printers`c=3`The total number of the gadgets is `10`, which means `n=10`Finally, solve for the permutationPermutation with Repetitions `=` $$\frac{\color{#9a00c7}{n}!}{\color{#004ec4}{a}!\color{#004ec4}{b}!\color{#004ec4}{c}!…}$$ `=` $$\frac{\color{#9a00c7}{10}!}{\color{#004ec4}{4}!\color{#004ec4}{3}!\color{#004ec4}{3}!}$$ Substitute `n,a,b` and `c` `=` $$\frac{10\cdot9\cdot8\cdot7\cdot\color{#CC0000}{6}\cdot5\color{#CC0000}{\cdot4\cdot3\cdot2\cdot1}}{\color{#CC0000}{4\cdot3\cdot2\cdot1\cdot3\cdot2\cdot1}\cdot3\cdot2\cdot1}$$ `=` $$\frac{25200}{6}$$ Cancel like terms and evaluate `=` $$4200$$ Therefore, there are `4200` ways to arrange ten gadgets if `4` are laptops, `3` are monitors and `3` are printers`4200`
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