x2
Introduction
( )
Types of Arrangements
=
Two More Methods
=
Some Applications

# Section 1.3 – Factorials, a Brief Diversion

In the last section, we had several situations where we started with a number and multiplied all of the integers less than that number together, e.g.

$5 \cdot 4 \cdot 3 \cdot 2 \cdot 1$ or $7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1$

This sort of situation comes up often enough (especially in combinatorics) that has its own notation and a special name.

Factorials

The factorial of a non-negative integer is all of the positive integers less than or equal to it multiplied together. We write a factorial using an !. For example, "five factorial" would be written 5! and is equal to $5 \cdot 4 \cdot 3 \cdot 2 \cdot 1$ or 120.

0! is defined to be equal to 1.

Be sure that you notice the part about 0!. That may seem kind of strange because we're used to thinking of 0 times any number as 0. You can think of this as just a special case that you have to remember. When we get into the next section you'll see why it's important that 0! not be equal to 0.

The table below lists the first nine factorial values.

n n!
0!1
1!1
2!2 · 1 = 2
3!3 · 2 · 1 = 6
4!4 · 3 · 2 · 1 = 24
5!5 · 4 · 3 · 2 · 1 = 120
6!6 · 5 · 4 · 3 · 2 · 1 = 720
7!7 · 6 · 5 · 4 · 3 · 2 · 1 = 5, 040
8!8 · 7 · 6 · 5 · 4 · 3 · 2 · 1 = 40, 320

One thing you should notice from those values is that factorials increase really fast. Even for a value as small as 8, the factorial is already over 40,000. This can make calculating factorial values challenging, even with a calculator. Fortunately for us, most of the situations that we're going to encounter will involve fractions where the numerator and denominator are a factorial. In these situations, we can take advantage of reducing the fraction to make the calculations a little easier. For example, if we needed to calculate

$$\frac{10!}{7!}$$

you could do it by multiplying out the two factorials

$$\frac{10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}$$

but notice how both the numerator and denominator have 7 · 6 · 5 · 4 · 3 · 2 · 1 in them. If we cancel out those factors, we're left with just 10 · 9 · 8 or 720.

# Example 1 – Simplifying a Factorial

Simplify $\frac{100}{96}$.

This is a situation like we saw in the previous paragraph. All of the factors that we get from the 96! are going to cancel out factors that we get from the 100! It would be tedious to try to write out all of the 100 factors in the numerator and the 96 factors in the denominator to see which ones cancel so I'm going to use our factorial notation to take a little shortcut.

$$\frac{100!}{96!} = \frac{100 \cdot 99 \cdot 98 \cdot 97 \cdot 96!}{96!}$$

On the right side, instead of writing out all 100 numbers in the numerator, I just wrote out the first four and then used the factorial notation to represent the last 96. This also helps to make it clear what's going to cancel with the 96! in the denominator. If we cancel out the two 96!'s then what's left will be

$$100 \cdot 99 \cdot 98 \cdot 97 = 94, 109,400$$

# Example 2 – Simplifying a Factorial

Simplify $\frac{18!}{12! \cdot 6!}$.

This kind of expression, where you have one factorial in the numerator and two in the denominator, is going to be a common situation in the following sections. To simplify this, I'll start by eliminating the 12! just like I eliminated the 96! in Example 1.

$$\frac{18 \cdot 17 \cdot 16 \cdot 15 \cdot 14 \cdot 13 \cdot 12!}{12!\cdot6!}$$

$$\frac{18 \cdot 17 \cdot 16 \cdot 15 \cdot 14 \cdot 13 \cdot 12!}{6!}$$

Once you've eliminated one of the factorials from the denominator, you'll have to fall back on writing out the remaining factorial and reducing just like you would with any other fraction.

$$\frac{18 \cdot 17 \cdot 16 \cdot 15 \cdot 14 \cdot 13}{6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}$$

$$\frac{3 \cdot 17 \cdot 16 \cdot 15 \cdot 14 \cdot 13}{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}$$

$$\frac{3 \cdot 17 \cdot 16 \cdot 3 \cdot 14 \cdot 13}{4 \cdot 3 \cdot 2 \cdot 1}$$

$$\frac{3 \cdot 17 \cdot 4 \cdot 3 \cdot 14 \cdot 13}{3 \cdot 2 \cdot 1}$$

$$\frac{17 \cdot 4 \cdot 3 \cdot 14 \cdot 13}{2 \cdot 1}$$

$$\frac{17 \cdot 2 \cdot 3 \cdot 14 \cdot 13}{1}$$

$$18, 564$$

I wrote out all of the reducing in a lot more detail than I normally would've, just so that you could see what I was doing.

# Exercises

### Exercises

= show solution = get hint

1. 10!
2. 15!
3. $\frac{12!}{5!}$
4. $\frac{42!}{41!}$
5. $\frac{6!}{0!}$
6. $\frac{12!}{10!}$
7. $\frac{99!}{92!}$
8. $\frac{16!}{12! \cdot 4!}$
9. $\frac{82!}{3! \cdot 7!}$
10. $\frac{12!}{6! \cdot 6!}$
11. $\frac{77!}{70! \cdot 7!}$
12. $\frac{20!}{12! \cdot 18!}$

### Analysis

1. An elementary school class has 18 students. How many ways can those students line up to go to lunch? How would you write your answer using a factorial?
2. If you wanted to graph the factorials from 0 through 5, what would the graph look like? What kind of scale would you use on the y-axis?

### Technical Writing

1. When you're reducing an expression like the one in exercise 9 above, why is it best to cancel out the 79! first and then simplify the 3!?