Counts the ways something can happen. Distinguishes the addition principle for alternatives from the multiplication principle for stages, then separates permutations from combinations by whether order matters — the single question that decides nearly every problem in the section.
Subject: Precalculus · 65 slides · symbolic lesson
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Title
Precalculus · Chapter 11 — Sequences, Probability and Counting Theory
§11.5 Counting Principles, pp. 1346-1357
Objectives
Five things, each one you can check yourself on paper.
OpenStax, Precalculus, §11.5 Counting Principles §11.5, pp. 1346-1357 — the pages these objectives are drawn from
Warm-up
Two questions about the same three people.
Discussion prompt
From Alice, Bob and Cara: how many ways to pick a president and a vice-president, against how many ways to pick a two-person committee?
Hint: Does swapping the two roles change anything?
Answer:
For the offices, Alice as president and Bob as vice differs from Bob as president and Alice as vice. Those are two outcomes.
For the committee, picking Alice and Bob gives the same committee whichever order you name them in. That is one outcome.
So the two questions have different answers — six and three. The difference is entirely whether order matters, and that single question runs through the whole section.
Concept
Counting arrangements and counting selections are different problems with different formulas, and the only thing distinguishing them is whether reordering produces a new outcome.
\[ P(n,r)=\frac{n!}{(n-r)!}, \quad C(n,r)=\frac{n!}{r!(n-r)!} \]
The two formulas differ by one factorial, which divides out the orderings. Everything else in the section is the addition and multiplication principles, which are simpler and less often confused.
Figure (svg): A card contrasting arrangements where order matters with selections where it does not, using the same three letters
OpenStax, Precalculus, §11.5 Counting Principles §11.5, pp. 1346-1350
Section
Section 1
Concept
Choosing one thing from separate groups means adding the counts. Making a choice at each of several stages means multiplying them.
The non-overlap requirement matters: if an item belongs to two groups, adding counts it twice. That case needs a correction, which is why the addition principle is usually stated for disjoint alternatives.
Figure (svg): A contrast between the addition principle, used for alternatives, and the multiplication principle, used for stages
OpenStax, Precalculus, §11.5 Counting Principles §11.5, pp. 1346-1350
Picture it
Read the problem for or against and.
Figure (svg): A contrast between the addition principle, used for alternatives, and the multiplication principle, used for stages
The second row on each side is the practical test. Those two words appear in almost every counting problem and settle which principle applies.
Worked example
A choice at each stage.
\[ \text{A menu has } 4 \text{ starters, } 6 \text{ mains and } 3 \text{ desserts. How many three-course meals?} \]
Identify the structure
Why: One from each course.
Apply multiplication
Why: Counts multiply.
\[ 4 \times 6 \times 3 \]
Compute
Why: The total.
\[ 72 \]
Note why
Why: Each starter pairs with every main.
Figure (svg): A tree diagram showing three choices at the first stage and two at the second, giving six outcomes
\[ 72 \]
Verify: check with a smaller case
Why: With 2 starters and 3 mains there would be 6 pairings, which is small enough to list. The multiplication principle is confirmed on a case you can check by hand before trusting it on a larger one.
OpenStax, Precalculus, §11.5 Counting Principles §11.5, pp. 1347-1349
Sorting
Alternatives or stages.
Sort into buckets
Sort each situation.
Worked example
One choice from separate groups.
\[ \text{A shop has } 5 \text{ novels and } 8 \text{ biographies. How many ways to buy one book?} \]
Identify the structure
Why: One book, from either group.
Check the groups are separate
Why: No book is both.
Apply addition
Why: Counts add.
\[ 5 + 8 \]
Compute
Why: The total.
\[ 13 \]
Figure (svg): The solution to Worked example add across alternatives shown as a ladder of expressions, one row per legal move
\[ 13 \]
Verify: check why multiplication would be wrong
Why: Multiplying would give 40, which counts pairs of books rather than single ones. Only one book is being bought, so the choices are alternatives rather than stages.
OpenStax, Precalculus, §11.5 Counting Principles §11.5, pp. 1349-1350
Trap
\[ 5\text{ novels},\;8\text{ biographies} \;\Longrightarrow\; 5\times 8=40 \]
Multiply because two numbers are given
Why: The structure is not read.
