Permutations with Repetitions 2

Question 1 of 5

How many ways can the number

$7 372 737$ be arranged?

(105)

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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 items

First, 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 permutation

Permutation 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

How many ways can the number

$9 041 513 194$ be arranged?

1. $151 200$
2. $6700$
3. $3 628 800$
4. $2400$

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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 items

First, 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 permutation

Permutation 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

Eight books are on a shelf.

$4$ are identical novels,

$3$ are identical self-help books and

$1$ is a cookbook. How many ways can these books be arranged?

(280)

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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 items

First, list down the books that are identical.

$4$ novels

$a=4$

$3$ self-help books

$b=3$

The total number of books on the shelf is

$8$ , which means $n=8$

Finally, solve for the permutation

Permutation 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 self-help books and

$1$ is a cookbook

$280$

Question 4 of 5

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)

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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 items

First, 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 permutation

Permutation 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

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)

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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 items

First, 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 permutation

Permutation 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

$4200$

Permutations with Repetitions 2

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1. Question

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Keep Going!

Permutation with Repetitions

2. Question

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Fantastic!

Permutation with Repetitions

3. Question

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Excellent!

Permutation with Repetitions

4. Question

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Nice Job!

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5. Question

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Well Done!

Permutation with Repetitions

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