L-PR-3: Solutions — Counting Techniques

A tag such as (C2 + C3) after a title names the Core Concepts parts the problem draws on, so you know where to look back.

Problem 1 — Counting-Probability Pipeline (C4 + C5 + C6)

The values change when the problem regenerates, but every version has the same structure. From people, choose an unordered group of . Of the people, have a stated trait, and exactly trait-holders are required.

Method: Use combinations because membership matters but selection order does not.

Total groups:

Favorable groups:

The two factors are multiplied because a favorable group must include both the trait-holders and the other members.

Probability:

Problem 2 — Combination Calculation and Symmetry (C4)

For the generated values and :

Choosing the objects to include uniquely determines the objects to exclude. Therefore:

For example:

Problem 3 — Fundamental Counting Principle (C1)

Each generated problem builds one complete outcome by making one choice at every stage. If the displayed stage counts are , then:

The Fundamental Counting Principle uses the number of choices actually available at each stage. Those counts may be constant or may shrink after earlier choices; probabilistic independence is not required.

For example, with 3 egg choices, 4 toast choices, and 2 drink choices:

Adding counts isolated menu options, not complete breakfasts.

Problem 4 — Permutation Calculation (C3)

When distinct positions are filled from objects without replacement and order matters, the available-choice counts shrink:

Apply the product rule:

For and :

Problem 5 — Factorial Arrangements (C2)

The generator asks for the number of ways to arrange all distinct objects, where .

For example:

A tag such as (C2 + C3) after a title names the Core Concepts parts the problem draws on, so you know where to look back.

Problem 1 — FCP Probability (C1 + C6)

Every generated version follows the same three-part plan:

1. Multiply the available-choice counts to find the total outcomes.
2. Replace the restricted stage count with its favorable count, then multiply again.
3. Divide favorable outcomes by total outcomes.

The three possible generated scenarios are:

Meal scenario

Route scenario

Outfit scenario

Problem 2 — Permutation vs. Combination Classification (C3 + C4 + C5)

• Ordered reading list, 4 from 9: permutation, because changing the order changes the list.

• Jury of 5 from 12, with no roles: combination, because only membership matters.

• Gold, silver, and bronze from 7 runners: permutation, because the medals are distinct.

• Two different scoops from 8, with stacking order declared irrelevant: combination.

• Four distinct password letters from 10: permutation, because changing the letter order changes the password.

Problem 3 — Combination-Based Probability (C4 + C6)

For people, including trait-holders, the probability that an unordered sample of contains exactly trait-holders is:

For the example , , , and :

Problem 4 — Find the Counting Error (C2 + C3 + C4 + C5)

The generator shows a three-step student solution. Choose the step where the first substantive error occurs; the feedback then explains why each step is correct or incorrect. It can generate any of these five error families:

1. Permutation used for an unordered group. Correct with because swapping selected members does not create a new group.
2. Repetition allowed when all code characters must differ. Replace with the shrinking product .
3. A correct combination multiplied by . The multiplication reintroduces ordering; the answer must remain .
4. All people treated as available for every one of chairs. Once a person is seated, that person is unavailable, so the correct count is , not .
5. “Choose nothing” treated as impossible. There is one empty selection, so .

Problem 5 — Hiring Committee Synthesis (C4 + C5 + C6)

(a) Total six-person groups:

(b) Exactly four experienced and two inexperienced candidates:

(c) Probability:

(d) Direct or complement? The direct calculation has three cases: exactly 4, 5, or 6 experienced candidates. The complement has four cases: exactly 0, 1, 2, or 3. Both methods are valid, but the direct method uses fewer terms here.

A tag such as (C2 + C3) after a title names the Core Concepts parts the problem draws on, so you know where to look back.

Question 1 — Feynman Test

The Fundamental Counting Principle always multiplies the number of choices actually available at each sequential stage. The counts do not have to remain constant, and probabilistic independence is not a requirement.

For a four-digit code made from digits 0–9 with no repeats, is wrong because it keeps all 10 digits available at every position. The pool shrinks after each chosen digit:

This is still an application of the Fundamental Counting Principle. The phrases “no repeats,” “distinct,” and “without replacement” signal shrinking available-choice counts.

Question 2 — Apply

Fix the president on the committee, then choose the remaining two members from the other seven:

The total number of unrestricted three-person committees is:

Therefore:

Complement check:

Question 3 — Error Analysis

The student used , which counts an unordered group. First, second, and third prizes are distinct positions, so changing who receives which prize creates a different outcome.

Relationship check:

A tag such as (C2 + C3) after a title names the Core Concepts parts the problem draws on, so you know where to look back.

Path A — The Cryptographer

The generator chooses a character-pool size from 30 through 36, a passcode length of 3 or 4, and a restricted first-position subset of size from 8 through 12.

Task 1 — Total with no repeated characters

The available counts shrink by one after every position:

Task 2 — First character from the -character subset

There are choices for the first character. After it is used, the remaining positions have choices:

Task 3 — Probability of a restricted first character

The interaction reports this value to three decimal places.

Task 4 — Total when repetition is allowed

The available-choice count stays at for all positions:

Model reflection: The Fundamental Counting Principle applies in both cases. With no repetition, each chosen character leaves one fewer character available, so the stage counts shrink. With repetition, a used character remains available, so every stage has choices. The distinction is replacement and the resulting stage-specific counts, not probabilistic independence.

Path B — The Tournament Director

Task 1 — Podium outcomes

Director B is correct. Gold, silver, and bronze are distinct ranks, so order matters:

Director A’s combination count ignores the internal medal order:

Task 2 — Three-team organizing committee

No roles are assigned, so order does not matter:

Task 3 — Falcons, Bears, and Eagles take all medals in any order

Task 4 — Falcons win gold and Bears win silver

Bronze can go to any of the remaining eight teams:

Model reflection: Tasks 1–2 differ because medals assign distinct ranks while committee members have equal roles. The swap test identifies the method: if swapping two selected teams changes the outcome, use a permutation; if it leaves the same group, use a combination. Tasks 3–4 then divide favorable ordered podiums by all 720 equally likely ordered podiums.

A tag such as (C2 + C3) after a title names the Core Concepts parts the problem draws on, so you know where to look back.

Problem 1 — Algebraic Proof + Combinatorial Argument

(a) Algebraic proof

(b) Combinatorial argument: Choosing the included objects uniquely determines the excluded objects, and choosing the excluded objects uniquely determines the included objects. This one-to-one pairing proves the two counts are equal.

For and , each pair from {A, B, C, D, E} maps to one complementary triple, so:

(c) Efficient evaluation

Problem 2 — Card Probability

There are 13 cards in the specified suit and 39 cards outside it. It is easier to count hands containing no cards from the suit:

Apply the complement rule:

Problem 3 — Restricted Arrangements

(a) Total seating arrangements

(b) Alex and Blake adjacent

Treat Alex and Blake as one block. Arrange the block and the other four people, then account for the two internal block orders:

(c) Alex and Blake not adjacent

(d) Probability