The Central Idea

Suppose five distinct objects are to be placed in a row.

There are \(5\) choices for the first position. After one object is used, \(4\) choices remain for the second position. Then \(3\) remain, then \(2\), then \(1\).

\[ 5\cdot4\cdot3\cdot2\cdot1=120. \]

This descending product occurs so often that mathematics gives it a special notation:

\[ \boxed{5!=120}. \]

The symbol \(5!\) is read five factorial.

Factorial Notation

For a positive integer \(n\), the factorial of \(n\) is

\[ \boxed{n!=n(n-1)(n-2)\cdots3\cdot2\cdot1}. \]

The exclamation mark is part of the mathematical notation. It does not indicate excitement or emphasis.

For example,

\[ 6!=6\cdot5\cdot4\cdot3\cdot2\cdot1=720. \]

Small Factorials

The first several factorial values are worth recognizing:

\[ 0!=1, \] \[ 1!=1, \] \[ 2!=2, \] \[ 3!=6, \] \[ 4!=24, \] \[ 5!=120, \] \[ 6!=720, \] \[ 7!=5040. \]

Factorials grow quickly because each new factorial multiplies the previous one by another positive integer.

The Recursive Pattern

Compare two consecutive factorials:

\[ 6!=6\cdot5\cdot4\cdot3\cdot2\cdot1 \]

and

\[ 5!=5\cdot4\cdot3\cdot2\cdot1. \]

Therefore,

\[ 6!=6(5!). \]

In general,

\[ \boxed{n!=n(n-1)!} \]

for positive integers \(n\).

Why \(0!=1\)

At first, \(0!=1\) may look unusual because there is no descending list of positive integers beginning at \(0\).

The recursive factorial rule explains the convention naturally. Since

\[ 1!=1(0!) \]

and \(1!=1\), we must have

\[ \boxed{0!=1}. \]

This value also makes counting formulas work correctly when there are no objects left to arrange.

The Empty Arrangement

There is exactly one way to arrange zero objects: choose nothing and make no placements.

This is sometimes called the empty arrangement.

Counting one empty arrangement may seem abstract, but it is the same idea that makes an empty product equal to \(1\). It gives counting formulas a consistent starting point.

\[ \boxed{0!=1}. \]

Factorials as Products of Sequential Choices

Factorials are closely connected to the fundamental counting principle.

If \(n\) distinct objects are arranged in \(n\) positions, then:

Multiplying the numbers of choices gives

\[ n(n-1)(n-2)\cdots2\cdot1=n!. \]

Example: Arranging Four Books

Four distinct books are placed on a shelf.

There are

\[ 4 \]

choices for the first position,

\[ 3 \]

for the second,

\[ 2 \]

for the third, and

\[ 1 \]

for the last.

Therefore,

\[ \boxed{4!=4\cdot3\cdot2\cdot1=24}. \]

Example: Ordering Seven Runners

Suppose seven runners finish a race with no ties.

Any of the seven runners could finish first. After that, six remain for second place, five for third place, and so on.

\[ 7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1=7!. \]

Thus the number of possible complete finishing orders is

\[ \boxed{7!=5040}. \]

Expanding a Factorial

A factorial can be expanded completely:

\[ 8!=8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1. \]

But complete expansion is often unnecessary.

For example,

\[ 8!=8\cdot7!. \]

We can also stop at any smaller factorial:

\[ 8!=8\cdot7\cdot6\cdot5!. \]

Partial Expansion

Partial expansion is one of the most useful techniques for simplifying factorial expressions.

For example,

\[ 10!=10\cdot9\cdot8!. \]

This makes it easy to simplify

\[ \frac{10!}{8!}. \]

Substitute the partial expansion:

\[ \frac{10\cdot9\cdot8!}{8!}. \]

Cancel \(8!\):

\[ \boxed{\frac{10!}{8!}=10\cdot9=90}. \]

Factorial Ratios

A ratio of factorials often simplifies dramatically because the smaller factorial is already contained inside the larger one.

For example,

\[ \frac{9!}{6!} \]

becomes

\[ \frac{9\cdot8\cdot7\cdot6!}{6!}. \]

Therefore,

\[ \boxed{\frac{9!}{6!}=9\cdot8\cdot7=504}. \]

A General Factorial Ratio

If \(n\) and \(r\) are nonnegative integers with \(r\le n\), then

\[ n!=n(n-1)(n-2)\cdots(r+1)r!. \]

Dividing by \(r!\) gives

\[ \boxed{\frac{n!}{r!}=n(n-1)(n-2)\cdots(r+1)}. \]

The product contains the integers from \(n\) down through \(r+1\).

