## 17Calculus Precalculus - Variable Factorials

##### 17Calculus

Factorials that involve variables are built the same way as basic factorials. However, you sometimes don't know what the final value is. So there are ways to write these factorials that build on the basic factorials concepts.

Variable Lengths

The trick when using factorials in calculus is that you may have factorials that look like this.
$$n! = n(n-1)(n-2). . . 3 \cdot 2 \cdot 1$$
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 an example.

Example

Simplify $$(n+2)! / n!$$

$$\displaystyle{\frac{(n+2)!}{n!} = }$$ $$\displaystyle{ \frac{(n+2)(n+1)(n)(n-1). . . 3 \cdot 2 \cdot 1}{(n)(n-1). . . 3 \cdot 2 \cdot 1} = }$$ $$\displaystyle{ (n+2)(n+1) }$$

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.

$$\displaystyle{\frac{(n-3)!}{n!}}$$

Problem Statement

Simplify $$\displaystyle{\frac{(n-3)!}{n!}}$$

$$\displaystyle{\frac{(n-3)!}{n!}}$$ $$\displaystyle{ = \frac{1}{(n-2)(n-1)n}}$$

Problem Statement

Simplify $$\displaystyle{\frac{(n-3)!}{n!}}$$

Solution

$$\displaystyle{\frac{(n-3)!}{n!}}$$ $$\displaystyle{ = \frac{(n-3)!}{(n-3)! \cdot (n-2) \cdot (n-1) \cdot n} = \frac{1}{(n-2) \cdot (n-1) \cdot n} }$$

$$\displaystyle{\frac{(n-3)!}{n!}}$$ $$\displaystyle{ = \frac{1}{(n-2)(n-1)n}}$$

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

Problem Statement

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

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

Problem Statement

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

Solution

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

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

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$$\displaystyle{\frac{(3n-1)!}{(3n+1)!}}$$

Problem Statement

$$\displaystyle{\frac{(3n-1)!}{(3n+1)!}}$$

$$\displaystyle{\frac{(3n-1)!}{(3n+1)!} = \frac{1}{(3n)(3n+1)}}$$

Problem Statement

$$\displaystyle{\frac{(3n-1)!}{(3n+1)!}}$$

Solution

$$\begin{array}{rcl} \displaystyle{\frac{(3n-1)!}{(3n+1)!}} & = & \displaystyle{\frac{(3n-1)!}{(3n-1)! \cdot (3n) \cdot (3n+1)}} \\ & = & \displaystyle{\frac{1}{(3n)(3n+1)}} \end{array}$$

$$\displaystyle{\frac{(3n-1)!}{(3n+1)!} = \frac{1}{(3n)(3n+1)}}$$

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$$\displaystyle{\frac{k!}{(k-2)!}}$$

Problem Statement

$$\displaystyle{\frac{k!}{(k-2)!}}$$

Solution

### MIP4U - 1528 video solution

video by MIP4U

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

Problem Statement

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

$$(n+2)(n+1)n = n^3 + 3n^2 + 2$$

Problem Statement

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

Solution

### 2756 video solution

$$(n+2)(n+1)n = n^3 + 3n^2 + 2$$

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

Problem Statement

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

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

Problem Statement

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

Solution

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

### 2757 video solution

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

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$$\displaystyle{ \frac{n!}{(n-2)!} }$$

Problem Statement

$$\displaystyle{ \frac{n!}{(n-2)!} }$$

$$n(n-1)$$

Problem Statement

$$\displaystyle{ \frac{n!}{(n-2)!} }$$

Solution

### 2758 video solution

$$n(n-1)$$

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$$\displaystyle{ \frac{(2m)!}{(2m+2)!} }$$

Problem Statement

$$\displaystyle{ \frac{(2m)!}{(2m+2)!} }$$

$$\displaystyle{ \frac{1}{(2m+2)(2m+1)} }$$

Problem Statement

$$\displaystyle{ \frac{(2m)!}{(2m+2)!} }$$

Solution

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

### 2759 video solution

$$\displaystyle{ \frac{1}{(2m+2)(2m+1)} }$$

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$$\displaystyle{ \frac{(n+3)!}{n!} }$$

Problem Statement

$$\displaystyle{ \frac{(n+3)!}{n!} }$$

$$(n+3)(n+2)(n+1)$$

Problem Statement

$$\displaystyle{ \frac{(n+3)!}{n!} }$$

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.

### 2760 video solution

$$(n+3)(n+2)(n+1)$$

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$$\displaystyle{ \frac{(3n+2)!}{(3n-1)!} }$$

Problem Statement

$$\displaystyle{ \frac{(3n+2)!}{(3n-1)!} }$$

$$(3n+2)(3n+1)(3n)$$

Problem Statement

$$\displaystyle{ \frac{(3n+2)!}{(3n-1)!} }$$

Solution

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

### 2761 video solution

$$(3n+2)(3n+1)(3n)$$

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$$\displaystyle{ \frac{(n^2-4)!}{(n-2)(n^2-5)!} }$$

Problem Statement

$$\displaystyle{ \frac{(n^2-4)!}{(n-2)(n^2-5)!} }$$

Hint

The last factorial in the denominator applies only to the $$(n^2-5)$$ term. So the denominator can be written more clearly as $$(n-2)[(n^2-5)!]$$.

Problem Statement

$$\displaystyle{ \frac{(n^2-4)!}{(n-2)(n^2-5)!} }$$

$$\displaystyle{ \frac{n^2-4}{n-2} }$$

Problem Statement

$$\displaystyle{ \frac{(n^2-4)!}{(n-2)(n^2-5)!} }$$

Hint

The last factorial in the denominator applies only to the $$(n^2-5)$$ term. So the denominator can be written more clearly as $$(n-2)[(n^2-5)!]$$.

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.

### 2762 video solution

$$\displaystyle{ \frac{n^2-4}{n-2} }$$

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

Problem Statement

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

$$\displaystyle{ \frac{n}{n+1} }$$

Problem Statement

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

Solution

### 2763 video solution

$$\displaystyle{ \frac{n}{n+1} }$$

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$$\displaystyle{ \frac{(n+3)!n!}{(n+2)!(n-1)!} }$$

Problem Statement

$$\displaystyle{ \frac{(n+3)!n!}{(n+2)!(n-1)!} }$$

$$n(n+3)$$

Problem Statement

$$\displaystyle{ \frac{(n+3)!n!}{(n+2)!(n-1)!} }$$

Solution

### 2764 video solution

$$n(n+3)$$

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

Problem Statement

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

$$n+2 - 1/(n+1)$$

Problem Statement

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

Solution

### 2765 video solution

$$n+2 - 1/(n+1)$$

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$$\displaystyle{\frac{(k+2)!}{(k-1)!}}$$

Problem Statement

$$\displaystyle{\frac{(k+2)!}{(k-1)!}}$$

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

### kbrescher - 1529 video solution

video by kbrescher

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