Explainer: Properties of Permutations

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

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. Notice that, for permutations,

  • order matters: BCA is not the same as BAC;
  • we do not allow repetition: AAB is not a valid permutation of ABC.

That is to say that a permutation represents counting without replacement in which order matters.

If we have a set of 𝑛 items, the total number of permutations is given by 𝑛×(𝑛1)×(𝑛2)××2×1. Using factorial notation, we can write this as 𝑛!.

We would like to generalize this idea to counting the number of permutations of length 𝑟 taken from a set of 𝑛, which is equivalent to the question of how many ways we can order 𝑟 elements from a set of 𝑛 elements with no repetition. We will denote this number 𝑃.

We begin by considering the number of options we have for each position. For the first position, we have 𝑛 choices, then for the second position, we have 𝑛1 because we have one less item to choose from. Similarly, for the third position, we have a choice of 𝑛2. We can continue this pattern until we have a sequence of length 𝑟.

First PositionSecond PositionThird Position(𝑟1)th Position𝑟th Position
𝑛 choices𝑛1 choices 𝑛2 choices𝑛(𝑟2) choices𝑛(𝑟1) choices

Applying the fundamental counting principle, the total number of distinct ways we can order 𝑟 elements from a set of 𝑛 is given by the product 𝑃=𝑛×(𝑛1)×(𝑛2)××(𝑛(𝑟2))×(𝑛(𝑟1)).

If we multiply this by (𝑛𝑟)!(𝑛𝑟)!, we can rewrite it as 𝑃=𝑛(𝑛1)(𝑛2)(𝑛(𝑟2))(𝑛(𝑟1))(𝑛𝑟)!(𝑛𝑟)!.

Repeatedly applying the property of factorials that 𝑛!=𝑛(𝑛1)!, we can simplify this expression to 𝑃=𝑛!(𝑛𝑟)!.

This derivation leads us to the general definition.

Definition: Permutations

The number of ways we can order 𝑟 elements from a set of 𝑛 elements with no repetition is given by 𝑃 (read as 𝑛-𝑃-𝑟) which is defined as 𝑃=𝑛!(𝑛𝑟)!. The permutations 𝑃 are sometimes referred to as 𝑟-permutations of 𝑛, variations, partial permutations, or sequences without repetition. Various forms of notation are used for 𝑃, the most common of which are 𝑃, 𝑃, 𝑃 and 𝑃(𝑛,𝑟).

Notice that when 𝑛=𝑟, we have 𝑃=𝑛!(𝑛𝑛)!=𝑛!0!.

Recall that, by definition, the factorial of zero is 1. Hence, 𝑃=𝑛!, which is exactly what we expect for the number of distinct permutations of 𝑛 items.

In this explainer, we will develop fluency in working with this definition and learn to solve equations including 𝑃. We will start with a simple example where we use the formula to evaluate 𝑃.

Example 1: Evaluating Permutations

Evaluate 𝑃.

Answer

By definition, 𝑃=𝑛!(𝑛𝑟)!.

In this case, we have 𝑛=9 and 𝑟=5. Therefore, 𝑃=9!(95)!=9!4!.

Expanding the factorial using the property that 𝑛!=𝑛(𝑛1)!, we can rewrite this as 𝑃=9×8×7×6×5×4!4!.

Canceling the 4! from the numerator and denominator gives =9×8×7×6×5=15,120.

Most scientific calculators have a function for evaluating permutations. So when asked to evaluate a permutation, oftentimes we can simply use a calculator to find the value. However, this is not always the case. Permutations get large very quickly so some calculators may not be able to handle numbers that large. For this reason, learning to use and manipulate the formula for permutations will be of great use. The next example will demonstrate a case where most calculators will fail to evaluate the numbers. However, by using the formula and the properties of factorials, we can evaluate the expression without the need for a calculator.

Example 2: Ratios of Permutations

Evaluate 𝑃𝑃.

Answer

Recall the definition of permutations: 𝑃=𝑛!(𝑛𝑟)!.

Using this definition, we can write 𝑃=74!(7468)!=74!6!,𝑃=73!(7367)!=73!6!.

