Lesson Explainer: Properties of Permutations Mathematics

In this explainer, we will learn how to use the properties of permutations to simplify expressions and solve equations.

We recall that a permutation is a rearrangement of a collection of items. For example, say we have the letters A, B, and C. We can arrange them as ABC, BCA, BAC, and so forth. Each different arrangement is an example of a permutation. The way to count the number of arrangements of ๐‘› items into ๐‘› different positions is ๐‘›. In the example of arrangements of ABC, this would mean 3=6 possibilities.

In addition to full permutations, we can also consider partial permutations. For instance, suppose we wanted to consider taking two letters out of A, B, and C and seeing how we could arrange them. We could have AB, BA, AC, and so on. Once again, there would be six possibilities, although the formula we use would be different.

Recall that when it comes to counting the number of ways we can order ๐‘Ÿ distinct elements from a set of ๐‘› elements, we have the following formula.

Definition: ๐‘Ÿ-Permutations of ๐‘›

The number of ways we can order ๐‘Ÿ elements from a set of ๐‘› elements is given by ๏Š๏Ž๐‘ƒ (read as ๐‘›-๐‘-๐‘Ÿ or ๐‘Ÿ-permutations of ๐‘›), which is defined as ๏Š๏Ž๐‘ƒ=๐‘›๐‘›โˆ’๐‘Ÿ, where ๐‘›=๐‘›ร—(๐‘›โˆ’1)ร—(๐‘›โˆ’2)ร—โ‹ฏร—2ร—1.

In this explainer, we will solely be dealing with problems involving partial permutations, meaning they will involve ๏Š๏Ž๐‘ƒ. We note that most calculators have a function for directly evaluating ๏Š๏Ž๐‘ƒ, so oftentimes we can simply use one to find the value if needed. However, this is not always the case. Permutations get large very quickly, so some calculators may not be able to handle certain evaluations. For this reason, learning to use and manipulate the above formula will be of great use.

Let us start by considering some basic properties of partial permutations that we can employ.

Properties: Partial Permutations

For ๐‘Ÿ-permutations of ๐‘›, where ๐‘›โ‰ฅ1, we have the following properties:

  1. ๏Š๏Šฆ๐‘ƒ=1
  2. ๏Š๏Šง๐‘ƒ=๐‘›
  3. ๏Š๏Š๐‘ƒ=๐‘›
  4. ๏Š๏Š๏Šฑ๏Šง๐‘ƒ=๐‘›

All of these properties can be obtained by applying the definition of ๏Š๏Ž๐‘ƒ. Let us prove each case in turn.

For (i), we take ๐‘Ÿ=0 in the formula to get ๏Š๏Šฆ๐‘ƒ=๐‘›๐‘›โˆ’0=๐‘›๐‘›=1.

For (ii), we take ๐‘Ÿ=1 in the formula and make use of the property ๐‘›=๐‘›๐‘›โˆ’1, giving us ๏Š๏Šง๐‘ƒ=๐‘›๐‘›โˆ’1=๐‘›๐‘›โˆ’1๐‘›โˆ’1=๐‘›.

For (iii), we take ๐‘Ÿ=๐‘› in the formula, resulting in ๏Š๏Š๐‘ƒ=๐‘›๐‘›โˆ’๐‘›=๐‘›0=๐‘›.

Finally, for (iv), we take ๐‘Ÿ=๐‘›โˆ’1, giving us ๏Š๏Š๏Šฑ๏Šง๐‘ƒ=๐‘›๐‘›โˆ’(๐‘›โˆ’1)=๐‘›1=๐‘›.

While these properties are fairly elementary to derive, it is worth keeping them in mind since we can use them to take shortcuts in doing calculations. Additionally, they can sometimes be used to solve problems that involve special cases of permutations, as we will see in the next example.

Example 1: Using the Properties of Permutations to Find the Value of an Unknown

Find ๐‘š such that ๏Šช๏Šฆ๏‰๏Šฐ๏Šง๏Šซ๏Šง๏Šง๏‰๏Šฐ๏Šง๏Šซ๐‘ƒ=๐‘ƒ.


The given equation is expressed in terms of partial permutations, which are defined by ๏Š๏Ž๐‘ƒ=๐‘›๐‘›โˆ’๐‘Ÿ.

Since the values of ๐‘› are different for each side of the equation, it may not seem at first as though there is any way to make the two sides equal. However, let us recall one of the properties of permutations: ๏Š๏Šฆ๐‘ƒ=1.

