Factorials are just a way to write the multiplication of positive integers using special notation, since multiplication of sequential integers shows up a lot in calculus.
For example, the idea is to write something like \( 1 \cdot 2 \cdot 3 \cdot 4 \) using special, more compact notation. Many times \(1\) is not included in the multiplication since it doesn't add anything, i.e. \( 1 \cdot 2 \cdot 3 \cdot 4 = 2 \cdot 3 \cdot 4 \).
We write this as \(1 \cdot 2 \cdot 3 \cdot 4 = 4!\), i.e. we multiply all positive integers starting with \(1\) up to and including whatever number appears with the exclamation point. So,
\(5! = 1 \cdot 2 \cdot 3 \cdot 4 \cdot 5\).
Since multiplication is commutative, we can also write the numbers in reverse order, like this
\(5! = 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1\).
The second way is sometimes easier to write since you start with the number with the exclamation point and go down.
Okay, let's work some practice problems.
Practice
Instructions  Unless otherwise instructed, determine the number these factorials represent.
Determine the number \(7!\) represents.
Problem Statement 

Determine the number \(7!\) represents.
Solution 

video by MIP4U 

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Determine the number \(4(5!)\) represents.
Problem Statement 

Determine the number \(4(5!)\) represents.
Solution 

video by PatrickJMT 

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Determine the number \(\displaystyle{\frac{5}{6!}}\) represents.
Problem Statement 

Determine the number \(\displaystyle{\frac{5}{6!}}\) represents.
Final Answer 

\(\displaystyle{\frac{5}{6!} = \frac{1}{144}}\)
Problem Statement 

Determine the number \(\displaystyle{\frac{5}{6!}}\) represents.
Solution 

\(\begin{array}{rcl} \displaystyle{\frac{5}{6!}} & = & \displaystyle{\frac{5}{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6}} \\ & = & \displaystyle{\frac{1}{1 \cdot 2 \cdot 3 \cdot 4 \cdot 6}} \\ & = & \displaystyle{\frac{1}{144}} \end{array} \)
Final Answer 

\(\displaystyle{\frac{5}{6!} = \frac{1}{144}}\) 
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Determine the number \(\displaystyle{6!\frac{(73)!}{3!}}\) represents.
Problem Statement 

Determine the number \(\displaystyle{6!\frac{(73)!}{3!}}\) represents.
Solution 

video by PatrickJMT 

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Building Factorials
The beauty of this, and something that will be important in calculus, is that factorials build on one another. So
\(
\begin{array}{rcl}
5! & = & 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 \\
& = & 5 \cdot ( 4 \cdot 3 \cdot 2 \cdot 1) \\
& = & 5 \cdot (4!)
\end{array}
\)
This is important since in calculus we will often have fractions with factorials in both the numerator and denominator and we need to cancel terms to simplify. Let's do an example.
Example 1
Simplify \(5! / 4! \)
\(\displaystyle{
\frac{5!}{4!} = \frac{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{ 4 \cdot 3 \cdot 2 \cdot 1} = 5
}\)
Notice that the terms, \( 1, 2, 3, 4 \) in both the numerator and denominator cancel leaving \(5\) in the numerator and \(1\) in the denominator.
video by PatrickJMT 

Okay, let's practice this concept on these problems.
Practice
Unless otherwise instructed, simplify these factorials.
Simplify the factorial \(\displaystyle{\frac{8!}{6!}}\).
Problem Statement 

Simplify the factorial \(\displaystyle{\frac{8!}{6!}}\).
Solution 

video by MIP4U 

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Simplify the factorial \(\displaystyle{\frac{10!}{7!3!}}\).
Problem Statement 

Simplify the factorial \(\displaystyle{\frac{10!}{7!3!}}\).
Solution 

video by MIP4U 

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Simplify the factorial \(\displaystyle{\frac{11!}{8!}}\).
Problem Statement 

Simplify the factorial \(\displaystyle{\frac{11!}{8!}}\).
Solution 

video by kbrescher 

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Simplify the factorial \(\displaystyle{\frac{10!}{8!2!}}\).
Problem Statement 

Simplify the factorial \(\displaystyle{\frac{10!}{8!2!}}\).
Solution 

video by kbrescher 

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Simplify the factorial \(\displaystyle{\frac{20!}{18!}}\).
Problem Statement 