That counts pairs of books when only one is being bought.
Multiply for stages and add for alternatives. Buying one book is a single choice from a combined pool.
The signal is the word: 'or' means alternatives and 'and' means stages.
Reading the structure before reaching for an operation is the whole discipline here, since both operations are available and only one fits.
Prediction
Three shirts and four pairs of trousers, making an outfit.
Predict first
How many outfits?
Correct: Twelve.
Why: An outfit needs one of each, which is a choice at each stage, so the counts multiply. Adding would count single garments rather than outfits.
Faded example
Four starters, six mains, three desserts.
Fill in the blanks
4\times 6\times3=72
Why: Each course is a stage and the choices are independent, so the counts multiply. A tree diagram shows why: every branch at one stage splits into the same number at the next.
Step zero
You are given a counting problem.
Discussion prompt
What do you read for before computing?
Hint: Two words.
Answer:
Or against and. Alternatives use the addition principle and stages use the multiplication principle.
If the problem says one item from either group, the counts add. If it says one from each group, they multiply.
Both operations are available and only one fits, so reading the structure is what decides the answer. The arithmetic afterwards is trivial by comparison.
Section
Section 2
Concept
Whether two selections with the same members but different orders count as one outcome or two is the question that decides which formula applies.
Words that signal order matters include arrangement, ranking, sequence, and any situation with distinct roles. Words signalling it does not include committee, group, team, selection, and hand of cards.
Figure (svg): A card contrasting arrangements where order matters with selections where it does not, using the same three letters
OpenStax, Precalculus, §11.5 Counting Principles §11.5, pp. 1350-1353
Picture it
Yes on the left, no on the right.
Figure (svg): A card contrasting arrangements where order matters with selections where it does not, using the same three letters
The examples under each side are worth memorising as patterns. Races and passwords have order; committees and card hands do not.
Worked example
Ask whether swapping changes anything.
\[ \text{Choosing } 3 \text{ people from } 10 \text{ for a committee: does order matter?} \]
Pick two arrangements
Why: Same three people, different order.
Compare the outcomes
Why: The same committee.
Conclude
Why: Order does not matter.
Choose the formula
Why: The one dividing out orderings.
\[ C(10, 3) \]
Figure (svg): A card contrasting arrangements where order matters with selections where it does not, using the same three letters
\[ \text{combination: }C(10,3) \]
Verify: contrast with a role-based version
Why: If the three were president, secretary and treasurer, the same three people in different roles would be different outcomes — a permutation. The people are the same and only the question changed, which is what makes the distinction subtle.
OpenStax, Precalculus, §11.5 Counting Principles §11.5, pp. 1351-1352
Sorting
Ask whether reordering changes the outcome.
Sort into buckets
Sort each situation.
Worked example
Distinct roles.
\[ \text{Choosing } 3 \text{ people from } 10 \text{ for president, secretary and treasurer: which type?} \]
Pick two arrangements
Why: Same three, roles swapped.
Compare the outcomes
Why: Different people in charge.
Conclude
Why: Order matters.
Choose the formula
Why: The arrangement one.
\[ P(10, 3) \]
Figure (svg): The solution to Worked example a case where order matters shown as a ladder of expressions, one row per legal move
\[ \text{permutation: }P(10,3) \]
Verify: compare the two answers
Why: The committee gives 120 and the offices give 720 — six times more, which is exactly the number of ways to arrange three people. That factor of six is the orderings the combination divides out.
OpenStax, Precalculus, §11.5 Counting Principles §11.5, pp. 1352-1353
Error analysis
A student counts committees.
Annotate
On: \( 3\text{ from }10\text{ for a committee}: \; P(10,3)=720 \)
The answer is wrong by a factor of exactly the arrangements being over-counted. Asking whether reordering changes the outcome, before choosing a formula, is what prevents it.
Prediction
Choosing three from ten, as arrangements or as selections.
Predict first
Which count is larger?
Correct: The arrangements, by a factor of six.
Why: Each selection of three people can be arranged in six ways, so the arrangement count is six times the selection count. That factor is three factorial, which is what the combination formula divides out.
Faded example
For a selection of three items.