That is, if ๐‘Ÿ=0, then both sides of the equation will just be equal to 1. Thus, we can consider what values of ๐‘š cause ๐‘Ÿ to be zero. That is, ๐‘š+15=0๐‘š=โˆ’15.

Another technique we should be aware of is how we can relate permutations with slightly different input values. Suppose we have ๏Šซ๏Šฉ๐‘ƒ, which is defined to be ๏Šซ๏Šฉ๐‘ƒ=55โˆ’3=5ร—4ร—3ร—2ร—12ร—1.

Notice that if we take the 5 out of the numerator, we can get a different number of permutations. That is, ๏Šซ๏Šฉ๏Šช๏Šจ๐‘ƒ=5๏€ผ4ร—3ร—2ร—12ร—1๏ˆ=5๏€ฟ44โˆ’2๏‹=5๏€น๐‘ƒ๏….

So, we have shown that ๏Šซ๏Šฉ๏Šช๏Šจ๐‘ƒ=5๏€น๐‘ƒ๏…. This is actually a general property that can be applied to any ๐‘Ÿ-permutations of ๐‘›, as we will describe below.

Property: Related Permutations (Part 1)

For ๐‘Ÿ-permutations of ๐‘›, where ๐‘›โ‰ฅ๐‘Ÿโ‰ฅ1, we have the following property: ๏Š๏Ž๏Š๏Šฑ๏Šง๏Ž๏Šฑ๏Šง๐‘ƒ=๐‘›๏€น๐‘ƒ๏….

This can be proved using the fact that ๐‘›=๐‘›๐‘›โˆ’1 as follows: ๏Š๏Ž๏Š๏Šฑ๏Šง๏Ž๏Šฑ๏Šง๐‘ƒ=๐‘›๐‘›โˆ’๐‘Ÿ=๐‘›๐‘›โˆ’1๐‘›โˆ’๐‘Ÿ=๐‘›๐‘›โˆ’1(๐‘›โˆ’1)โˆ’(๐‘Ÿโˆ’1)=๐‘›๏€น๐‘ƒ๏….

In the next example, we will examine a use of this property to find a missing unknown.

Example 2: Using the Properties of Permutations to Solve Equations

If ๏Š๏Šง๏Šซ๏Š๏Šฑ๏Šง๏Šง๏Šช๐‘ƒ=23๏€น๐‘ƒ๏…, find ๐‘›.


Looking closely at the expression we have been given, we can see that the ๏Š๏Ž๐‘ƒ terms on both sides are closely related. Namely, going from the left side to the right side, ๐‘› goes to ๐‘›โˆ’1 and 15 goes to 14. This suggests to us that we can use the property of related permutations: ๏Š๏Ž๏Š๏Šฑ๏Šง๏Ž๏Šฑ๏Šง๐‘ƒ=๐‘›๏€น๐‘ƒ๏….

Indeed, if we take ๐‘Ÿ=15, we can see that this is almost exactly the same equation as the one given in the question, except with ๐‘›=23 filled out on the right-hand side. Thus, we can see that ๐‘› must be 23.

Our first example was relatively quick to solve by virtue of the related permutation property, but often things will not be so simple. Let us explore how we might have to be a bit more creative with our solutions.

Example 3: Evaluating Permutations to Find the Values of Unknowns

Find the solution set of the equation 240๐‘ƒ=๐‘ƒ๏—๏Šฐ๏Šจ๏Šจ๏—๏Šฐ๏Šช๏Šช.


As a starting point to finding the solution set, we can compare ๏—๏Šฐ๏Šจ๏Šจ๐‘ƒ and ๏—๏Šฐ๏Šช๏Šช๐‘ƒ on either side of the equation. We can see that the indices on the left have both been decreased by 2 compared to the right. Recall that we have the following property for relating ๏Ž๏Š๐‘ƒ expressions with similar indices: ๏Š๏Ž๏Š๏Šฑ๏Šง๏Ž๏Šฑ๏Šง๐‘ƒ=๐‘›๏€น๐‘ƒ๏….

Although this only applies to decreasing the indices of ๏Ž๏Š๐‘ƒ by one, there is nothing to stop us applying it multiple times in succession. If we do this once more, then we get ๏Š๏Ž๏Š๏Šฑ๏Šจ๏Ž๏Šฑ๏Šจ๐‘ƒ=๐‘›(๐‘›โˆ’1)๏€น๐‘ƒ๏….

