Permutations – Example and Practice Problems

With permutations, we can count the number of different ways of choosing objects from a set if the order of the objects does matter. This is different from combinations, where the order of the objects does not matter.

Here, we will start with a summary of permutations and look at their formula. Then, we will see several examples with answers to understand the application of the permutations formula.

ALGEBRA
examples of permutations

Relevant for

Learning about permutations with solved examples.

See examples

ALGEBRA
examples of permutations

Relevant for

Learning about permutations with solved examples.

See examples

Summary of permutations

A permutation is a list of objects, in which the order is important. Permutations are used when we are counting without replacing objects and order does matter. If the order doesn’t matter, we use combinations.

In general P(n, k) means the number of permutations of n objects from which we take k objects. Alternatively, the permutations formula is expressed as follows:

$latex _{n}{{P}_{k}}=\frac{{n!}}{{( {n-k} )!}}$

where:

  • n represents the total number of elements in a set
  • k represents the number of selected objects
  • ! is the factorial symbol

To solve permutations problems, we have to remember that the factorial (denoted as “!”) is equal to the product of all positive integers less than or equal to the number preceding the factorial. For example, $latex 4!=1 \times 2 \times 3 \times 4 = 24$.


Permutations – Examples with answers

In the following examples, we will see the application of the permutations formula. Each example has its respective detailed solution, which can be used to understand the reasoning in the answer to each exercise.

EXAMPLE 1

Find the result of the permutation $latex _{7}P_{3}$.

We have to use the permutations formula $latex _{n}{{P}_{k}}=\frac{{n!}}{{(n-k)!}}$, and substitute $latex n=7$ and $latex k=3$:

$latex _{7}{{P}_{3}}=\frac{{7!}}{{(7-3)!}}$

$latex =\frac{{7!}}{{(4)!}}$

We can simplify this by writing to 7! as $latex 7\times 6\times 5\times 4!$:

$latex\frac{{7!}}{{(4)!}}=\frac{{7\times 6\times 5\times 4!}}{{(4)!}}$

$latex =7\times 6\times 5=210$

EXAMPLE 2

Find the result of the permutation $latex _{8}P_{7}$.

We use the permutations formula $latex _{n}{{P}_{k}}=\frac{{n!}}{{(n-k)!}}$ using the values $latex n=8$ and $latex k=7$:

$latex _{8}{{P}_{7}}=\frac{{8!}}{{(8-7)!}}$

$latex =\frac{{8!}}{{(1)!}}$

In this case, we have nothing to simplify, so we have to calculate 8!:

$latex\frac{{8!}}{{(1)!}}=8!$

$latex =40 320$

EXAMPLE 3

Find the number of permutations $latex _{9}P_{6}$.

In this case, we have the values $latex n=9$ and $latex k=6$:

$latex _{9}{{P}_{9}}=\frac{{9!}}{{(9-6)!}}$

$latex =\frac{{9!}}{{(3)!}}$

We recognize that we can write 9! as $latex 9\times 8\times 7\times 6\times 5\times 4\times 3!$.Therefore, we simplify the 3!:

$latex\frac{{9!}}{{(3)!}}=\frac{{9\times 8\times 7\times 6\times 5\times 4\times 3!}}{{(3)!}}$

$latex =9\times 8\times 7\times 6\times 5\times 4=60480$

EXAMPLE 4

How many permutations are there with 4 objects and 2 places?

We can recognize the values $latex n=4$ and $latex k=2$. Therefore, we use the formula substituting these values:

$latex _{4}{{P}_{2}}=\frac{{4!}}{{(4-2)!}}$

$latex =\frac{{4!}}{{(2)!}}$

We rewrite the 4! as $latex 4\times 3\times 2!$ and we simplify:

$latex\frac{{4!}}{{(2)!}}=\frac{{4\times 3\times 2!}}{{(2)!}}$

$latex =4\times 3=12$

EXAMPLE 5

In how many ways can a president, treasurer, and secretary be chosen from 7 candidates?

The problem involves 7 candidates, of which we chose 3. Therefore, we have the values $latex n=7$ and $latex k=3$:

$latex _{7}{{P}_{3}}=\frac{{7!}}{{(7-3)!}}$

$latex =\frac{{7!}}{{(4)!}}$

Now, we write 7! as $latex 7\times 6\times 5\times 4!$. Then, we simplify the 4!:

$latex\frac{{7!}}{{(4)!}}=\frac{{7\times 6\times 5\times 4!}}{{(4)!}}$

$latex =7\times 6\times 5=210$

EXAMPLE 6

In how many ways can the first 3 places be awarded in a race with 5 participants?

We recognize the values $latex n=5$ and $latex k=3$:

$latex _{5}{{P}_{3}}=\frac{{5!}}{{(5-3)!}}$

$latex =\frac{{5!}}{{(2)!}}$

We rewrite the factorial 5! as $latex 5 \times 4 \times 3 \times 2!$. Then, we simplify the 2!:

$latex\frac{{5!}}{{(2)!}}=\frac{{5\times 4\times 3\times 2!}}{{(2)!}}$

$latex =5\times 4\times 3=60$


Permutations – Practice problems

Practice and test your knowledge of permutations. Select an answer and check it to see if you chose the correct answer. If you need help, you can look at the solved examples above.

How many permutations are there with 6 objects and 2 places?

Choose an answer






How many permutations are there with 5 objects and 3 places?

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In how many ways can the positions of president and vice president be assigned from a group of 8 people?

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Calculate the permutation $latex_{10}P_{2}$.

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Calcula la permutación $latex_{12}P_{3}$.

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See also

Interested in learning more about factorials, permutations, and combinations? Take a look at these pages:

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