Fill in the blanks
3!=3\cdot 2\cdot1=6 \text___
Why: Three items can be arranged in three factorial ways. That is exactly the over-counting factor when a permutation formula is used for a selection problem.
Explain it
Two formulas, one question.
Discussion prompt
Explain to a classmate how to choose between them.
Hint: What test settles it?
Answer:
Pick two orderings of the same members and ask whether they are the same outcome or different ones.
Different means order matters and the permutation formula applies. Same means it does not, and the combination formula applies.
The test takes one sentence and settles the whole problem. A good explanation stresses doing it explicitly rather than relying on which formula the chapter is currently about, since both appear together in exercises.
Section
Section 3
Concept
A permutation counts the ways to arrange a chosen number of items from a group, where different orders count separately.
The descending product is often easier than the formula: for three from ten it is ten times nine times eight, which is the multiplication principle applied directly. The factorial quotient is the same thing written compactly.
Figure (svg): Two cards giving the permutation and combination formulas and the relationship between them
OpenStax, Precalculus, §11.5 Counting Principles §11.5, pp. 1353-1355
Picture it
The first card is this idea.
Figure (svg): Two cards giving the permutation and combination formulas and the relationship between them
The third card connects them. A combination is a permutation with the orderings divided out, which makes the extra factorial in its formula inevitable rather than arbitrary.
Worked example
A descending product.
\[ \text{How many ways to award gold, silver and bronze among } 8 \text{ runners?} \]
Count the gold choices
Why: Any of the eight.
\[ 8 \]
Count the silver choices
Why: One is taken.
\[ 7 \]
Count the bronze choices
Why: Two are taken.
\[ 6 \]
Multiply
Why: The multiplication principle.
\[ 336 \]
Figure (svg): Two cards giving the permutation and combination formulas and the relationship between them
\[ P(8,3)=336 \]
Verify: check against the formula
Why: Eight factorial over five factorial cancels down to eight times seven times six, which is 336 — the same descending product. The formula and the direct count agree, as they must, since the formula is just that product written compactly.
OpenStax, Precalculus, §11.5 Counting Principles §11.5, pp. 1354-1355
Prediction
Arranging four items from nine.
Predict first
What product counts the arrangements?
Correct: Nine times eight times seven times six.
Why: Each position has one fewer choice than the last, and there are four positions to fill. The product stops after four factors, which is what the formula's denominator accomplishes.
Worked example
The whole group.
\[ \text{How many ways to arrange all } 6 \text{ books on a shelf?} \]
Count the first position's choices
Why: Any of the six.
\[ 6 \]
Continue down
Why: One fewer each time.
\[ 5, 4, 3, 2, 1 \]
Multiply
Why: The descending product.
\[ 720 \]
Recognise it
Why: Six factorial.
\[ 6! \]
Figure (svg): The solution to Worked example arrange everything shown as a ladder of expressions, one row per legal move
\[ 6!=720 \]
Verify: check with the formula
Why: Arranging all six is P of six from six, which is six factorial over zero factorial. With zero factorial defined as one, that is just six factorial — which is why the convention was chosen as it was.
OpenStax, Precalculus, §11.5 Counting Principles §11.5, pp. 1355-1355
Trap
\[ 3\text{ medals from }8\text{ runners}: \; 8!=40320 \]
Take the factorial of the group size
Why: The number being chosen is ignored.
That counts arrangements of all eight runners, not of three medal positions.
Stop the descending product after three factors. Only three positions are being filled.
The formula's denominator does exactly that, cancelling the factors below the stopping point.
The descending product is the safer form for hand computation, since it makes the stopping point explicit rather than implicit in a cancellation.
Faded example
Three medals among eight runners.
Fill in the blanks
8\times 7\times6=336
Why: Each medal is a stage with one fewer runner available, so the multiplication principle gives a descending product. Three medals means three factors.
Sorting
One per position filled.
Sort into buckets
Sort each situation by the number of factors.
Explain it to yourself
Each factor is one less than the last.
Discussion prompt
Explain why the count descends.
Hint: What happens after each choice?
Answer:
Each position is a stage, so the multiplication principle applies and the counts multiply.
