Explainer: Factorials

In this explainer, we will learn how to find the factorial of any number 𝑛, which is the product of all integers less than or equal to 𝑛 and greater than or equal to one, and we will learn how to find factorials to solve problems.

When considering the number of different 4-digit numbers that we can form from the digits 3, 5, 7, and 9, we find that there are a total of 4×3×2×1 distinct possible numbers. More generally, if we want to know how many ways we can rearrange a set of 𝑛 items, we find that the total number is 𝑛×(𝑛1)×(𝑛2)××2×1. This calculation, of finding the product of all of the positive integers less than or equal to 𝑛, comes up sufficiently regularly in various areas of mathematics that mathematicians have a given it a name: the factorial of 𝑛.

Definition: Factorial

The factorial of a positive integer 𝑛 is the product of all the positive integers less than or equal to 𝑛. We use the notation 𝑛!, which is read as 𝑛 factorial, to denote the factorial. Hence, 𝑛!=𝑛×(𝑛1)×(𝑛2)××2×1.

We define the factorial of zero to be equal to one; that is, 0!=1.

From the definition, it is not difficult to see that, for any integer 𝑛1, 𝑛!=𝑛(𝑛1)!.

In many ways, this is the key property of the factorial and, as we will see, it will be applied time and again to solve problems involving factorials.

Example 1: Finding Factorials

Evaluate 4!

Answer

Recall that the definition of 𝑛! is the product of all the positive integers less than or equal to 𝑛, or equivalently 𝑛!=𝑛×(𝑛1)×(𝑛2)××2×1.

Hence, 4!=4×3×2×1=24.

Most scientific calculators have a button for calculating the factorial of a number. In examples like the one above, it would be completely legitimate to simply use a calculator to evaluate the expressions. However, this is not always possible. In fact, factorials become large so quickly that most calculators are unable to evaluate factorials of numbers larger than 69. However, this does not mean that we are unable to work with them. Instead, using the properties of factorials will enable us to solve problems which involve numbers that are too large for our calculators.

Example 2: Properties of Factorials

Without using a calculator, evaluate the expression 210(209)!210!.

Answer

Using the following property of the factorial: 𝑛!=𝑛(𝑛1)!, we see that 210(209!)=210!

Hence, 210(209!)210!=210!210!=0.

We can also apply this property when working with expressions involving the factorial of an unknown number.

Example 3: Simplifying Expressions Involving Factorials

Simplify (𝑛+2)!𝑛!.

Answer

Using the following property of factorials: 𝑟!=𝑟(𝑟1)!,

we can rewrite (𝑛+2)!=(𝑛+2)(𝑛+1)!.

Applying the same property again, we can write (𝑛+2)!=(𝑛+2)(𝑛+1)𝑛!.

Substituting this in the given expression we have (𝑛+2)!𝑛!=(𝑛+2)(𝑛+1)𝑛!𝑛!.

Canceling the common terms in the numerator and denominator, we get (𝑛+2)!𝑛!=(𝑛+2)(𝑛+1).

Example 4: Solving Problems Involving Factorials of Zero

Find the solution set of (𝑛26)!=0!.

Answer

We might be tempted to reason as follows: Since (𝑛26)!=0!, 𝑛26=0. Hence, 𝑛=26. Unfortunately, this is not the complete answer. Instead, we need to remember that 0!=1 and zero is not the only number whose factorial is equal to one. In particular, the factorial of one is also equal to one: 1!=1. Hence, to solve (𝑛26)!=0!, we have to consider both cases: where 𝑛26=0 and where 𝑛26=1. Hence, we find that 𝑛=26 and 𝑛=27 are both possible solutions. Therefore, the solution set is {26, 27}.

Example 5: Using the Properties of Factorials to Solve Problems

Find the value of 𝑛 which satisfies the equation (𝑛+48)!(𝑛+47)!=65.

Answer

Using the following property of factorials: 𝑟!=𝑟(𝑟1)!, we can rewrite (𝑛+48)!=(𝑛+48)(𝑛+47)!.

Substituting this into the given equation, we have 6=(𝑛+48)!(𝑛+47)!=(𝑛+48)(𝑛+47)!(𝑛+47)!.

Canceling the common factors in the numerator and denominator, we can rewrite this as 65=𝑛+48.

Rearranging gives 𝑛=17.

