Counting questions ask how many distinct outcomes are possible without listing them one by one. The essential SAT decision is whether to multiply stage choices directly, count ordered arrangements, or count unordered groups.
Learning objectives
- Use the Fundamental Counting Principle for sequential stages.
- Evaluate factorial expressions and cancel factorial ratios efficiently.
- Recognize when order, rank, position, or role makes a permutation necessary.
- Recognize when only group membership matters and use a combination.
- State whether repetition is allowed before choosing a method.
Fundamental Counting Principle
Multiply choices across sequential stages
A three-stage access-code process has 3 symbol choices, 4 digit choices, and 2 color choices, producing 24 distinct codes.
- Stage 1Choose a symbol3 choices
- Star
- Circle
- Triangle
- Stage 2Choose a digit4 choices
- 2
- 4
- 6
- 8
- Stage 3Choose a color2 choices
- Mint
- Peach
Multiply the stages3 × 4 × 2 = 2424 total outcomes
Build a meal bundle
A café offers \(5\) sandwiches, \(3\) sides, and \(2\) drinks. How many bundles contain one choice from each category?
- Identify stages
The three stages have \(5\), \(3\), and \(2\) choices.
- Multiply
Apply the counting principle: \(5\times3\times2=30\).
Factorial notation
- Factorial
- For the positive whole-number context of this chapter, \(n!\) is the product of \(n\) and every positive whole number below it through \(1\).
For example, \(5!=5\cdot4\cdot3\cdot2\cdot1=120\).
Cancel a factorial ratio
Evaluate \(\frac{8!}{5!}\).
- Expand only what is needed
Write \(8!=8\cdot7\cdot6\cdot5!\).
- Cancel
Then \(\frac{8\cdot7\cdot6\cdot5!}{5!}=8\cdot7\cdot6\).
Permutations: order matters
- Permutation
- A permutation is an arrangement in which changing order, rank, position, or assigned role creates a different outcome.
Here \(n\) is the available number and \(r\) is the number of ordered positions filled.
Assign distinct roles
From \(8\) volunteers, a coordinator, recorder, and presenter are assigned. How many assignments are possible?
- Test order
The three roles are different, so changing who holds each role changes the outcome.
- Calculate
Use \({}_8P_3=8\cdot7\cdot6=336\).
Combinations: order does not matter
- Combination
- A combination is a selection in which only membership matters. Selecting Kai and Mira is the same group as selecting Mira and Kai.
The factor \(r!\) removes the different orders of each selected group.
Choose an unordered team
A design club has \(9\) members. How many \(4\)-member review teams are possible?
- Test order
The prompt assigns no roles, so rearranging the same four members changes nothing.
- Calculate
Use \({}_9C_4=\frac{9!}{4!5!}=126\).
Choose the method
Permutation or combination?
A two-branch decision identifies the correct counting method by asking whether changing order creates a new outcome.
Would changing the order or assigned roles create a different outcome?
| Concept | Does order matter? | Formula or rule | Typical situation |
|---|---|---|---|
| Fundamental Counting Principle | Depends on the stages | Multiply legal choices at each stage | Build a code or select one item from each category |
| Permutation | Yes | \({}_nP_r=\frac{n!}{(n-r)!}\) | Assign ranks, seats, or distinct roles |
| Combination | No | \({}_nC_r=\frac{n!}{r!(n-r)!}\) | Choose a committee, team, or subset |
A reliable counting checklist
- Describe one outcome
Ask exactly what would make two outcomes different.
- Check repetition
Determine whether a selected object can be used again.
- Test order
If swapping selected objects changes the result, order matters.
- Calculate
Use stage multiplication, \({}_nP_r\), or \({}_nC_r\) and simplify carefully.
Mixed reasoning
Combine an unordered and ordered decision
A panel chooses \(3\) of \(7\) proposals for review, then labels one selected proposal as the lead. How many results are possible?
- Choose the group
Select \(3\) proposals without order: \({}_7C_3=35\).
- Assign the lead
Any of the \(3\) selected proposals can be lead.
- Multiply stages
The total is \(35\times3=105\).
Common mistakes and traps
- Using a combination when ranks, seats, or assigned roles make order matter.
- Using a permutation for an unordered team and counting the same group repeatedly.
- Keeping the same number of choices after selecting an object that cannot repeat.
- Expanding every factorial instead of canceling common factors first.
- Reversing \(n\) and \(r\); \(n\) is available and \(r\) is selected or arranged.
- Assuming every counting problem requires \({}_nP_r\) or \({}_nC_r\) when direct multiplication is simpler.
Order test
Ten finalists are available. In how many ways can judges select a winner and runner-up?
- \({}_{10}C_2\)
- \({}_{10}P_2\)
- \(10!\)
- \(2^{10}\)
Show answer and explanation
Answer: \({}_{10}P_2=90\)
Winner and runner-up are distinct ranks, so reversing two finalists creates a different result.
What to remember
- Multiply the number of legal choices at each sequential stage.
- A permutation counts ordered arrangements; a combination counts unordered selections.
- Ask whether changing order creates a different outcome before choosing a formula.
- State repetition conditions clearly and decrease choices when reuse is forbidden.
- Cancel factorials before multiplying large numbers.
Put these notes into practice
Apply the ideas with SAT-style questions, then reinforce key details with flashcards.