L–PR–3 · GUIDED INQUIRY Introduction to Probability and Statistics

L–PR–3 · Guided inquiry

How can we count without listing?

Counting Techniques

What changes when a sample space is too large to write out?

Anonymous check

Which comparison is correct?

A six-digit PIN allows repeated digits. A six-letter password uses distinct letters from the alphabet.

  1. \(10^6=1{,}000{,}000\) PINs; \({}_{26}P_6=165{,}765{,}600\) passwords

  2. \({}_{10}P_6\) PINs; \(26^6\) passwords

  3. Both counts are powers because both are sequences

  4. Both counts are combinations because symbols are selected

Inquiry map

What decisions build a counting model?

Six concepts in the approved order

C1 Multiply stages

Use the available number of continuations at each stage.

C2 Compress products

Factorials package descending products.

C3–C4 Order or membership?

Permutations preserve order; combinations remove it.

C5–C6 Choose and apply

Select the model, then count favourable and total outcomes consistently.

Every formula answers a prior modelling decision.

Retrieval from L-PR-1

What must be true before we count?

A bag contains five red and three blue marbles.

Investigate: What is the probability of drawing red, and why is the denominator eight?

  1. Favourable
    \(n(A)=5\)

    Five outcomes satisfy “red.”

  2. Total
    \(n(S)=5+3=8\)

    The sample space contains every marble.

  3. Probability
    \(P(A)=\frac{5}{8}\)

    Favourable outcomes divided by equally likely total outcomes.

Counting methods change how we find the counts—not what probability means.

Predict

C1 · Why multiply the stages?

A restaurant offers three starters, five mains, and four desserts.

Investigate: How many distinct three-course meals are possible?

  1. Stage one
    \(3\)

    Choose a starter.

  2. Stage two
    \(5\)

    For each starter, choose a main.

  3. Stage three
    \(4\)

    For each partial meal, choose a dessert.

Each complete path uses one choice from every stage.

Reveal

C1 · A complete path is one outcome.

Every starter–main pair can continue to any dessert.

Investigate: What operation counts all complete paths exactly once?

Product \(3\times5\times4\)

Multiply available continuations along a path.

Count \(3\times5\times4=60\)

There are sixty complete meals.

Multiplication counts combinations of stage choices; addition would count incomplete alternatives.

Fundamental Counting Principle

C1 · What does the product rule actually require?

A sequential process has a known number of continuations at every stage.

  1. Stage counts
    \(m_1,m_2,\ldots,m_k\)

    Record the available choices at each stage.

  2. Total
    \(m_1m_2\cdots m_k\)

    Multiply the stage-specific counts.

  3. Boundary
    Not an independence test

    Probabilistic independence is not required.

The factors may stay constant, shrink, or vary by branch.

Misconception check

C1 · No repeats still means multiply.

A four-digit code uses distinct digits.

Investigate: What changes after each digit is used?

  1. Available choices
    \(10,9,8,7\)

    One digit leaves the pool after each choice.

  2. Correct count
    \(10\times9\times8\times7=5040\)

    Use the shrinking stage counts.

  3. Incorrect model
    \(10^4=10000\)

    This silently allows repetition.

Replacement determines the factors; it does not determine whether multiplication is allowed.

Predict before naming the notation

C2 · How many arrangements of three books?

Arrange the distinct books A, B, and C in a row.

Investigate: How do the available choices change from position to position?

  1. First position
    \(3\)

    Any book may go first.

  2. Second position
    \(2\)

    Two books remain.

  3. Third position
    \(1\)

    The final book is forced.

The count is \(3\times2\times1=6\).

Factorials

C2 · A descending product gets its own symbol.

Factorials count arrangements of all distinct objects.

  1. Definition
    \(n!=n(n-1)\cdots2\cdot1\)

    Multiply every positive integer down to one.

  2. Example
    \(5!=120\)

    Five distinct objects have one hundred twenty arrangements.

  3. Growth
    \(10!=3{,}628{,}800\)

    Factorials grow quickly.

Factorial notation compresses the full descending product.

Boundary case

C2 · Why is doing nothing one way?

The empty arrangement is still one complete outcome.

Investigate: What value keeps the counting formulas consistent?

  1. Convention
    \(0!=1\)

    There is one way to arrange no objects.

  2. Combination check
    \(\binom{n}{0}=1\)

    There is one way to choose nothing.

