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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 7377 372 737 be arranged?- (105)
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Permutation with Repetitions
n!a!b!c!…n!a!b!c!…nn==total number of items
a,b,ca,b,c==count of each repeated itemsFirst, list down the digits that are repeated.7 372 7377 372 73777 is repeated 44 timesa=4a=433 is repeated 22 timesb=2b=2The total number of digits in 7 372 7377 372 737 is 77, which means n=7n=7Finally, solve for the permutationPermutation with Repetitions == n!a!b!c!…n!a!b!c!… == 7!4!2!7!4!2! Substitute n,an,a and bb == 7⋅6⋅5⋅4⋅3⋅2⋅14⋅3⋅2⋅1⋅2⋅17⋅6⋅5⋅4⋅3⋅2⋅14⋅3⋅2⋅1⋅2⋅1 == 21022102 Cancel like terms == 105105 Therefore, there are 105105 ways to arrange 7 372 7377 372 737105105 -
Question 2 of 5
2. Question
How many ways can the number 9 041 513 1949 041 513 194 be arranged?- 1.
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Permutation with Repetitions
n!a!b!c!…n!a!b!c!…nn==total number of items
a,b,ca,b,c==count of each repeated itemsFirst, list down the digits that are repeated.9 041 513 1949 041 513 19499 is repeated 22 timesa=2a=244 is repeated 22 timesb=2b=211 is repeated 33 timesc=3c=3The total number of digits in 9 041 513 1949 041 513 194 is 1010, which means n=10n=10Finally, solve for the permutationPermutation with Repetitions == n!a!b!c!…n!a!b!c!… == 10!2!2!3!10!2!2!3! Substitute n,a,bn,a,b and cc == 10⋅9⋅8⋅7⋅6⋅5⋅4⋅3⋅2⋅12⋅1⋅2⋅1⋅3⋅2⋅110⋅9⋅8⋅7⋅6⋅5⋅4⋅3⋅2⋅12⋅1⋅2⋅1⋅3⋅2⋅1 == 15120011512001 Cancel like terms == 151200151200 Therefore, there are 151 200151 200 ways to arrange 9 041 513 1949 041 513 194151 200151 200 -
Question 3 of 5
3. Question
Eight books are on a shelf. 44 are identical novels, 33 are identical self-help books and 11 is a cookbook. How many ways can these books be arranged?- (280)
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Permutation with Repetitions
n!a!b!c!…n!a!b!c!…nn==total number of items
a,b,ca,b,c==count of each repeated itemsFirst, list down the books that are identical.44 novelsa=4a=433 self-help booksb=3b=3The total number of books on the shelf is 88, which means n=8n=8Finally, solve for the permutationPermutation with Repetitions == n!a!b!c!…n!a!b!c!… == 8!4!3!8!4!3! Substitute n,an,a and bb == 8⋅7⋅6⋅5⋅4⋅3⋅2⋅13⋅2⋅1⋅4⋅3⋅2⋅18⋅7⋅6⋅5⋅4⋅3⋅2⋅13⋅2⋅1⋅4⋅3⋅2⋅1 == 280280 Cancel like terms and evaluate Therefore, there are 280280 ways to arrange eight books if 44 are identical novels, 33 are identical self-help books and 11 is a cookbook280280 -
Question 4 of 5
4. Question
Ten marbles are in a jar. 55 of them are blue, 33 are black and 22 are red. How many ways can these marbles be arranged if they are lined up straight on a table?- (2520)
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Chapters- Chapters
Permutation with Repetitions
n!a!b!c!…n!a!b!c!…nn==total number of items
a,b,ca,b,c==count of each repeated itemsFirst, list down the marbles that are identical.55 blue marblesa=5a=533 black marblesb=3b=322 red marblesc=2c=2The total number of marbles is 1010, which means n=10n=10Finally, solve for the permutationPermutation with Repetitions == n!a!b!c!…n!a!b!c!… == 10!5!3!2!10!5!3!2! Substitute n,a,bn,a,b and cc == 10⋅9⋅8⋅7⋅6⋅5⋅4⋅3⋅2⋅15⋅4⋅3⋅2⋅1⋅3⋅2⋅1⋅2⋅110⋅9⋅8⋅7⋅6⋅5⋅4⋅3⋅2⋅15⋅4⋅3⋅2⋅1⋅3⋅2⋅1⋅2⋅1 == 5040250402 Cancel like terms and evaluate == 25202520 Therefore, there are 25202520 ways to arrange ten marbles if 55 are blue, 33 are black and 22 are red25202520 -
Question 5 of 5
5. Question
An electronics shop has 44 laptops, 33 monitors and 33 printers. The owner wants to know how many arrangements can be made with these gadgets.- (4200)
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Need TextPlayCurrent Time 0:00/Duration Time 0:00Remaining Time -0:00Stream TypeLIVELoaded: 0%Progress: 0%0:00Fullscreen00:00MutePlayback Rate1x- 2x
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- 1x
- 0.75x
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Subtitles- subtitles off
Captions- captions off
- English
Chapters- Chapters
Permutation with Repetitions
n!a!b!c!…n!a!b!c!…nn==total number of items
a,b,ca,b,c==count of each repeated itemsFirst, list down the gadgets that are identical.44 laptopsa=4a=433 monitorsb=3b=333 printersc=3c=3The total number of the gadgets is 1010, which means n=10n=10Finally, solve for the permutationPermutation with Repetitions == n!a!b!c!…n!a!b!c!… == 10!4!3!3!10!4!3!3! Substitute n,a,bn,a,b and cc == 10⋅9⋅8⋅7⋅6⋅5⋅4⋅3⋅2⋅14⋅3⋅2⋅1⋅3⋅2⋅1⋅3⋅2⋅110⋅9⋅8⋅7⋅6⋅5⋅4⋅3⋅2⋅14⋅3⋅2⋅1⋅3⋅2⋅1⋅3⋅2⋅1 == 252006252006 Cancel like terms and evaluate == 42004200 Therefore, there are 42004200 ways to arrange ten gadgets if 44 are laptops, 33 are monitors and 33 are printers42004200
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