Example: Simplifying \(\frac{12!}{9!}\)

Expand only as far as \(9!\):

\[ 12!=12\cdot11\cdot10\cdot9!. \]

Then

\[ \frac{12!}{9!} = \frac{12\cdot11\cdot10\cdot9!}{9!}. \]

Cancel:

\[ \boxed{\frac{12!}{9!}=12\cdot11\cdot10=1320}. \]

Example: A More Complicated Ratio

Simplify

\[ \frac{8!}{3!\,5!}. \]

Expand \(8!\) only far enough to expose \(5!\):

\[ 8!=8\cdot7\cdot6\cdot5!. \]

Then

\[ \frac{8!}{3!\,5!} = \frac{8\cdot7\cdot6\cdot5!}{3!\,5!}. \]

Cancel \(5!\):

\[ \frac{8\cdot7\cdot6}{3!}. \]

Since \(3!=6\),

\[ \boxed{\frac{8!}{3!\,5!}=56}. \]

Do Not Cancel Through Addition or Subtraction

Factorials are factors only when they are being multiplied.

In an expression such as

\[ \frac{8!+7!}{7!}, \]

we cannot simply cancel \(7!\) across the addition sign.

First factor \(7!\) from the numerator:

\[ 8!+7! = 8(7!)+7! = 7!(8+1). \]

Therefore,

\[ \frac{8!+7!}{7!} = \frac{7!(9)}{7!} = \boxed{9}. \]

Factoring a Factorial Expression

The recursive rule can reveal a common factorial factor.

Consider

\[ n!+(n+1)!. \]

Since

\[ (n+1)!=(n+1)n!, \]

we have

\[ n!+(n+1)! = n!+(n+1)n!. \]

Factor out \(n!\):

\[ \boxed{n!+(n+1)!=n!(n+2)}. \]

Factorials and Multiplication Are Not the Same Operation

The notation

\[ 5! \]

means

\[ 5\cdot4\cdot3\cdot2\cdot1. \]

It does not mean \(5\cdot5\), \(5^5\), or \(5+4+3+2+1\).

Factorial is a distinct operation defined by a particular product pattern.

Factorials Are Defined Here for Nonnegative Integers

In elementary counting, factorial notation is used for nonnegative integers:

\[ 0,1,2,3,\ldots \]

Expressions such as

\[ (-3)! \]

are not defined by the elementary factorial definition.

More advanced mathematics extends factorial-like behavior through the gamma function, but that extension is not needed for the counting methods in this section.

Why Factorials Grow So Quickly

Each successive factorial satisfies

\[ (n+1)!=(n+1)n!. \]

So moving from \(n!\) to \((n+1)!\) multiplies the previous value by \(n+1\).

For example,

\[ 8!=40320, \] \[ 9!=362880, \] \[ 10!=3628800. \]

Even modest values of \(n\) therefore produce very large factorials.

Factorials and Ordered Outcomes

The key counting interpretation of \(n!\) is:

\[ \boxed{\text{\(n!\) counts the complete orderings of \(n\) distinct objects.}} \]

The word distinct matters. If some objects are identical, then different placements may produce the same visible arrangement.

Repeated objects require an adjustment that will be developed later in this section.

Example: Seating Five People in a Row

Five distinct people are assigned five seats in a row.

There are \(5\) choices for the first seat, \(4\) for the second, \(3\) for the third, \(2\) for the fourth, and \(1\) for the fifth.

\[ 5\cdot4\cdot3\cdot2\cdot1=5!. \]

Therefore,

\[ \boxed{5!=120} \]

different seating orders are possible.

Example: Assigning Every Position

Suppose six distinct photographs are placed into six labeled positions.

Because every photograph is used exactly once and every position matters, the number of assignments is

\[ \boxed{6!=720}. \]

This is the same factorial structure as arranging six objects in a row: each completed assignment is an ordering of the six distinct items.

Factorial Structure Inside a Larger Counting Problem

A factorial may appear as only one stage of a larger sequence of choices.

Suppose a committee first chooses one of \(4\) rooms and then arranges \(5\) distinct speakers in speaking order.

The room can be chosen in \(4\) ways, and the speakers can be ordered in \(5!\) ways.

By the multiplication principle,

\[ 4\cdot5! = 4\cdot120 = \boxed{480}. \]

Factorials in Algebraic Formulas

Counting formulas often contain factorials instead of fully expanded products.

For example,

\[ \frac{n!}{(n-r)!} \]

is a compact way to write the product

\[ n(n-1)(n-2)\cdots(n-r+1). \]

This expression will become the standard formula for permutations without repetition.