  3. Permutation check
    \({}_{n}P_n=\frac{n!}{0!}=n!\)

    Arranging every object still gives the expected count.

\(0!\) is one, not zero.

Bridge to permutations

C2 · What survives factorial cancellation?

Only the factors needed for the selected positions remain.

Investigate: Which factors remain when three positions are filled from eight objects?

  1. Ratio
    \(\frac{8!}{5!}\)

    The shared tail cancels.

  2. Remaining product
    \(8\times7\times6\)

    Three descending factors remain.

  3. Result
    \(336\)

    The result counts ordered selections.

Factorial cancellation exposes the same shrinking product as the FCP.

Predict

C3 · Do three roles create distinct positions?

A club has eight members and elects a president, vice-president, and secretary.

Investigate: Does swapping two elected people create a different outcome?

  1. President
    \(8\)

    Eight choices for the first role.

  2. Vice-president
    \(7\)

    Seven members remain.

  3. Secretary
    \(6\)

    Six members remain.

Distinct roles make the selection ordered.

Derivation

C3 · Build the ordered count from the FCP.

Choose and place one distinct object at each position.

Falling product \(n(n-1)\cdots(n-r+1)\)

There are exactly r factors.

Closed form \(\frac{n!}{(n-r)!}\)

The unused factorial tail cancels.

A permutation is the FCP written for ordered selection without replacement.

Definition

C3 · When does a permutation fit?

A different order of the same selected objects must count as a different outcome.

  1. Notation
    \({}_{n}P_r\)

    Select and arrange r distinct objects from n.

  2. Formula
    \({}_{n}P_r=\frac{n!}{(n-r)!}\)

    Keep r descending factors.

  3. Decision
    Order matters

    Rankings, roles, schedules, and sequences are common signals.

Name the outcome before choosing the formula.

Worked example

C3 · Check the formula against the stages.

Elect three distinct officers from twelve club members.

Investigate: Can the compact formula and the stage product agree?

  1. Stage product
    \(12\times11\times10\)

    One factor for each distinct role.

  2. Permutation
    \({}_{12}P_3=\frac{12!}{9!}\)

    Cancel the unused tail.

  3. Count
    \(1320\)

    Both representations agree.

Use the expanded product as a reasonableness check.

Predict

C4 · What changes when the roles disappear?

Choose a three-person committee from eight candidates.

Investigate: How many times does an ordered count repeat the same committee?

  1. Ordered count
    \({}_{8}P_3=336\)

    Every listing order is counted.

  2. Orders per group
    \(3!=6\)

    Each committee appears in six orders.

  3. Unordered count
    \(\frac{336}{6}=56\)

    Collapse equivalent orders.

Removing internal order means dividing by the number of arrangements within each group.

Reveal

C4 · A committee is membership, not sequence.

The sets contain the same people even when their names are listed differently.

  1. Same committee
    \(\{A,B,C\}\)

    Listing C, A, B does not create a new group.

  2. Overcount removed
    \(\frac{{}_{8}P_3}{3!}\)

    Divide out all internal orders.

  3. Result
    \(\binom{8}{3}=56\)

    Count each membership set once.

The outcome is the group itself.

Definition

C4 · When does a combination fit?

Changing only the listing order does not create a new outcome.

  1. Notation
    \({}_{n}C_r=\binom{n}{r}\)

    Select r objects from n without order.

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

    Remove the r-factorial internal overcount.

  3. Decision
    Order does not matter

    Groups, teams, committees, and subsets are common signals.

A combination counts distinct memberships.

Symmetry

C4 · Why do both counts match?

Choosing eight from ten determines exactly which two are excluded.

Investigate: Why must the two counts match?

  1. Choose included
    \(\binom{10}{8}\)

    Identify the eight objects kept.

  2. Choose excluded
    \(\binom{10}{2}\)

    Identify the two objects left out.

  3. Same count
    \(\binom{10}{8}=\binom{10}{2}=45\)

    Each description determines the other.

Use symmetry as both a shortcut and a self-check.

Anonymous check

Which classification is correct?

Twelve members are available. Scenario A assigns president, vice-president, and secretary. Scenario B forms a three-person committee with no roles.

  1. A uses \({}_{12}P_3\); B uses \(\binom{12}{3}\)

  2. A uses \(\binom{12}{3}\); B uses \({}_{12}P_3\)

  3. Both use permutations because three people are selected

  4. Both use combinations because the same people may appear

Decision rule

C5 · What question chooses the formula?