But each choice uses up one item, so the next stage has one fewer available. That is why the factors descend by one each time.
The product stops when all the required positions are filled. A good explanation notes that the factorial formula is just this product written compactly, with the denominator cancelling the factors below the stopping point.
Section
Section 4
Concept
A combination counts selections where order is irrelevant, obtained by counting arrangements and then dividing by the number of ways each selection could be arranged.
Seeing the combination formula as a permutation divided by a factorial makes the extra term inevitable. Memorising two independent formulas invites confusing them; deriving one from the other does not.
Figure (svg): Two cards giving the permutation and combination formulas and the relationship between them
OpenStax, Precalculus, §11.5 Counting Principles §11.5, pp. 1355-1357
Picture it
The third card is the derivation.
Figure (svg): Two cards giving the permutation and combination formulas and the relationship between them
The relationship also gives a check: a combination count times the factorial of the number chosen must give the permutation count for the same numbers.
Worked example
Arrangements, then divide.
\[ \text{How many } 3\text{-person committees from } 10 \text{ people?} \]
Count the arrangements
Why: A descending product.
\[ 10 \times 9 \times 8 = 720 \]
Count the orderings of three
Why: Three factorial.
\[ 6 \]
Divide
Why: To remove the over-counting.
\[ \frac{720}{6} \]
Compute
Why: The selection count.
\[ 120 \]
Figure (svg): Two cards giving the permutation and combination formulas and the relationship between them
\[ C(10,3)=120 \]
Verify: check the relationship
Why: Multiplying 120 by six gives 720, the permutation count — which is the relationship between the two formulas. That check confirms both computations at once.
OpenStax, Precalculus, §11.5 Counting Principles §11.5, pp. 1356-1357
Prediction
Converting arrangements of four items into selections.
Predict first
What do you divide by?
Correct: Four factorial.
Why: Each selection of four items can be arranged in four factorial ways, so that is how many times each selection was counted. Dividing by it removes exactly that over-counting.
Worked example
Choosing what to leave out.
\[ \text{Why does choosing } 3 \text{ from } 10 \text{ give the same count as choosing } 7? \]
Note what a choice determines
Why: The three chosen fix the seven left.
Note the reverse
Why: The seven left fix the three chosen.
Conclude
Why: The two counts match.
Check the formula
Why: The denominators swap.
Figure (svg): The solution to Worked example a symmetry shown as a ladder of expressions, one row per legal move
\[ C(10,3)=C(10,7) \]
Verify: check the formula's symmetry
Why: The formula's denominator is three factorial times seven factorial, which is unchanged by swapping the two — so the two counts are equal by the formula as well as by the argument. That symmetry can halve the arithmetic when the number chosen is more than half the group.
OpenStax, Precalculus, §11.5 Counting Principles §11.5, pp. 1357-1357
Error analysis
A student converts a permutation count to a combination count.
Annotate
On: \( C(10,3)=\frac{P(10,3)}{10!} \)
The divisor counts the orderings being treated as equivalent, which is the arrangements of the chosen subset rather than of the whole group. Naming what is being divided out prevents the confusion.
Faded example
From 720 arrangements of three chosen items.
Fill in the blanks
C=\frac6120=\frac______}=___
Why: Three items arrange in six ways, so each committee was counted six times among the arrangements. Dividing by six gives the number of distinct committees.
Sorting
Arrangements or selections.
Sort into buckets
Sort each description.
Explain it
The combination formula has one more factorial than the permutation formula.
Discussion prompt
Explain what it is doing.
Hint: What is being over-counted?
Answer:
The permutation formula counts every ordering separately. For a selection problem, all the orderings of the same members are the same outcome.
Each selection of r items has r factorial orderings, so it was counted that many times. Dividing by r factorial removes exactly that over-counting.
So the extra factorial is not a separate fact to memorise — it is the number of orderings being pooled. A good explanation notes that deriving the combination formula from the permutation one makes confusing them much harder.
Section
Section 5
Concept
Every counting problem is settled by asking whether the structure is alternatives or stages, and then whether order matters.