Seeing the Cancellation Pattern

Consider

\[ \frac{n!}{(n-3)!}. \]

Expand \(n!\) only until \((n-3)!\) appears:

\[ n!=n(n-1)(n-2)(n-3)!. \]

Therefore,

\[ \frac{n!}{(n-3)!} = \frac{n(n-1)(n-2)(n-3)!}{(n-3)!}. \]

Cancel:

\[ \boxed{\frac{n!}{(n-3)!}=n(n-1)(n-2)}. \]

Counting Only the First Few Positions

Suppose there are \(8\) distinct contestants, but we care only about first, second, and third place.

The sequential-choice count is

\[ 8\cdot7\cdot6. \]

Factorial notation writes the same product as

\[ \frac{8!}{5!}. \]

Indeed,

\[ \frac{8!}{5!} = \frac{8\cdot7\cdot6\cdot5!}{5!} = \boxed{336}. \]

The next lesson will formalize this kind of partial arrangement.

Factorial Form versus Expanded Form

Both of the following can represent the same quantity:

\[ 9! \]

and

\[ 9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1. \]

Factorial form is compact and makes structure visible. Expanded form is useful when computing a numerical value or exposing factors for cancellation.

When Not to Expand Completely

Suppose we want to simplify

\[ \frac{20!}{18!}. \]

Computing \(20!\) and \(18!\) separately would create enormous numbers for no reason.

Instead,

\[ 20!=20\cdot19\cdot18!. \]

So

\[ \boxed{\frac{20!}{18!}=20\cdot19=380}. \]

Preserve factorial notation until you know which factors are actually needed.

Factorials with Variables

Factorial expressions can contain variables as long as those variables represent allowable integer values.

For example,

\[ (n+2)! \]

can be expanded as

\[ (n+2)!=(n+2)(n+1)n!. \]

Likewise,

\[ (n+1)!=(n+1)n!. \]

Example: Simplifying a Variable Factorial Ratio

Simplify

\[ \frac{(n+2)!}{n!}. \]

Expand:

\[ (n+2)!=(n+2)(n+1)n!. \]

Therefore,

\[ \frac{(n+2)!}{n!} = \frac{(n+2)(n+1)n!}{n!}. \]

Cancel \(n!\):

\[ \boxed{\frac{(n+2)!}{n!}=(n+2)(n+1)}. \]

Example: Simplifying \(\frac{(n+1)!}{(n-1)!}\)

Expand the larger factorial:

\[ (n+1)!=(n+1)n(n-1)!. \]

Then

\[ \frac{(n+1)!}{(n-1)!} = \frac{(n+1)n(n-1)!}{(n-1)!}. \]

Therefore,

\[ \boxed{\frac{(n+1)!}{(n-1)!}=n(n+1)}. \]

Factorials and Parentheses

Parentheses determine what quantity the factorial applies to.

Compare

\[ (n+1)! \]

with

\[ n!+1. \]

These are generally not equal.

For \(n=3\),

\[ (3+1)!=4!=24, \]

while

\[ 3!+1=6+1=7. \]

Common Error: Treating \(n!\) as \(n\cdot n\)

Factorial does not mean multiply a number by itself.

For example,

\[ 4!=4\cdot3\cdot2\cdot1=24, \]

not

\[ 4\cdot4=16. \]

Common Error: Forgetting \(0!=1\)

Many counting formulas produce \(0!\) at boundary cases.

For example,

\[ \frac{5!}{5!} = 1. \]

The permutation formula introduced next will sometimes contain \((n-n)!=0!\). The value

\[ \boxed{0!=1} \]

keeps the formula consistent.

Common Error: Adding Instead of Multiplying

The factorial

\[ 5! \]

is

\[ 5\cdot4\cdot3\cdot2\cdot1, \]

not

\[ 5+4+3+2+1. \]

Factorials encode a product of sequential numbers.

Common Error: Expanding Too Far

In

\[ \frac{15!}{13!}, \]

there is no need to calculate either large factorial.

Write

\[ 15!=15\cdot14\cdot13! \]

and cancel:

\[ \boxed{\frac{15!}{13!}=210}. \]

Common Error: Canceling Terms Instead of Factors

In

\[ \frac{7!+6!}{6!}, \]

the denominator cannot be canceled with only one term of the numerator.

First factor:

\[ 7!+6! = 7(6!)+6! = 6!(7+1). \]

Then

\[ \boxed{\frac{7!+6!}{6!}=8}. \]

Common Error: Confusing \((n+1)!\) with \(n!+1\)

The recursive rule gives

\[ \boxed{(n+1)!=(n+1)n!}. \]

It does not give

\[ (n+1)!=n!+1. \]

Factorial growth is multiplicative, not additive.