Simplify the factorial \(\displaystyle{\frac{20!}{18!}}\).
Final Answer 

\(\displaystyle{\frac{20!}{18!} = 380}\)
Problem Statement 

Simplify the factorial \(\displaystyle{\frac{20!}{18!}}\).
Solution 

\(\begin{array}{rcl} \displaystyle{\frac{20!}{18!}} & = & \displaystyle{\frac{18! \cdot 19 \cdot 20}{18!} } \\ & = & 19 \cdot 20 = 380 \end{array} \)
Final Answer 

\(\displaystyle{\frac{20!}{18!} = 380}\) 
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Simplify the factorial \(\displaystyle{\frac{13!}{16!}}\).
Problem Statement 

Simplify the factorial \(\displaystyle{\frac{13!}{16!}}\).
Final Answer 

\(\displaystyle{\frac{13!}{16!} = \frac{1}{3360}}\)
Problem Statement 

Simplify the factorial \(\displaystyle{\frac{13!}{16!}}\).
Solution 

\(\begin{array}{rcl} \displaystyle{\frac{13!}{16!}} & = & \displaystyle{\frac{13!}{13! \cdot 14 \cdot 15 \cdot 16}} \\ & = & \displaystyle{\frac{1}{14 \cdot 15 \cdot 16} = \frac{1}{3360}} \end{array} \)
Final Answer 

\(\displaystyle{\frac{13!}{16!} = \frac{1}{3360}}\) 
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Simplify the factorial \(\displaystyle{\frac{23!}{5!20!}}\).
Problem Statement 

Simplify the factorial \(\displaystyle{\frac{23!}{5!20!}}\).
Final Answer 

\(\displaystyle{\frac{23!}{5!20!} = \frac{1771}{20}}\)
Problem Statement 

Simplify the factorial \(\displaystyle{\frac{23!}{5!20!}}\).
Solution 

\(\begin{array}{rcl} \displaystyle{\frac{!23}{5!20!}} & = & \displaystyle{\frac{20! \cdot 21 \cdot 22 \cdot 23}{20! \cdot 2 \cdot 3 \cdot 4 \cdot 5}} \\ & = & \displaystyle{\frac{7 \cdot 11 \cdot 23}{4 \cdot 5}} \\ & = & \displaystyle{\frac{1771}{20}} \end{array} \)
Final Answer 

\(\displaystyle{\frac{23!}{5!20!} = \frac{1771}{20}}\) 
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Simplify the factorial \(\displaystyle{\frac{(7+2)!}{6!2!}}\).
Problem Statement 

Simplify the factorial \(\displaystyle{\frac{(7+2)!}{6!2!}}\).
Final Answer 

\(\displaystyle{\frac{(7+2)!}{6!2!} = 252}\)
Problem Statement 

Simplify the factorial \(\displaystyle{\frac{(7+2)!}{6!2!}}\).
Solution 

\(\begin{array}{rcl} \displaystyle{\frac{(7+2)!}{6!2!}} & = & \displaystyle{\frac{9!}{6!2!}} \\ & = & \displaystyle{\frac{6! \cdot 7 \cdot 8 \cdot 9}{6! \cdot 2}} \\ & = & \displaystyle{\frac{7 \cdot 8 \cdot 9}{2}} \\ & = & \displaystyle{\frac{7 \cdot 4 \cdot 9}{1} = 252} \end{array} \)
Final Answer 

\(\displaystyle{\frac{(7+2)!}{6!2!} = 252}\) 
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Simplify the factorial \(\displaystyle{\frac{5!(43)!}{4!}}\).
Problem Statement 

Simplify the factorial \(\displaystyle{\frac{5!(43)!}{4!}}\).
Final Answer 

\(\displaystyle{\frac{5!(43)!}{4!} = 5}\)
Problem Statement 

Simplify the factorial \(\displaystyle{\frac{5!(43)!}{4!}}\).
Solution 

\(\begin{array}{rcl} \displaystyle{\frac{5!(43)!}{4!}} & = & \displaystyle{\frac{4! \cdot 5 \cdot 1!}{4!}} \\ & = & 5 \cdot 1 = 5 \end{array} \)
Final Answer 

\(\displaystyle{\frac{5!(43)!}{4!} = 5}\) 
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Simplify \(\displaystyle{ \frac{3! 7!}{6! 4!} }\)
Problem Statement 