From here, we have two options. We could either substitute the two equations into the given expression and simplify or we could use the property of the factorial that 𝑛!=𝑛(𝑛1)! to rewrite 𝑃=74×73!6!=74(𝑃).

Then, we can substitute this into the expression to get 𝑃𝑃=74(𝑃)𝑃=74.

In the previous example, we found that 𝑃=74(𝑃). This is actually an example of a general property of permutations that 𝑃=𝑛(𝑃).

Using this property of permutations, we can simplify expressions as the next example will demonstrate.

Example 3: Using the Properties of Permutations to Solve Equations

If 𝑃=23(𝑃), find 𝑛.

Answer

Using the property of permutations that 𝑃=𝑛(𝑃), we can rewrite 𝑃=𝑛(𝑃).

Substituting this back into the equation, we have 𝑛(𝑃)=23(𝑃). Canceling 𝑃 from both sides, we are left with 𝑛=23 as required.

We now turn our attention to a class of problems where we are given the value of 𝑛 in 𝑃, but the unknown is in the 𝑟 position. Through looking at a few examples, we will highlight the techniques required to solve this class of problems.

Example 4: Finding Unknowns in the 𝑟 Position

Find the value of 𝑚 such that 𝑃=4,080.

Answer

Remember that the number of permutations, 𝑃, of size 𝑟 taken from a set of size 𝑛 is given by 𝑃=𝑛!(𝑛𝑟)!.

Hence, we can write 𝑃=17!(17𝑚)!.

Expanding 17! by successively using the property that 𝑛!=𝑛(𝑛1)!, we can rewrite this as 𝑃=17×16××(17𝑚+1)×(17𝑚)!(17𝑚)!.

Canceling the common factor of (17𝑚)!, we have

𝑃=17×16××(17𝑚+1).(1)

Notice that this is a product of consecutive integers starting at 17 and progressively getting smaller. Hence, we would like to express 4,080 as a product of decreasing consecutive integers starting at 17. We can do this by dividing 4,080 by 17, then by 16, then by each successive smaller integer. Starting with 17, we have 4,08017=240.

Hence, 4,080=17×240.

We now divide 240 by 16 which gives 15. Hence, 4,080=17×16×15, which is a product of successive integers. Therefore, we have expressed 𝑃 as a product of consecutive integers as 𝑃=17×16×15.

Comparing this with equation (1), we can equate the smallest integers to get 17𝑚+1=15. By rearranging, we find that 𝑚=3.

We can double check that we have the correct answer by using a calculator to confirm that 𝑃=4,080.

Example 5: Solving Permutation Equations for Unknowns in the 𝑟 Position

Solve the following equation for 𝑟: 𝑃=120.

Answer

Recall the definition of permutations: 𝑃=𝑛!(𝑛𝑟)!. Using this, we can write 𝑃=5!(5𝑟)!.

Using the property that 𝑛!=𝑛(𝑛1)!, we can expand 5! to get 𝑃=5×4××(5𝑟+1)×(5𝑟)!(5𝑟)!.

Canceling the common factor of (5𝑟)!, we have that

𝑃=5×4××(5𝑟+1),(2)

which is a product of consecutive decreasing integers starting with 5. We would, therefore, like to express 120 in the same form. We can do this by dividing 120 by 5, then 4, then each consecutive smaller integer as follows: 120=5×24=5×4×6=5×4×3×2=5×4×3×2×1.

Hence, we have 𝑃=5×4×3×2×1.

If we compare this with equation (2) and try to equate the smallest integers, we find that the smallest integer could be either 2 or 1. Hence, we have two possible cases, when 5𝑟+1=2, or when 5𝑟+1=1. Rearranging, we get that 𝑟=4 or 𝑟=5.

We can double check both of these values by using a calculator to confirm that 𝑃=120 and 𝑃=120.

In the next couple of examples, we use the definition of 𝑃 and the properties of factorials to solve permutation problems by simplifying the equations to a linear or quadratic form.

Example 6: Solving Permutation Equations

If (𝑥47)!×𝑃=3,906(𝑥2)!, find the value of 𝑥.

Answer

Recall that 𝑃=𝑛!𝑛𝑟!. Using this, we can write 𝑃=(𝑥)!(𝑥47)!.

Substituting this into the given equation, we get (𝑥47)!(𝑥)!(𝑥47)!=3,906(𝑥2)!.