The last row is worth noting: when items can be reused, each stage has the full count available and the product does not descend. A four-digit code from ten digits allows repeats, giving ten to the fourth rather than a descending product.
| the situation | the tool |
|---|---|
| one choice from separate groups | add the counts |
| a choice at each stage | multiply the counts |
| arranging a subset, order relevant | permutations |
| selecting a subset, order irrelevant | combinations |
| stages with repetition allowed | multiply, without descending |
Figure (svg): A contrast between the addition principle, used for alternatives, and the multiplication principle, used for stages
OpenStax, Precalculus, §11.5 Counting Principles §11.5, pp. 1347-1357
Picture it
Alternatives or stages, before anything else.
Figure (svg): A contrast between the addition principle, used for alternatives, and the multiplication principle, used for stages
Once the structure is settled, the order question follows. Both are readings of the problem rather than computations, which is where the difficulty of this section genuinely lies.
Worked example
The product does not descend.
\[ \text{How many } 4\text{-digit codes using digits } 0 \text{ to } 9, \text{ repeats allowed?} \]
Count the first digit's options
Why: All ten.
\[ 10 \]
Count the second's
Why: Repeats allowed, so all ten again.
\[ 10 \]
Continue
Why: Four positions.
\[ 10, 10 \]
Multiply
Why: The total.
\[ 10000 \]
Figure (svg): A contrast between the addition principle, used for alternatives, and the multiplication principle, used for stages
\[ 10^4=10000 \]
Verify: contrast with no repeats
Why: Without repeats the product would descend: ten times nine times eight times seven, which is 5040 — about half as many. Whether repetition is allowed roughly doubles the count here, so it has to be read from the problem.
OpenStax, Precalculus, §11.5 Counting Principles §11.5, pp. 1348-1352
Prediction
A four-digit code where digits may repeat.
Predict first
What does the product look like?
Correct: Ten multiplied by itself four times.
Why: Repetition means each position has all ten digits available, so no factor is used up. Without repetition the product would descend, giving about half as many codes.
Worked example
Combinations at each stage, then multiply.
\[ \text{Choose } 2 \text{ from } 5 \text{ women and } 3 \text{ from } 6 \text{ men. How many groups?} \]
Count the women's selections
Why: Order irrelevant.
\[ C(5, 2) = 10 \]
Count the men's selections
Why: Order irrelevant.
\[ C(6, 3) = 20 \]
Identify the structure
Why: Both selections needed.
Multiply
Why: The two counts.
\[ 200 \]
Figure (svg): The solution to Worked example a two-stage problem shown as a ladder of expressions, one row per legal move
\[ 200 \]
Verify: check both decisions were made
Why: Each selection is a combination, since a group is unordered, and the two selections are stages rather than alternatives, so they multiply. Both readings were needed, and getting either wrong would give a very different answer.
OpenStax, Precalculus, §11.5 Counting Principles §11.5, pp. 1352-1357
Trap
\[ 4\text{-digit code from }10\text{ digits}: \; 10\times 9\times 8\times 7 \]
Descend the product automatically
Why: Repetition is assumed to be disallowed.
That forbids codes like 1123, which are perfectly valid unless the problem says otherwise.
Read whether repetition is allowed. If it is, each stage has the full count and the product does not descend.
Digits, letters and dice rolls usually allow repeats; drawing people or cards without replacement does not.
The two answers differ substantially, so the reading matters as much as the order question.
Sorting
Some situations reuse items and some do not.
Sort into buckets
Sort each situation.
Matching
Two readings settle every problem.
Match the pairs
Why: The first two are readings of the structure and the last two of the order question. Both readings have to be made, and either one wrong gives a wrong answer regardless of the arithmetic.
Explain it to yourself
The arithmetic in this section is easy.
Discussion prompt
Explain what actually makes these problems hard.
Hint: What has to be decided?
Answer:
The arithmetic is multiplication and factorials, neither of which is difficult. The difficulty is in the readings.
Three decisions have to be made from the wording: alternatives or stages, order relevant or not, and repetition allowed or not. Each has two possibilities and each changes the answer.
So a wrong answer here is usually a misreading rather than a miscalculation. A good explanation notes that writing the three decisions down explicitly before computing is what makes these problems reliable.
Comparison
Fill the blanks from memory. One question separates them.