Simplify \(\displaystyle{ \frac{3! 7!}{6! 4!} }\)
Final Answer 

\( 7/4 \)
Problem Statement 

Simplify \(\displaystyle{ \frac{3! 7!}{6! 4!} }\)
Solution 

Final Answer 

\( 7/4 \) 
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Simplify \(\displaystyle{ \frac{12! 7!}{9! 6!} }\)
Problem Statement 

Simplify \(\displaystyle{ \frac{12! 7!}{9! 6!} }\)
Final Answer 

\( 12 \cdot 11 \cdot 10 \cdot 7 = 9240 \)
Problem Statement 

Simplify \(\displaystyle{ \frac{12! 7!}{9! 6!} }\)
Solution 

Final Answer 

\( 12 \cdot 11 \cdot 10 \cdot 7 = 9240 \) 
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Simplify \(\displaystyle{ \frac{10! 6!}{8! 7!} }\)
Problem Statement 

Simplify \(\displaystyle{ \frac{10! 6!}{8! 7!} }\)
Final Answer 

\( 90/7 \)
Problem Statement 

Simplify \(\displaystyle{ \frac{10! 6!}{8! 7!} }\)
Solution 

Final Answer 

\( 90/7 \) 
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Variable Lengths
The trick when using factorials in calculus is that you may have factorials that look like this.
\(n! = n(n1)(n2). . . 3 \cdot 2 \cdot 1\)
and you don't know what \(n\) is. In this case, the terms on the right are all you can write. However, in most cases, you will have fractions with various factorials in both the numerator and denominator and you will need to simplify. Here is another example.
Example 2
Simplify \( (n+2)! / n! \)
\(\displaystyle{\frac{(n+2)!}{n!} = }\) \(\displaystyle{
\frac{(n+2)(n+1)(n)(n1). . . 3 \cdot 2 \cdot 1}{(n)(n1). . . 3 \cdot 2 \cdot 1} = }\) \(\displaystyle{
(n+2)(n+1)
}\)
There is another way to write this using the fact that factorials build on one another.
\(
\begin{array}{rcl}
\displaystyle{\frac{(n+2)!}{n!}} & = & \displaystyle{\frac{(n+2)(n+1)(n!)}{n!}} \\
& = & (n+2)(n+1)
\end{array}
\)
This last way of writing the factorials may be easier to see what cancels.
Things To Watch For
1. Parentheses are very important in factorials. For example, \((2n)! \neq 2n!\) since \(2n! = 2(n!)\).
2. When in doubt, write out the first few terms to make it obvious what is canceling. If it helps, use the idea that factorials build to rewrite factorials.
Time for some practice problems.
Practice
Unless otherwise instructed, simplify these factorials.
Simplify the factorial \(\displaystyle{\frac{(n3)!}{n!}}\).
Problem Statement 

Simplify the factorial \(\displaystyle{\frac{(n3)!}{n!}}\).
Final Answer 

\(\displaystyle{\frac{(n3)!}{n!} = \frac{1}{(n2)(n1)n}}\)
Problem Statement 

Simplify the factorial \(\displaystyle{\frac{(n3)!}{n!}}\).
Solution 

\(\begin{array}{rcl} \displaystyle{\frac{(n3)!}{n!}} & = & \displaystyle{\frac{(n3)!}{(n3)! \cdot (n2) \cdot (n1) \cdot n}} \\ & = & \displaystyle{\frac{1}{(n2) \cdot (n1) \cdot n}} \end{array} \)
Final Answer 

\(\displaystyle{\frac{(n3)!}{n!} = \frac{1}{(n2)(n1)n}}\) 
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Simplify the factorial \(\displaystyle{\frac{(2n+1)!}{(2n)!}}\).
Problem Statement 

Simplify the factorial \(\displaystyle{\frac{(2n+1)!}{(2n)!}}\).
Final Answer 

\(\displaystyle{\frac{(2n+1)!}{(2n)!} = (2n+1)}\)
Problem Statement 

Simplify the factorial \(\displaystyle{\frac{(2n+1)!}{(2n)!}}\).
Solution 

\(\begin{array}{rcl} \displaystyle{\frac{(2n+1)!}{(2n)!}} & = & \displaystyle{\frac{1 \cdot 2 \cdot . . . \cdot (2n1) \cdot (2n) \cdot (2n+1)}{1 \cdot 2 \cdot . . . \cdot (2n1) \cdot (2n)}} \\ & = & \displaystyle{\frac{(2n)! \cdot (2n+1)}{(2n)!} = (2n+1)} \end{array} \)
Final Answer 