Canceling common factors gives us (𝑥)!=3,906(𝑥2)!.

Using the property of factorials that 𝑛!=𝑛(𝑛1)!, we can rewrite (𝑥)!=𝑥(𝑥1)!=𝑥(𝑥1)(𝑥2)!.

Substituting this back into our equation gives 𝑥(𝑥1)(𝑥2)!=3,906(𝑥2)!.

Dividing both sides of the equation by (𝑥2)! yields 𝑥(𝑥1)=3,906.

Rearranging, we get the following quadratic equation: 𝑥𝑥3,906=0.

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

Example 7: Solving Permutation Equations

Find the value of 𝑥 given that 𝑃3𝑥𝑃=0.

Answer

Using the definition of 𝑃, we rewrite the given equation as 235!(235𝑥)!3𝑥235!(235(𝑥1))!=0.

Dividing by 235!, we can rewrite this as 1(235𝑥)!3𝑥1(235𝑥+1)!=0.

Multiplying by (235𝑥+1)!, we get (235𝑥+1)!(235𝑥)!3𝑥=0.

Using the fact that (235𝑥+1)!=(235𝑥+1)(235𝑥)!, we rewrite this as (235𝑥+1)(235𝑥)!(235𝑥)!3𝑥=0.

Canceling the common of (235𝑥)! from the numerator and denominator gives (235𝑥+1)3𝑥=0.

By gathering like terms and rearranging, we get 4𝑥=236.

Dividing by 4, we have 𝑥=59.

In the final example, we look at the case where we are given the value of 𝑟 in 𝑃, but the unknown is in the 𝑛 position. This case is less obvious as to how to solve it. However, we will use an example to show a technique that can be applied in cases like this.

Example 8: Solving Permutation Equations for Unknowns in the 𝑛 Position

Find the value of 𝑛 such that 𝑃=32,736.

Answer

Remember that the number of permutations, 𝑃, of size 𝑟 taken from a set of size 𝑛 is given by 𝑃=𝑛!(𝑛𝑟)!.

Therefore, using the property of factorials that 𝑛!=𝑛(𝑛1)!, we can write 𝑃=𝑛!(𝑛3)!=𝑛(𝑛1)(𝑛2)(𝑛3)!(𝑛3)!=𝑛(𝑛1)(𝑛2).

Therefore, we are looking for three consecutive integers whose product equals 32,736. There are two ways to do this, for smaller numbers, we can consider the prime factors of the number and then rearrange them to form three consecutive numbers. This can work for larger numbers too, as we will show. However, in general, this process could prove increasingly difficult for larger and larger numbers. However, there is a useful trick we can use: taking the cube root of 32,736 gives us a number that when multiplied by itself twice gives 32,736. Therefore, the value of this number will be close to the value of the three consecutive numbers we are looking for.

For 32,736, we have 32,736=31.9895.

We therefore divide 32,736 by the two integers on either side of this value: 31 and 32. When we do this we find 32,736=31×1,056=31×32×33, which is the product of three successive integers as required. Hence, we see that 𝑛=33.

As mentioned above, we can consider the prime factors of 32,736 and try to rearrange them into three consecutive integers. For 32,736, we find that one of the prime factors is 31; hence, we are able to easily organize the other factors into two other consecutive integers as the calculation below demonstrates: 𝑛(𝑛1)(𝑛2)=32,736=2×3×11×31=(3×11)×2×31=33×32×31, from which we can conclude 𝑛=33.

Key Points

  1. The number of permutations of size 𝑟 taken from a set of size 𝑛 is given by 𝑃=𝑛!(𝑛𝑟)!.
  2. Some of the key techniques to help solve problems involving 𝑃 are as follows:
    • Using the property of factorials that 𝑛!=𝑛(𝑛1)!
    • Using the property of permutations that 𝑃=𝑛𝑃
    • We often need to divide a number by consecutive integers. If we are given 𝑛 in 𝑃, we often divide by consecutive decreasing integers starting from 𝑛. However, if we are given 𝑟 in 𝑃, we know the number of consecutive integers but not where to start. Therefore, we can either consider prime factors or use the integers on either side of the 𝑛th root as a starting point.

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