If we apply this formula to the right-hand side of the given equation, we get 240๐‘ƒ=๐‘ƒ=(๐‘ฅ+4)(๐‘ฅ+3)๐‘ƒ.๏—๏Šฐ๏Šจ๏Šจ๏—๏Šฐ๏Šช๏Šช๏—๏Šฐ๏Šจ๏Šจ

Then, dividing by ๏—๏Šฐ๏Šจ๏Šจ๐‘ƒ on each side gives us 240=(๐‘ฅ+4)(๐‘ฅ+3).

As this is a quadratic equation, we can solve it for ๐‘ฅ by expanding the parentheses and rearranging everything to one side, before factoring it: (๐‘ฅ+4)(๐‘ฅ+3)=240๐‘ฅ+4๐‘ฅ+3๐‘ฅ+12โˆ’240=0๐‘ฅ+7๐‘ฅโˆ’228=0(๐‘ฅโˆ’12)(๐‘ฅ+19)=0.๏Šจ๏Šจ

We have factored this by noting that 12 and 19 are factors of 228, but note that this can also be solved using the quadratic equation or by completing the square if it is easier.

Finally, we can solve this equation by setting the factors to zero, giving us ๐‘ฅโˆ’12=0๐‘ฅ+19=0๐‘ฅ=12,๐‘ฅ=โˆ’19.

Out of these two solutions, note that only the positive one is valid, since we require ๐‘ฅ+2 to be nonnegative for ๏—๏Šฐ๏Šจ๏Šจ๐‘ƒ to be a valid expression. Thus, the solution set is {12}.

As we saw in the previous example, the property of related permutations can be extended further by applying it multiple times in succession. In addition to this, there are also various other properties that relate similar permutations, which we will consider below

Properties: Related Permutations (Part 2)

For ๐‘Ÿ-permutations of ๐‘›, where ๐‘›โ‰ฅ๐‘Ÿโ‰ฅ1, we have the following properties:

  1. ๏Š๏Ž๏Š๏Šฑ๏Šง๏Ž๏Šฑ๏Šง๐‘ƒ=๐‘›๏€น๐‘ƒ๏…
  2. ๏Š๏Ž๏Š๏Šฑ๏Šง๏Ž๏Š๏Šฑ๏Šง๏Ž๏Šฑ๏Šง๐‘ƒ=๐‘ƒ+๐‘Ÿ๏€น๐‘ƒ๏…
  3. ๏Š๏Ž๏Š๏Ž๏Šฑ๏Šง๐‘ƒ๐‘ƒ=๐‘›โˆ’๐‘Ÿ+1

As discussed, the first property is just an extension of the property we had already stated.

For (ii), this equality can be proved by considering the right-hand side and simplifying it: ๏Š๏Šฑ๏Šง๏Ž๏Š๏Šฑ๏Šง๏Ž๏Šฑ๏Šง๐‘ƒ+๐‘Ÿ๏€น๐‘ƒ๏…=๐‘›โˆ’1๐‘›โˆ’1โˆ’๐‘Ÿ+๐‘Ÿ๐‘›โˆ’1๐‘›โˆ’1โˆ’(๐‘Ÿโˆ’1)=๐‘›โˆ’1๐‘›โˆ’๐‘Ÿโˆ’1+๐‘Ÿ๐‘›โˆ’1๐‘›โˆ’๐‘Ÿ.

To add the two fractions, we need to give them the same denominator. We can do this by multiplying the numerator and denominator of the first fraction by (๐‘›โˆ’๐‘Ÿ) and using the property of factorials that ๐‘›โˆ’๐‘Ÿ=(๐‘›โˆ’๐‘Ÿ)๐‘›โˆ’๐‘Ÿโˆ’1: ๏Š๏Šฑ๏Šง๏Ž๏Š๏Šฑ๏Šง๏Ž๏Šฑ๏Šง๏Š๏Ž๐‘ƒ+๐‘Ÿ๏€น๐‘ƒ๏…=(๐‘›โˆ’๐‘Ÿ)๐‘›โˆ’1(๐‘›โˆ’๐‘Ÿ)๐‘›โˆ’๐‘Ÿโˆ’1+๐‘Ÿ๐‘›โˆ’1๐‘›โˆ’๐‘Ÿ=(๐‘›โˆ’๐‘Ÿ)๐‘›โˆ’1+๐‘Ÿ๐‘›โˆ’1๐‘›โˆ’๐‘Ÿ=๐‘›๐‘›โˆ’1๐‘›โˆ’๐‘Ÿ=๐‘›๐‘›โˆ’๐‘Ÿ=๐‘ƒ.