A Reliable Procedure for Evaluating a Factorial

  1. Identify the nonnegative integer inside the factorial notation.
  2. Write the descending positive integers down to \(1\).
  3. Multiply the factors.
  4. Remember the special value \(0!=1\).

For example,

\[ 5!=5\cdot4\cdot3\cdot2\cdot1=120. \]

A Reliable Procedure for Simplifying Factorial Ratios

  1. Identify the larger factorial.
  2. Expand it only until the smaller factorial appears.
  3. Cancel the common factorial factor.
  4. Multiply or simplify the remaining factors.
  5. Avoid calculating large factorial values unless necessary.

Example: Full Simplification Workflow

Simplify

\[ \frac{11!}{8!}. \]

Expand \(11!\) down to \(8!\):

\[ 11!=11\cdot10\cdot9\cdot8!. \]

Substitute:

\[ \frac{11!}{8!} = \frac{11\cdot10\cdot9\cdot8!}{8!}. \]

Cancel \(8!\):

\[ 11\cdot10\cdot9. \]

Therefore,

\[ \boxed{\frac{11!}{8!}=990}. \]

Example: Factorial Structure in an Arrangement

Six distinct cards are arranged from left to right.

The sequential choices are

\[ 6,\ 5,\ 4,\ 3,\ 2,\ 1. \]

Their product is

\[ 6!. \]

Therefore,

\[ \boxed{6!=720} \]

complete arrangements are possible.

Practice 1: Evaluate a Factorial

Evaluate:

\[ 7!. \]

Practice 2: The Zero Factorial

Evaluate:

\[ 0!. \]

Practice 3: Partial Expansion

Rewrite \(9!\) using \(7!\) without expanding completely.

Practice 4: Factorial Ratio

Simplify:

\[ \frac{10!}{7!}. \]

Practice 5: Factorial Ratio with Two Denominator Factors

Simplify:

\[ \frac{7!}{5!\,2!}. \]

Practice 6: Variable Factorials

Simplify:

\[ \frac{(n+3)!}{(n+1)!}. \]

Practice 7: Factor a Factorial Expression

Simplify:

\[ \frac{9!+8!}{8!}. \]

Practice 8: Complete Arrangements

How many ways can \(6\) distinct books be arranged on a shelf?

Practice 9: Partial Arrangement Product

Simplify:

\[ \frac{12!}{10!}. \]

Practice 10: Identify the Error

A student writes

\[ 5!=5+4+3+2+1=15. \]

Explain the error and give the correct value.

Practice Answers

Practice 1

\[ 7! = 7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1 = \boxed{5040}. \]

Practice 2

\[ \boxed{0!=1}. \]

Practice 3

\[ \boxed{9!=9\cdot8\cdot7!}. \]

Practice 4

\[ \frac{10!}{7!} = \frac{10\cdot9\cdot8\cdot7!}{7!} = \boxed{720}. \]

Practice 5

\[ \frac{7!}{5!\,2!} = \frac{7\cdot6\cdot5!}{5!\cdot2} = \frac{42}{2} = \boxed{21}. \]

Practice 6

\[ (n+3)! = (n+3)(n+2)(n+1)!. \]

Therefore,

\[ \boxed{\frac{(n+3)!}{(n+1)!}=(n+3)(n+2)}. \]

Practice 7

\[ 9!+8! = 9(8!)+8! = 10(8!). \]

Therefore,

\[ \boxed{\frac{9!+8!}{8!}=10}. \]

Practice 8

Six distinct books have

\[ \boxed{6!=720} \]

possible complete arrangements.

Practice 9

\[ \frac{12!}{10!} = \frac{12\cdot11\cdot10!}{10!} = \boxed{132}. \]

Practice 10

Factorial requires multiplication, not addition:

\[ 5! = 5\cdot4\cdot3\cdot2\cdot1 = \boxed{120}. \]

What to Remember

For a positive integer \(n\),

\[ \boxed{n!=n(n-1)(n-2)\cdots2\cdot1}. \]

Also,

\[ \boxed{0!=1} \]

and

\[ \boxed{n!=n(n-1)!}. \]

The main counting interpretation is:

\[ \boxed{\text{\(n!\) counts the complete orderings of \(n\) distinct objects.}} \]

When simplifying factorial ratios, expand only as far as needed to expose a common factorial factor. This avoids unnecessary arithmetic and makes the underlying counting structure easier to see.

Where We Go Next

Factorials count complete arrangements in which every distinct object is used. Many counting problems use only some of the available objects while still caring about order.

The next lesson, Permutations without Repetition, develops the formula

\[ \boxed{P(n,r)=\frac{n!}{(n-r)!}} \]

for ordered selections of \(r\) distinct objects chosen from \(n\) distinct possibilities.