2.6 Use counting rules, factorials, and combinations | Athabasca University MATH 215: Introduction to Statistics

Athabasca University MATH 215: Introduction to Statistics

Probability

Counting possible outcomes when choices happen in stages or order does not matter

Counting rules answer questions such as “How many different outcomes are possible?” without requiring you to list every outcome. First decide what counts as a different outcome. A group containing Ana, Ben, and Chen is the same group regardless of the order in which the names are listed. A first-place and second-place ranking is different if the people switch positions. A count is a whole number of possibilities, not a probability. Fundamental algebra will help you simplify expressions: when you divide or cancel a common factor, you must keep the value unchanged.

What you will learn

Use multiplication and addition counting rules in suitable situations.
Evaluate factorial expressions, including zero factorial.
Recognize when a selection is a combination and calculate the number of possible groups.
Explain why order does or does not matter in a counting problem.

Counting choices by stages and alternatives

The multiplication rule applies when an outcome is formed by making a choice at each stage of a process. If one stage can be completed in

a

ways and, for each such choice, the next stage can be completed in

b

ways, the two stages produce

ab

outcomes. For more stages, multiply the number of choices at each stage. This rule requires the stated number of choices at each stage to be valid for every preceding choice. If choices are removed or the available options change, count the choices at each stage carefully.

For example, if a lunch order offers 4 sandwiches and 3 drinks, and one of each is chosen, there are

4\times3=12

possible orders. Each order includes a choice from both categories.

The addition rule applies when an outcome is formed by choosing one option from separate alternatives. If one alternative has

a

outcomes and another has

b

, and the alternatives share no outcomes, there are

a+b

outcomes altogether. The alternatives must not overlap, or some outcomes would be counted twice. For example, choosing either one of 5 bus routes or one of 2 train routes gives 7 choices if the two lists have no route in common.

The rules can be combined. Describe how one outcome is formed before choosing a rule: are there several stages, or is there one choice from separate alternatives? This helps prevent adding when multiplication is needed, or multiplying when the alternatives should be added.

N=a_1a_2\cdots a_k

Multiply counts for successive stages of a process.
Add counts for separate alternatives that have no outcomes in common.
Check what counts as a distinct outcome and whether the choices at each stage change.

Factorials and ordered selections

A factorial is a product that starts with a positive whole number and multiplies down through every positive whole number to 1. The symbol n! is read “

n

factorial.” For example, CAD 5!

means

5\times4\times3\times2\times1, which equals 120. By definition,

0!=1

. This convention makes factorial formulas work consistently, including when there are no items left to arrange.

Factorials can count ways to arrange distinct objects when every object is used once. For 4 different books on a shelf, there are 4 choices for the first position, then 3 remaining choices for the second, then 2, then 1. The multiplication rule gives CAD 4! arrangements. Changing the order creates a different arrangement.

More generally, arranging all

n

distinct objects in

n

positions gives n! arrangements. If only some objects are placed in ordered positions, use the multiplication rule to count the choices for each position. For example, selecting a first and second finisher from 6 people gives 6 choices for first and 5 remaining choices for second.

Before using a factorial or an ordered count, decide what makes outcomes different. If exchanging two selected people's positions creates a new outcome, order matters. If the task is simply to form a group, listing its members in a different order does not create a new group. That distinction determines whether a combination is appropriate.

n!=n(n-1)\cdots2\times1

Evaluate a factorial by multiplying down to 1; use $0!=1$ .
Factorials count arrangements of all distinct objects.
Use successive choice counts for ordered selections when not every object is selected.

Combinations: selecting a group when order does not matter

A combination is a selection of items in which the order of selection does not matter. Selecting 3 people for a committee is a combination: the same three people form the same committee whichever person is named first. Assigning those people to three different offices is different because their roles matter, so that situation is not counted using combinations alone.

The notation

\binom{n}{r}

means the number of ways to choose

r

items from

n

distinct items when order does not matter. Here,

n

is the total number available and

r

is the number selected. The items must be distinct, the selection must have a fixed size, and each group is counted once regardless of the order of its members. Usually, an item cannot be selected more than once.