\(\displaystyle{\frac{(2n+1)!}{(2n)!} = (2n+1)}\) 
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Simplify the factorial \(\displaystyle{\frac{(3n1)!}{(3n+1)!}}\).
Problem Statement 

Simplify the factorial \(\displaystyle{\frac{(3n1)!}{(3n+1)!}}\).
Final Answer 

\(\displaystyle{\frac{(3n1)!}{(3n+1)!} = \frac{1}{(3n)(3n+1)}}\)
Problem Statement 

Simplify the factorial \(\displaystyle{\frac{(3n1)!}{(3n+1)!}}\).
Solution 

\(\begin{array}{rcl} \displaystyle{\frac{(3n1)!}{(3n+1)!}} & = & \displaystyle{\frac{(3n1)!}{(3n1)! \cdot (3n) \cdot (3n+1)}} \\ & = & \displaystyle{\frac{1}{(3n)(3n+1)}} \end{array} \)
Final Answer 

\(\displaystyle{\frac{(3n1)!}{(3n+1)!} = \frac{1}{(3n)(3n+1)}}\) 
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Simplify the factorial \(\displaystyle{\frac{k!}{(k2)!}}\).
Problem Statement 

Simplify the factorial \(\displaystyle{\frac{k!}{(k2)!}}\).
Solution 

video by MIP4U 

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Simplify \(\displaystyle{ \frac{(n+1)!}{(n2)!} }\)
Problem Statement 

Simplify \(\displaystyle{ \frac{(n+1)!}{(n2)!} }\)
Final Answer 

\( (n+1)n(n1) = n^3  n \)
Problem Statement 

Simplify \(\displaystyle{ \frac{(n+1)!}{(n2)!} }\)
Solution 

Although this instructor multiplies out the factors, check with your instructor to see if they require you to do that.
Final Answer 

\( (n+1)n(n1) = n^3  n \) 
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Simplify \(\displaystyle{ \frac{(n+2)!}{(n1)!} }\)
Problem Statement 

Simplify \(\displaystyle{ \frac{(n+2)!}{(n1)!} }\)
Final Answer 

\( (n+2)(n+1)n = n^3 + 3n^2 + 2 \)
Problem Statement 

Simplify \(\displaystyle{ \frac{(n+2)!}{(n1)!} }\)
Solution 

Check with your instructor to see if they want you to multiply out your answer.
Final Answer 

\( (n+2)(n+1)n = n^3 + 3n^2 + 2 \) 
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Simplify \(\displaystyle{ \frac{(n+1)!}{(n+3)!} }\)
Problem Statement 

Simplify \(\displaystyle{ \frac{(n+1)!}{(n+3)!} }\)
Final Answer 

\(\displaystyle{ \frac{1}{(n+3)(n+2)} }\)
Problem Statement 

Simplify \(\displaystyle{ \frac{(n+1)!}{(n+3)!} }\)
Solution 

Check with your instructor to see if they want you to multiply out the denominator.
Final Answer 

\(\displaystyle{ \frac{1}{(n+3)(n+2)} }\) 
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Simplify \(\displaystyle{ \frac{n!}{(n2)!} }\)
Problem Statement 

Simplify \(\displaystyle{ \frac{n!}{(n2)!} }\)
Final Answer 

\( n(n1) \)
Problem Statement 

Simplify \(\displaystyle{ \frac{n!}{(n2)!} }\)
Solution 

Check with your instructor to see if they want you to multiply out your answer.
Final Answer 

\( n(n1) \) 
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Simplify \(\displaystyle{ \frac{(2m)!}{(2m+2)!} }\)
Problem Statement 

Simplify \(\displaystyle{ \frac{(2m)!}{(2m+2)!} }\)
Final Answer 

\(\displaystyle{ \frac{1}{(2m+2)(2m+1)} }\)
Problem Statement 

Simplify \(\displaystyle{ \frac{(2m)!}{(2m+2)!} }\)
Solution 

Check with your instructor to see if they want you to multiply out the denominator.
Final Answer 