For property (iii), we can use the properties of factorials to simplify: ๏Š๏Ž๏Š๏Ž๏Šฑ๏Šง๐‘ƒ๐‘ƒ=๐‘›๐‘›โˆ’๐‘Ÿรท๐‘›๐‘›โˆ’(๐‘Ÿโˆ’1)=๐‘›๐‘›โˆ’๐‘Ÿ+1๐‘›โˆ’๐‘Ÿ๐‘›=(๐‘›โˆ’๐‘Ÿ+1)๐‘›โˆ’๐‘Ÿ๐‘›โˆ’๐‘Ÿ=๐‘›โˆ’๐‘Ÿ+1.

We should keep all of these formulas in mind, although we should note that, for many problems, it may be more convenient to directly use the definition of ๏Š๏Ž๐‘ƒ and the properties of factorials to solve the problem.

In particular, the main property of factorials that we will continue to make thorough use of is ๐‘›=๐‘›๐‘›โˆ’1.

When in doubt, we can always try rewriting things in terms of factorials and seeing if it simplifies things. Let us see an example of this.

Example 4: Finding Unknowns by Considering the Ratio between Two Permutations

If ๏Šจ๏Š๏Šฐ๏Šง๏Šฌ๏Šจ๏Š๏Šฑ๏Šง๏Šซ๐‘ƒโˆถ๐‘ƒ=272โˆถ11, find ๐‘›.


To solve this problem, the best way to start is by recalling that ๏Š๏Ž๐‘ƒ=๐‘›๐‘›โˆ’๐‘Ÿ, so we have ๏Šจ๏Š๏Šฐ๏Šง๏Šฌ๐‘ƒ=2๐‘›+12๐‘›+1โˆ’6=2๐‘›+12๐‘›โˆ’5, and we have ๏Šจ๏Š๏Šฑ๏Šง๏Šซ๐‘ƒ=2๐‘›โˆ’12๐‘›+1โˆ’5=2๐‘›โˆ’12๐‘›โˆ’4.

We are interested in the ratio between these expressions, which means we need to compare the two. To do this, we can make use of the property of factorials that ๐‘›=๐‘›๐‘›โˆ’1 (and by extension, ๐‘›=๐‘›(๐‘›โˆ’1)๐‘›โˆ’2). Using this, we can rewrite the first expression as follows: ๏Šจ๏Š๏Šฐ๏Šง๏Šฌ๏Šจ๏Š๏Šฑ๏Šง๏Šซ๐‘ƒ=2๐‘›+12๐‘›โˆ’5=(2๐‘›+1)2๐‘›2๐‘›โˆ’1(2๐‘›โˆ’5)2๐‘›โˆ’6=(2๐‘›+1)2๐‘›2๐‘›โˆ’5๏€น๐‘ƒ๏….

Therefore, by considering the given ratio of the two expressions, we have (2๐‘›+1)2๐‘›2๐‘›โˆ’5=27211.

This may appear quite daunting to solve; however, since ๐‘› has to be a positive integer, this limits the possibilities we have to consider. In fact, if we consider the possibility that the numerators and denominators of these fractions are equal, then we just have 2๐‘›โˆ’5=11,(2๐‘›+1)2๐‘›=272.

Solving the first equation by rearranging, we get ๐‘›=8, which can be substituted into the second equation to confirm that it satisfies both the denominator and the numerator.

Finally, we can calculate ๐‘›=8 using a calculator, giving us 40โ€Žโ€‰โ€Ž320.

In the next couple of examples, we will continue using the definition of ๏Š๏Ž๐‘ƒ and the properties of factorials to solve permutation problems by simplifying the equations to a linear or quadratic form.

Example 5: Solving Permutation Equations

If ๐‘ฅโˆ’47ร—๐‘ƒ=3906๐‘ฅโˆ’2๏—๏Šช๏Šญ, find the value of ๐‘ฅ.


Recall that ๏Š๏Ž๐‘ƒ=๐‘›๐‘›โˆ’๐‘Ÿ. Using this, we can write ๏—๏Šช๏Šญ๐‘ƒ=๐‘ฅ๐‘ฅโˆ’47.