Example 6: Solving Factorial Equations

Find the solution set of 1(𝑛+7)!+1(𝑛+8)!=256(𝑛+9)!.

Answer

We begin by multiplying both sides of the equation by (𝑛+9)! which gives (𝑛+9)!(𝑛+7)!+(𝑛+9)!(𝑛+8)!=256.

Using the following property of factorials: 𝑟!=𝑟(𝑟1)!, we can rewrite (𝑛+9)!=(𝑛+9)(𝑛+8)! and (𝑛+9)!=(𝑛+9)(𝑛+8)(𝑛+7)!. Substituting these into the equation in the numerators, we have (𝑛+9)(𝑛+8)(𝑛+7)!(𝑛+7)!+(𝑛+9)(𝑛+8)!(𝑛+8)!=256.

Canceling the common factors in the numerators and denominators, we can rewrite this as (𝑛+9)(𝑛+8)+(𝑛+9)=256.

Expanding the brackets, we have 𝑛+17𝑛+72+𝑛+9=256.

Rearranging, we arrive at the quadratic 𝑛+18𝑛175=0.

By factoring, or applying the quadratic formula, we can express this as (𝑛+25)(𝑛7)=0.

Hence, 𝑛=25 or 𝑛=7. Since factorials are only defined for nonnegative integers, we can discard the solutions 𝑛=25. Hence, the solution set of the equation is {7}.

Thus far, we have been able to use the properties of factorials to simplify equations and isolate any unknowns as solutions to linear or quadratic equations. However, these techniques will not help us when we need to find an unknown number given its factorial. For this, we use the fact that a factorial is the product of positive integers less than or equal to a particular number. Therefore, given a number, we can successively divide by consecutive positive integers until we are left with an integer. The next example will demonstrate this process.

Example 7: Finding an Unknown Number Given its Factorial

Find the value of 𝑛 such that 𝑛!=720.

Answer

Since a factorial is a product of consecutive positive integers, we can divide 720 by consecutive positive integers as follows. Starting with 1: since 7201=720, we can rewrite 720=720×1.

Next we divide 720 by 2 which gives us 7202=360.

Hence, 720=360×2×1.

Dividing 360 by 3 gives 120. Therefore, we can write 720=120×3×2×1.

Similarly, by dividing 120 by 4, we get 30. Hence, 720=30×4×3×2×1.

Finally, dividing 30 by 5, we get 6 which gives 720=6×5×4×3×2×1.

Using this process, we have expressed 720 as a product of the first 6 consecutive integers. Therefore, we have 720=6!. We can write this more succinctly as 720=720×1=360×2×1=120×3×2×1=30×4×3×2×1=6×5×4×3×2×1=6!.

Hence, our final answer is 𝑛=6.

We finish by looking at another example where we can apply all the techniques we have learned to solve one final factorial problem.

Example 8: Problem-Solving with Factorials

Find the value of 𝑛 such that 𝑛(8𝑛1)!=5,040.

Answer

First let us consider the value 5,040. Since we have the product of a factorial and an integer on the left-hand side of the equation, we would like to express 5,040 as a factorial or as the product of a factorial and another integer. To do this, we can divide it consecutively by the natural numbers as follows: 5,040=5,040×1=2,520×2×1=840×3×2×1=210×4×3×2×1=42×5×4×3×2×1=7×6×5×4×3×2×1=7!.

Now we can consider the other side of the equation. Remember that for a positive integer 𝑛𝑛!=𝑛(𝑛1)!.

Currently, we do not have two consecutive numbers 𝑛 and 𝑛1 to apply this formula. However, by multiplying and dividing by 8, we can produce two consecutive numbers 8𝑛 and 8𝑛1 and apply the formula as follows: 𝑛(8𝑛1)!=188𝑛(8𝑛1)!=18(8𝑛)!.

Therefore, 18(8𝑛)!=7!(8𝑛)!=8×7!=8!.

Hence, 𝑛=1.

Key Points

  1. The factorial of a positive integer 𝑛 is defined as the product of all positive integers less than or equal to 𝑛. We write it as 𝑛!.
  2. The key property of the factorial is that 𝑛!=𝑛(𝑛1)!. Using this, we can often simplify expressions involving factorials and solve factorial equations.
  3. When trying to find an unknown integer given its factorial, we divide by consecutive positive integers.

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