The combination formula starts with the count of ordered selections, then divides by the number of different orders in which each group can be listed. Each group of

r

distinct items has r! such orders, so dividing removes repeated listings. When calculating, cancel common factors where possible before multiplying; this simplifies the arithmetic without changing the value.

Check that the answer is a whole number and interpret it in context. A combination count is a number of groups, committees, or other outcomes, not a probability.

\binom{n}{r}=\frac{n!}{r!(n-r)!}

Use combinations when selecting a fixed-size group and order does not matter.
Identify $n$ as the available total and $r$ as the number selected.
Divide the ordered-selection count by the number of orders within each group.

Worked example

Forming a committee

A community garden has 8 volunteers. How many different committees of 3 volunteers can be formed if every committee has three distinct members and committee roles are not assigned?

Identify the counting situation
The 8 volunteers are the available choices, and the outcome is a committee of 3. Since the committee has no assigned roles, changing the order in which its members are listed does not create a new committee. The members are distinct and each volunteer is selected at most once, so a combination applies.
Set up the combination
Let $n=8$ be the number available and $r=3$ the number selected. Use the combination formula because order does not matter.
$\binom{8}{3}=\frac{8!}{3!(8-3)!}$
Substitute and simplify
Replace $8-3$ with 5. Cancel the common factor CAD 5! from the numerator and denominator, then evaluate the remaining factorials.
$\frac{8!}{3!5!}=\frac{8\times7\times6}{3\times2\times1}=56$
Interpret the count
Each possible committee is counted once, even though its three members could be listed in six different orders. Therefore, 56 is the number of distinct committees that can be formed from these volunteers.

Answer: There are 56 possible committees.

Check: The count is a whole number, and the calculation accounts for the different orders that would otherwise repeat the same committee.

Common mistakes and how to avoid them

Multiplying when the outcomes are separate alternatives, or adding when choices occur in stages.

Correction: Describe how one outcome is formed. Multiply the choices across stages; add counts for alternatives that do not overlap.

Treating two differently ordered lists of the same committee as different committees.

Correction: Ask whether swapping the selected people changes the outcome. If not, use combinations.

Using n! for a selection of only some of the available items.

Correction: A factorial by itself counts arrangements of all

n

distinct items. For a group of

r

from

n

when order does not matter, use the combination formula.

Forgetting that

0!=1

or evaluating a factorial as a sum.

Correction: A factorial is a product, and CAD 0! is defined to equal 1.

Lesson summary

Use multiplication for successive stages and addition for non-overlapping alternatives.
A factorial multiplies a positive whole number down to 1; $0!=1$ .
Use combinations to count fixed-size selections of distinct items when order does not matter.
For a combination, identify the total $n$ and selection size $r$ , then evaluate $\binom{n}{r}=\frac{n!}{r!(n-r)!}$ .
Interpret a count in the context of the outcomes being counted.

Check your understanding

Question 1

A club has 7 members. How many different two-person groups can it form if the order of the two members does not matter?

14
21
42
49

Show answer and explanation

21

This is a combination with

n=7

and

r=2

. Thus

\binom{7}{2}=\frac{7!}{2!5!}=\frac{7\times6}{2}=21

groups.

Question 2

A code is made by choosing one letter from 3 choices and then one digit from 4 choices. How many codes are possible?

7
12
24
1

Show answer and explanation

12

The code has two successive stages, so multiply the choices:

3\times4=12

.

Key terms

Counting rule: A method for finding the number of possible outcomes without listing them all.
Factorial: For a positive whole number $n$ , the product $n(n-1)\cdots1$ , written n!; by definition, $0!=1$ .
Combination: A selection of items in which changing the order does not create a different selection.
Distinct: Different from one another, so that each item can be identified separately.
Outcome: One possible result of a counting situation, such as one particular committee or code.

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2.6 · Use counting rules, factorials, and combinations