Focus on the defined outcome—not on a keyword alone.

Investigate: Does swapping two selected items create a different outcome?

  1. Yes
    Permutation

    The positions, ranks, or sequence are part of the outcome.

  2. No
    Combination

    Only the selected membership matters.

  3. Stages instead
    FCP

    Use stage-specific counts when the process is naturally sequential.

Describe one outcome precisely; the method usually follows.

Misconception check

C5 · Everyday language can mislead.

A “combination lock” opens with an ordered code.

Investigate: Is three–seven–four the same outcome as four–seven–three?

  1. Order changes
    Different code

    Changing positions changes the outcome.

  2. Repeats allowed
    \(m^r\)

    Use a constant stage count.

  3. No repeats
    \({}_{m}P_r\)

    Use shrinking stage counts.

Mathematical combinations are unordered; a lock code usually is not.

Counting-based probability

C6 · Do both counts use the same outcome?

Both counts must describe outcomes at the same level of detail.

  1. Favourable
    \(n(A)\)

    Count outcomes satisfying the event.

  2. Total
    \(n(S)\)

    Count all equally likely outcomes.

  3. Probability
    \(P(A)=\frac{n(A)}{n(S)}\)

    Divide compatible counts.

Never mix ordered outcomes in one count with unordered outcomes in the other.

Worked example

C6 · Count the whole sample space first.

Choose three students from a class of twenty.

Investigate: If order does not matter, how many equally likely groups are possible?

  1. Total groups
    \(\binom{20}{3}\)

    Choose any three students.

  2. Calculation
    \(\frac{20\times19\times18}{3\times2\times1}\)

    Cancel before multiplying.

  3. Result
    \(1140\)

    This is the denominator.

Define the sample-space outcome before counting the event.

Worked example continued

C6 · Now count the favourable groups.

Eight of the twenty students are athletes. All three selected students must be athletes.

Investigate: Which pool supplies the favourable count?

  1. Favourable
    \(\binom{8}{3}=56\)

    Choose all three from the eight athletes.

  2. Probability
    \(\frac{56}{1140}=\frac{14}{285}\)

    Use the same unordered-group outcome model.

  3. Decimal
    \(\frac{14}{285}\approx0.049\)

    About four point nine percent.

The numerator and denominator are both counts of three-person committees.

Complement strategy

C6 · When is counting the opposite shorter?

“At least one athlete” includes several direct cases, but “none” is one case.

Investigate: Which route uses fewer counts?

  1. Complement event
    \(A^\prime=\text{no athletes}\)

    Choose all three from the twelve non-athletes.

  2. Complement count
    \(\binom{12}{3}\)

    One favourable case to count.

  3. Return to event
    \(P(A)=1-P(A^\prime)\)

    Subtract the simpler opposite event.

“At least” suggests checking the complement—but use it only when it is actually simpler.

Worked prediction

How large should an unordered group count be?

Choose a four-person committee from ten candidates, with no roles.

Investigate: Is \(\binom{10}{4}\) closer to two hundred or five thousand?

  1. Ordered comparison
    \({}_{10}P_4=5040\)

    This counts every internal order.

  2. Orders per committee
    \(4!=24\)

    Each membership appears twenty-four times.

  3. Unordered result
    \(\binom{10}{4}=\frac{5040}{24}=210\)

    The answer is near two hundred.

A magnitude prediction can expose a missing factorial division.

Find the error

Can you diagnose the overcount?

A student uses \({}_{9}P_4=3024\) for a four-person project team with no roles.

Investigate: What exactly has been counted more than once?

  1. Error
    Order was invented

    The team outcome contains no internal ranking.

  2. Overcount
    \(4!=24\)

    Every team appears once for each listing order.

  3. Correction
    \(\binom{9}{4}=\frac{3024}{24}=126\)

    Count each membership once.

Correct the outcome model before correcting the arithmetic.

Anonymous check

Which expression is favourable?

A class has thirty students: twelve athletes and eighteen non-athletes. Five students are selected. Exactly three must be athletes.

  1. \(\binom{12}{3}\binom{18}{2}\)

  2. \(\binom{30}{5}\)

  3. \({}_{12}P_3\,{}_{18}P_2\)

  4. \(\binom{12}{3}+\binom{18}{2}\)

Guided transfer

Build the hiring-committee sample space.