Substituting this into the given equation, we get ๐‘ฅโˆ’47๐‘ฅ๐‘ฅโˆ’47=3906๐‘ฅโˆ’2.

Canceling common factors gives us ๐‘ฅ=3906๐‘ฅโˆ’2.

Using the property of factorials that ๐‘›=๐‘›๐‘›โˆ’1, we can rewrite ๐‘ฅ=๐‘ฅ๐‘ฅโˆ’1=๐‘ฅ(๐‘ฅโˆ’1)๐‘ฅโˆ’2.

Substituting this back into our equation gives ๐‘ฅ(๐‘ฅโˆ’1)๐‘ฅโˆ’2=3906๐‘ฅโˆ’2.

Dividing both sides of the equation by ๐‘ฅโˆ’2 yields ๐‘ฅ(๐‘ฅโˆ’1)=3906.

Rearranging, we get the following quadratic equation: ๐‘ฅโˆ’๐‘ฅโˆ’3906=0.๏Šจ

Using the quadratic formula or factoring, we can solve this to find that ๐‘ฅ=63 or ๐‘ฅ=โˆ’62. Since factorials and permutations are only defined for positive integers, we disregard the ๐‘ฅ=โˆ’62 solution. Thus, ๐‘ฅ=63.

Example 6: Solving Permutation Equations

Find the value of ๐‘ฅ given that ๏Šจ๏Šฉ๏Šซ๏—๏Šจ๏Šฉ๏Šซ๏—๏Šฑ๏Šง๐‘ƒโˆ’3๐‘ฅ๐‘ƒ=0.


Using the definition of ๏Š๏Ž๐‘ƒ, we rewrite the given equation as 235235โˆ’๐‘ฅโˆ’3๐‘ฅ235235โˆ’(๐‘ฅโˆ’1)=0.

Dividing by 235, we can rewrite this as 1235โˆ’๐‘ฅโˆ’3๐‘ฅ1235โˆ’๐‘ฅ+1=0.

Multiplying by 235โˆ’๐‘ฅ+1, we get 235โˆ’๐‘ฅ+1235โˆ’๐‘ฅโˆ’3๐‘ฅ=0.

Using the fact that 235โˆ’๐‘ฅ+1=(235โˆ’๐‘ฅ+1)235โˆ’๐‘ฅ, we rewrite this as (235โˆ’๐‘ฅ+1)235โˆ’๐‘ฅ235โˆ’๐‘ฅโˆ’3๐‘ฅ=0.

Canceling the common term 235โˆ’๐‘ฅ from the numerator and denominator gives (235โˆ’๐‘ฅ+1)โˆ’3๐‘ฅ=0.

By gathering like terms and rearranging, we get 4๐‘ฅ=236.

Dividing by 4, we have ๐‘ฅ=59.

Let us finish by summarizing the main things we have learned in this explainer.

Key Points

  • To solve equations involving the number of permutations of ๐‘Ÿ elements of a set with ๐‘› elements, it is often helpful to rewrite instances of ๏Š๏Ž๐‘ƒ using the formula ๏Š๏Ž๐‘ƒ=๐‘›๐‘›โˆ’๐‘Ÿ.
  • We can use the following properties of permutations to help us solve problems:
    1. ๏Š๏Šฆ๐‘ƒ=1
    2. ๏Š๏Šง๐‘ƒ=๐‘›
    3. ๏Š๏Š๐‘ƒ=๐‘›
    4. ๏Š๏Š๏Šฑ๏Šง๐‘ƒ=๐‘›
    5. ๏Š๏Ž๏Š๏Šฑ๏Šง๏Ž๏Šฑ๏Šง๐‘ƒ=๐‘›๏€น๐‘ƒ๏…
    6. ๏Š๏Ž๏Š๏Šฑ๏Šง๏Ž๏Š๏Šฑ๏Šง๏Ž๏Šฑ๏Šง๐‘ƒ=๐‘ƒ+๐‘Ÿ๏€น๐‘ƒ๏…
    7. ๏Š๏Ž๏Š๏Ž๏Šฑ๏Šง๐‘ƒ๐‘ƒ=๐‘›โˆ’๐‘Ÿ+1
  • The most important general rule we can use is the property of factorials that ๐‘›=๐‘›๐‘›โˆ’1.

Download the Nagwa Classes App

Attend sessions, chat with your teacher and class, and access class-specific questions. Download the Nagwa Classes app today!

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.