# Factorials – Example and Practice Problems

Factorials are simply products, indicated by an exclamation point. The factorials indicate that there is a multiplication of all the numbers from 1 to that number. Algebraic expressions with factorials can be simplified by expanding the factorials and looking for common factors.

Here, we will look at a summary of factorials. Also, we will look at examples of factorials and simplification of factorials with answers to understand the reasoning used when solving these types of exercises.

##### ALGEBRA

Relevant for

Solving factorial and factorial simplification problems.

See examples

##### ALGEBRA

Relevant for

Solving factorial and factorial simplification problems.

See examples

## Summary of factorials

We represent the factorials with the exclamation point “!” placed after the number or variable. The exclamation point means that we have to multiply all the integers that are between the number and 1.

For example:

Generally, we read this as “5 factorial”, although we can also read it as “factorial of 5”.

For various reasons, 0! is defined as being equal to 1, not 0, so it is recommended to memorize this.

### Factorial simplification

When we have factorials in both the numerator and denominator, we can easily simplify this by expanding the factorials and simplifying the corresponding numbers.

To simplify factorial expressions with variables in both the numerator and denominator, we want to form common factors so that we can cancel. The bottom line is to compare the factorials and determine which one has a greater value.

For example, if we have the factorials $latex (n + 3)!$ and $latex (n + 1)!$, we easily know that $latex (n + 3)!$ is greater, so we expand it until $latex (n + 1)!$ appears in the sequence to then simplify:

$latex (n+3)!=(n+3)(n+2)(n+1)!$

## Factorials – Examples with answers

The following examples indicate the simplification of factorials. Each exercise has its respective solution, which details the reasoning used to solve the problem. Try to solve the exercises yourself before looking at the solution.

### EXAMPLE 1

Find the result of the factorial 8!.

We can evaluate this by fully expanding the factorial:

$latex 8!=1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7 \times 8$

$latex =40 320$

We can see that we got a large number. Factorials grow rapidly.

### EXAMPLE 2

Simplify the factorial expression $latex \frac{6!}{4!}$.

We have to expand the factorials to simplify:

$latex \frac{6!}{4!}=\frac{1 \times 2\times 3\times 4 \times 5\times 6}{1 \times 2\times 3\times 4}$

We see that the numbers from 1 to 4 are repeated both in the numerator and in the denomination, so we can eliminate them:

$latex \frac{1 \times 2\times 3\times 4 \times 5\times 6}{1 \times 2\times 3\times 4}= 5\times 6$

$latex =30$

### EXAMPLE 3

Simplify the factorial expression $latex \frac{5!}{2!3!}$.

Expanding these factorials, we have:

$latex \frac{5!}{2!3!}=\frac{5\times 4\times 3\times 2\times 1}{(2\times 1)(3\times 2\times 1)}$

We can simplify the numbers from 1 to 3. We can also simplify the 4 with the 2:

$latex \frac{5\times 4\times 3\times 2\times 1}{(2\times 1)(3\times 2\times 1)}=10$

### EXAMPLE 4

Simplify the expression: $latex \frac{17!}{14!3!}$.

We start by expanding the factorials:

$latex \frac{17!}{14!3!}=\frac{1\times 2\times 3\times 4…14\times 15 \times 16 \times 17}{(1\times 2\times 3\times 4 …14)(1\times 2 \times 3)}$

Now, we can simplify the numbers from 1 to 14 that are common to both 14! as for 17 !:

$latex \frac{1\times 2\times 3\times 4…14\times 15 \times 16 \times 17}{(1\times 2\times 3\times 4 …14)(1\times 2 \times 3)}=\frac{ 15 \times 16 \times 17}{1\times 2 \times 3}$

Now, we can simplify to 2 and 3 with 16 and 15:

$latex \frac{ 15 \times 16 \times 17}{1\times 2 \times 3}=5\times 8 \times 17=680$

In this exercise, we use the three dots in the middle. This is useful to simplify the factorial expression a bit.

### EXAMPLE 5

Simplify the expression $latex \frac{n!}{(n-2)!}$.

The factorial expression in the numerator is greater than the factorial expression in the denominator, so we expand n! partially until the expression $latex (n-2)!$ appears:

$latex \frac{n!}{(n-2)!}=\frac{n(n-1)(n-2)!}{(n-2)!}$

Now, we can cancel the common factors:

$latex \frac{n(n-1)(n-2)!}{(n-2)!}=n(n-1)$

We can apply the distributive property to simplify:

$latex n(n-1)={{n}^2}-n$

### EXAMPLE 6

Simplify the expression $latex \frac{(k+1)!}{(k+3)!}$.

In this case, the denominator is clearly greater than the numerator since 3 is being added to n as opposed to the numerator which is only being added by 1.

We keep the numerator unchanged while expanding the denominator until the expression $latex (k + 1)!$ appears:

$latex \frac{(k+1)!}{(k+3)!}= \frac{(k+1)!}{(k+3)(k+2)(k+1)!}$

Now, we cancel the common factors:

$latex \frac{(k+1)!}{(k+3)(k+2)(k+1)!}=\frac{1}{(k+3)(k+2)}$

We multiply the binomials in the denominator to finish:

$latex \frac{1}{(k+3)(k+2)}=\frac{1}{{{k}^2}+5k+6}$

### EXAMPLE 7

Simplify the expression $latex \frac{(n+2)!}{(n-1)!}$.

Here, the numerator is greater since we are adding 2 and in the denominator we are subtracting 1. Therefore, we partially expand the numerator until the expression $latex (n-1)!$ appears:

$latex \frac{(n+2)!}{(n-1)!}= \frac{(n+2)(n+1)(n)(n-1)!}{(n-1)!}$

We simplify $latex (n-1)!$ both in the numerator and in the denominator:

$$\frac{(n+2)(n+1)(n)(n-1)!}{(n-1)!}=(n+2)(n+1)(n)$$

Now, we expand the expression to simplify:

$latex (n+2)(n+1)(n)={{n}^3}+3{{n}^2}+2n$

## Factorials – Practice problems

Use the following problems to test your knowledge of factorials and factorial simplification. Solve the problems, select an answer and check it to verify that you chose the correct one.