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 items into different positions is . In the example of arrangements of ABC, this would mean 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 distinct elements from a set of elements, we have the following formula.

### Definition: 𝑟-Permutations of 𝑛

The number of ways we can order elements from a set of elements is given by (read as -- or -permutations of ), which is defined as where .

In this explainer, we will solely be dealing with problems involving
*partial permutations*, meaning they will involve
. We note that most calculators have a
function for directly evaluating ,
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 -permutations of , where , we have the following properties:

All of these properties can be obtained by applying the definition of . Let us prove each case in turn.

For (i), we take in the formula to get

For (ii), we take in the formula and make use of the property , giving us

For (iii), we take in the formula, resulting in

Finally, for (iv), we take , giving us

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 such that .

### Answer

The given equation is expressed in terms of partial permutations, which are defined by

Since the values of 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:

That is, if , then both sides of the equation will just be equal to 1. Thus, we can consider what values of cause to be zero. That is,

Another technique we should be aware of is how we can relate permutations with slightly different input values. Suppose we have , which is defined to be

Notice that if we take the 5 out of the numerator, we can get a different number of permutations. That is,

So, we have shown that . This is actually a general property that can be applied to any -permutations of , as we will describe below.

### Property: Related Permutations (Part 1)

For -permutations of , where , we have the following property:

This can be proved using the fact that as follows:

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 , find .

### Answer

Looking closely at the expression we have been given, we can see that the terms on both sides are closely related. Namely, going from the left side to the right side, goes to and 15 goes to 14. This suggests to us that we can use the property of related permutations:

Indeed, if we take , we can see that this is almost exactly the same equation as the one given in the question, except with filled out on the right-hand side. Thus, we can see that 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 .

### Answer

As a starting point to finding the solution set, we can compare and 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 expressions with similar indices:

Although this only applies to decreasing the indices of by one, there is nothing to stop us applying it multiple times in succession. If we do this once more, then we get

If we apply this formula to the right-hand side of the given equation, we get

Then, dividing by on each side gives us

As this is a quadratic equation, we can solve it for by expanding the parentheses and rearranging everything to one side, before factoring it:

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

Out of these two solutions, note that only the positive one is valid, since we require to be nonnegative for to be a valid expression. Thus, the solution set is .

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 -permutations of , where , we have the following properties:

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:

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 and using the property of factorials that :

For property (iii), we can use the properties of factorials to simplify:

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 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

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 , find .

### Answer

To solve this problem, the best way to start is by recalling that , so we have and we have

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 (and by extension, ). Using this, we can rewrite the first expression as follows:

Therefore, by considering the given ratio of the two expressions, we have

This may appear quite daunting to solve; however, since 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

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

Finally, we can calculate using a calculator, giving us 40 320.

In the next couple of examples, we will continue using the definition of 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 , find the value of .

### Answer

Recall that . Using this, we can write

Substituting this into the given equation, we get

Canceling common factors gives us

Using the property of factorials that , we can rewrite

Substituting this back into our equation gives

Dividing both sides of the equation by yields

Rearranging, we get the following quadratic equation:

Using the quadratic formula or factoring, we can solve this to find that or . Since factorials and permutations are only defined for positive integers, we disregard the solution. Thus, .

### Example 6: Solving Permutation Equations

Find the value of given that .

### Answer

Using the definition of , we rewrite the given equation as

Dividing by , we can rewrite this as

Multiplying by , we get

Using the fact that , we rewrite this as

Canceling the common term from the numerator and denominator gives

By gathering like terms and rearranging, we get

Dividing by 4, we have .

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 elements of a set with elements, it is often helpful to rewrite instances of using the formula
- We can use the following properties of permutations
to help us solve problems:
- The most important general rule we can use is the property of factorials that