Fifteen candidates include nine with experience and six without. Six are selected for interviews.

Investigate: What is one equally likely outcome?

  1. Outcome
    One six-person group

    No interview roles are assigned.

  2. Total
    \(\binom{15}{6}=5005\)

    Count all unordered groups.

  3. Favourable
    \(\binom{9}{4}\binom{6}{2}=126\times15=1890\)

    Choose four experienced and two inexperienced candidates together.

Lock the outcome model, then multiply the required subgroup counts.

Anonymous check

Which route uses fewer cases?

The same six-person group is selected from nine experienced and six inexperienced candidates.

  1. Direct: exactly four, five, or six experienced

  2. Complement: exactly zero, one, two, or three experienced

  3. Both always require the same number of cases

  4. Complement is mandatory whenever the phrase “at least” appears

Boss path · Cryptographer

Transfer · What changes when repetition returns?

Compare a distinct-letter password with a repeated-digit PIN.

Investigate: Which condition controls whether the factors shrink?

No repetition \({}_{26}P_6=26\times25\times24\times23\times22\times21\)

Used letters leave the pool.

Repetition allowed \(10^6\)

All ten digits remain available at every position.

The FCP handles both models; availability determines the factors.

Boss path · Tournament director

Transfer · Same teams, different outcomes.

Ten teams compete for medals, then form an organizing committee.

Investigate: Why do the two counts differ by a factorial?

  1. Podium
    \({}_{10}P_3=720\)

    Gold, silver, and bronze are distinct ranks.

  2. Committee
    \(\binom{10}{3}=120\)

    Only membership matters.

  3. Relationship
    \(720=120\times3!\)

    Each committee membership has six medal orders.

A context can contain both ordered and unordered outcomes—define each one separately.

Objective evidence · Explain

Exit 1 · Explain shrinking factors.

Use a four-digit code with no repeated digits.

Investigate: Why does the FCP still apply even though the factors shrink?

  1. Name the stages
    \(10,9,8,7\)

    State what each factor counts.

  2. Separate ideas
    Replacement ≠ independence

    Explain why these are different claims.

  3. Conclude
    \(10\times9\times8\times7=5040\)

    Connect the model to the count.

Write or explain your reasoning before revealing the checklist.

Objective evidence · Apply

Exit 2 · Apply a required-member condition.

A three-person committee is selected from eight members. One specific president must be included.

Investigate: How many valid committees exist, and what is the probability a random committee includes the president?

  1. Fix the president
    \(\binom{7}{2}=21\)

    Choose the other two members from the remaining seven.

  2. Total committees
    \(\binom{8}{3}=56\)

    Count all equally likely groups.

  3. Probability
    \(\frac{21}{56}=\frac{3}{8}\)

    Compatible favourable and total counts.

A fixed required member reduces the number of remaining choices.

Objective evidence · Diagnose

Exit 3 · Diagnose a method error.

A student claims that three ranked prizes among eight students give \(\binom{8}{3}=56\) outcomes.

Investigate: What is the precise modelling error?

  1. Diagnosis
    Ranks were ignored

    First, second, and third are distinct positions.

  2. Correction
    \({}_{8}P_3=8\times7\times6=336\)

    Use an ordered selection.

  3. Cross-check
    \(56\times3!=336\)

    Restore all internal rankings.

A correct calculation with the wrong outcome model is still wrong.

Closing synthesis

A count is a model of an outcome.

Carry this decision sequence into every counting probability problem.

1 · Define What is one outcome?

A path, an arrangement, a group, or something else?

2 · Decide Does order matter?

Use stages, a permutation, or a combination.

3 · Count What is available?

Track replacement, restrictions, and subgroup requirements.

4 · Compare \(P(A)=\frac{n(A)}{n(S)}\)

Keep favourable and total outcomes compatible.

The formula is trustworthy only when the outcome model is visible.

Anonymous exit reflection · confidence, not mastery

How confident are you now?

Use your exit responses as evidence. Rate every statement from \(0\) to \(5\), then submit once.

C1

I can count a multi-stage process using the available choices at each stage.

C2

I can evaluate factorials and explain why zero factorial equals one.

C3–C5

I can decide whether order matters and use a permutation or combination correctly.

C6

I can count compatible favourable and total outcomes to find a probability.

0 Not yet 1 Very unsure 2 Beginning 3 Mostly understand 4 Confident 5 I could explain it