Comparison matrix
| permutations | combinations | |
|---|---|---|
| order | matters | does not matter |
| counts | arrangements | selections |
| which is larger | this one | smaller, by r factorial |
| typical words | rank, arrange, code, race | committee, team, hand, group |
The last row is worth memorising as patterns. Those words appear repeatedly and each reliably signals which formula applies.
Pattern
Five steps, and the first three are readings rather than computations.
Steps 1 to 3 are three independent yes-or-no readings, and each changes the answer. Writing them down before computing is what makes the section reliable.
OpenStax Algebra and Trigonometry 2e, §13.5 Counting Principles §13.5
Check
The two principles.
Check your understanding
You choose one item from each of three groups. What do you do with the counts?
Answer: A
Why: One from each group means a choice at each stage, so every option at one stage pairs with every option at the next and the counts multiply. Adding would apply if only one item were being chosen from the combined pool.
Check
The order question.
Check your understanding
Choosing five cards for a hand: does order matter?
Answer: A
Why: A hand is the same whichever order the cards were dealt in, so orderings are pooled and the combination formula applies. Using the permutation formula would over-count by 120, the arrangements of five cards.
Check
The relationship.
Check your understanding
How does the combination count relate to the permutation count for the same numbers?
Answer: A
Why: Each selection of r items has r factorial orderings, all counted separately by the permutation formula. Dividing by that removes exactly the over-counting, which is where the combination formula's extra factorial comes from.
Real world
Password strength is a counting problem.
Discussion prompt
Why is a longer password disproportionately stronger than a more varied one?
Hint: Where does each appear in the count?
Answer:
The number of possible passwords is the character-set size raised to the length. The set size is the base and the length is the exponent.
Growing the base multiplies the count by a fixed factor; growing the exponent multiplies it by the whole base. So each extra character multiplies the possibilities by the full character-set size.
Which is why security advice favours length over complexity: adding one character to a lowercase password multiplies the possibilities by twenty-six, where adding capitals only doubles the base. The counting principle makes the advice quantitative rather than a matter of opinion.
Commit first
State your confidence along with your answer.
Predict first
What single question decides between permutations and combinations?
Correct: Whether reordering a selection gives a different outcome.
Why: If it does, orderings are counted separately and the permutation formula applies. If it does not, they are pooled and the combination formula divides them out. Repetition is a separate question that affects both.
Explain it
The test of understanding is being able to say why, not just what.
Discussion prompt
A classmate used the permutation formula for a committee problem. Explain the error.
Hint: How many times was each committee counted?
Answer:
A committee is the same whichever order its members are named, but the permutation formula counts each ordering separately.
With three members there are six orderings, so every committee was counted six times. Dividing by six — three factorial — gives the right count.
That division is exactly the extra factorial in the combination formula. A good explanation notes that deriving it this way makes the two formulas impossible to confuse, since one is visibly the other with the orderings removed.
Exit ticket
One honest answer, so the next lesson can start in the right place.
Predict first
Which idea from this lesson would you most want to see again?
Correct: Any of these is a legitimate answer; the useful one is the honest one.
Why: There is no correct choice here. The second is where nearly every wrong answer originates, since it is a reading rather than a computation. The fourth makes the two formulas impossible to confuse once seen as a derivation.
Connect it up
One page, drawn from memory, is worth more than rereading the section.
Draw it
Write the three yes-or-no readings every counting problem needs, with the tool each answer selects. Beside them, work the same numbers as a permutation and as a combination and show the factor between the answers. Underneath, draw a small tree diagram illustrating the multiplication principle.
If your two answers differ by exactly the factorial of the number chosen, and your three readings are written as questions rather than rules, the section's structure is on the page.
Recap
Five things, and three of them are readings rather than computations.
| if you remember one thing | it should be this |
|---|---|
| about the principles | or means add, and means multiply |
| about the order question | ask whether reordering changes the outcome |
| about the formulas | a combination is a permutation with orderings divided out |
| about repetition | if items are reused, the product does not descend |
Section 11.6 uses combinations to expand a binomial power, where the coefficients turn out to be exactly the selection counts from this section.
OpenStax, Precalculus, §11.5 Counting Principles §11.5, pp. 1346-1357 — everything on these slides traces back here
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