\(\displaystyle{ \frac{1}{(2m+2)(2m+1)} }\) 
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Simplify \(\displaystyle{ \frac{(n+3)!}{n!} }\)
Problem Statement 

Simplify \(\displaystyle{ \frac{(n+3)!}{n!} }\)
Final Answer 

\( (n+3)(n+2)(n+1) \)
Problem Statement 

Simplify \(\displaystyle{ \frac{(n+3)!}{n!} }\)
Solution 

Although he says at the end of the problem that you need to foil it out, most instructors will not require you to do that for these types of problems.
Final Answer 

\( (n+3)(n+2)(n+1) \) 
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Simplify \(\displaystyle{ \frac{(3n+2)!}{(3n1)!} }\)
Problem Statement 

Simplify \(\displaystyle{ \frac{(3n+2)!}{(3n1)!} }\)
Final Answer 

\( (3n+2)(3n+1)(3n) \)
Problem Statement 

Simplify \(\displaystyle{ \frac{(3n+2)!}{(3n1)!} }\)
Solution 

Again, do not multiply out the answer unless specifically told to do so by your instructor.
Final Answer 

\( (3n+2)(3n+1)(3n) \) 
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Simplify \(\displaystyle{ \frac{(n^24)!}{(n2)(n^25)!} }\)
Problem Statement 

Simplify \(\displaystyle{ \frac{(n^24)!}{(n2)(n^25)!} }\)
Hint 

The last factorial in the denominator applies only to the \((n^25)\) term. So the denominator can be written more clearly as \((n2)[(n^25)!]\).
Problem Statement 

Simplify \(\displaystyle{ \frac{(n^24)!}{(n2)(n^25)!} }\)
Final Answer 

\(\displaystyle{ \frac{n^24}{n2} }\)
Problem Statement 

Simplify \(\displaystyle{ \frac{(n^24)!}{(n2)(n^25)!} }\)
Hint 

The last factorial in the denominator applies only to the \((n^25)\) term. So the denominator can be written more clearly as \((n2)[(n^25)!]\).
Solution 

Although they give the final answer as \(n+2\), you know from the domain and range page that they should have also said that \( n \neq 2 \) as part of the final answer.
Final Answer 

\(\displaystyle{ \frac{n^24}{n2} }\) 
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Simplify \(\displaystyle{ \frac{(n!)^2}{(n1)!(n+1)!} }\)
Problem Statement 

Simplify \(\displaystyle{ \frac{(n!)^2}{(n1)!(n+1)!} }\)
Final Answer 

\(\displaystyle{ \frac{n}{n+1} }\)
Problem Statement 

Simplify \(\displaystyle{ \frac{(n!)^2}{(n1)!(n+1)!} }\)
Solution 

Final Answer 

\(\displaystyle{ \frac{n}{n+1} }\) 
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Simplify \(\displaystyle{ \frac{(n+3)!n!}{(n+2)!(n1)!} }\)
Problem Statement 

Simplify \(\displaystyle{ \frac{(n+3)!n!}{(n+2)!(n1)!} }\)
Final Answer 

\(n(n+3)\)
Problem Statement 

Simplify \(\displaystyle{ \frac{(n+3)!n!}{(n+2)!(n1)!} }\)
Solution 

Check with your instructor to see if they want you to multiply out your answer.
Final Answer 

\(n(n+3)\) 
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Simplify \(\displaystyle{ \frac{(n+2)!n!}{(n+1)!} }\)
Problem Statement 

Simplify \(\displaystyle{ \frac{(n+2)!n!}{(n+1)!} }\)
Final Answer 

\( n+2  1/(n+1) \)
Problem Statement 

Simplify \(\displaystyle{ \frac{(n+2)!n!}{(n+1)!} }\)
Solution 

Final Answer 

\( n+2  1/(n+1) \) 
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Simplify the factorial \(\displaystyle{\frac{(k+2)!}{(k1)!}}\).
Problem Statement 

Simplify the factorial \(\displaystyle{\frac{(k+2)!}{(k1)!}}\).
Solution 

This problem is solved in two separate videos by two different instructors. Watching both of them will help you better understand these concepts.
video by MIP4U 

video by kbrescher 

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