Discrete Mathematics for BSAI

01 · Opening Scenario: How Many Options?

A student can take one of three routes to campus. For each route, there are two available travel modes. How many route-and-mode plans are possible?

Predict: Sketch the choices first. Route A can pair with either mode, giving two plans. The same is true for routes B and C. There are 2+2+2=6 complete plans, also written 3×2=6.

02 · Learning Objectives and Three-Hour Guide

By the end of Week 11, you should be able to add counts for separate cases, multiply choices across stages, correct double counting for two overlapping sets, give a small pigeonhole argument, count ordered arrangements of distinct objects, and explain when order changes an outcome.

Classroom essentials Sections 01–11 and Practice 1–6. Optional self-study Colab and Practice 7–8. The feature-set case continues in Week 12; this week establishes the distinction between selecting and ranking.

03 · Add Counts for Separate Cases

A menu offers three soups or two salads as a single starter. No option is both a soup and a salad. To count one starter, list the disjoint cases: three soups plus two salads gives 3+2=5 choices.

Here A is the set of soup choices and B the set of salad choices. The notation says the combined set has the sum of their sizes when the groups share no item. This is the addition rule for separate cases.

04 · Multiply Counts Across Stages

Now a meal consists of one of three soups and then one of two drinks. Each soup pairs with two drinks, so there are 3×2=6 meals. We count complete pairs, not individual menu items.

This is the product rule: if stage 1 has m choices and each stage-1 choice permits n stage-2 choices, there are m×n complete outcomes. We need the same number of second-stage choices for every first-stage choice in this simple form.

05 · Which Structure Is Present?

Words such as “or” and “and” can help, but the actual outcomes decide the method. Ask whether one outcome comes from choosing a single case, or from completing several stages.

For the third row, a branch can itself use multiplication. The whole plan is counted by cases around stages. Check that cases do not overlap before adding.

06 · Correct a Double Count

In a small club survey, 12 students chose coding, 9 chose design, and 4 chose both. How many chose coding or design, including those who chose both?

Adding 12+9=21 counts each of the four “both” students twice. Subtract one extra copy of each:

The regions are 12−4=8 coding only, 4 both, and 9−4=5 design only. This is the inclusion–exclusion rule for two sets. It counts “at least one,” not “exactly one.” Exactly one would be 8+5=13.

07 · When Sharing Is Unavoidable

Place four cards into three boxes. Is it possible for every box to hold at most one card? Three boxes could then hold at most three cards. The fourth card forces some box to have at least two.

This is the pigeonhole principle: more objects than boxes force at least one shared box. It guarantees that a shared box exists; it does not say which box, or that every box is occupied.

Discuss: Does four cards in three boxes guarantee that exactly two share a box? No. All four cards could be in one box. The guarantee is “at least two.”

08 · Does Order Change the Outcome?

Suppose the three distinct cards are A,B,C. Choosing cards A and B for a set gives the same selection as choosing B and A. But ranking A first and B second differs from ranking B first and A second.

We will count ordered placements this week. Week 12 develops selection without order in more detail and compares the two systematically.

09 · Arrange Distinct Objects

Arrange three distinct cards A, B, C in a row. There are three choices for the first position, then two remaining choices for the second, and one for the third:

They are ABC, ACB, BAC, BCA, CAB, CBA. This is an ordered arrangement, or permutation, of distinct objects. We have counted them by stages, without needing a new formula.

For all four distinct cards in a row, the counts fall as the cards are used: 4×3×2×1=24. These stages represent different positions, and no card is reused.

10 · Counting Detective

Predict → Test → Explain Read the entire outcome, decide how to count it, and select a number. After each answer, read the explanation before moving on.

11 · Major Case Study 3 Begins: Exploring Possible Feature Sets

Scenario. A learning-support team considers eight candidate measurements for a future dashboard. The task here is purely combinatorial: count possible choices and rankings under stated rules. We have no evidence that any feature predicts learning well, and counting possibilities cannot assess predictive quality.

Understand: what is one outcome?

If the team selects three measurements, {Q,E,A} and {A,Q,E} represent the same chosen set. If the team ranks the top three, (Q,E,A) and (A,Q,E) are different assignments to first, second, and third place.

Represent: count a ranking by stages

For three ranked positions with no repeated measurement, there are eight choices for first, seven for second, and six for third. Thus 8×7×6=336 rankings. Do not use 336 as the number of three-feature sets: each selected set appears in several orders.

Reason with a small sample

Restrict attention temporarily to Q,E,A. For a pair, (Q,E) and (E,Q) are different ordered pairs but the same unordered selection {Q,E}. Write all ordered pairs from those three codes without repetition, then group pairs that select the same two codes. This prepares the Week 12 count from all eight features.

We will resolve these counts and restrictions in Week 12. For now the deliverable is a clear written distinction between selection and ranking, plus the small-pair grouping.

12 · Optional Colab: Enumerate Tiny Outcomes

Self-study · mathematics first Hand-list the six route-and-mode pairs and the six ordered two-card placements from A, B, C before using Python. These cells reuse lists and loops from Week 9; there are no imports.

1 · Complete two stages

The outer loop picks a route. For each route, the indented inner loop picks a mode. One printed line is one complete plan.

routes = ["A", "B", "C"]
modes = ["walk", "bus"]
for route in routes:
    for mode in modes:
        print(route, mode)

Predict six lines: A walk, A bus, B walk, B bus, C walk, C bus. The second stage has two possibilities for every first-stage choice.

2 · Ordered placements without reuse

Now the two roles are “first” and “second.” The if condition skips using the same card twice.

cards = ["A", "B", "C"]
for first in cards:
    for second in cards:
        if first != second:
            print(first, second)

Predict 3×2=6 printed placements: AB, AC, BA, BC, CA, CB. Explain why the outputs AB and BA are different rankings but describe the same two-card set. The code checks only tiny hand counts, not feature quality.

13 · Practice and Short Counting-Method Quiz

For each problem, state what one complete outcome is. Show a short reason, not only an arithmetic expression.

1. A menu has 4 teas or 3 juices as a single drink, with no drink in both groups. How many drink choices?
2. Choose one of 4 teas and one of 3 snacks. How many tea-snack pairs?
3. At a workshop, 15 students chose Python, 11 chose R, and 6 chose both. How many chose at least one? How many chose exactly one?
4. Eight objects go into three boxes. Explain why some box contains at least three objects.
5. Arrange three distinct books P, Q, R on a shelf. Count all orders, then list them.
6. Assign a first and second speaker from five distinct volunteers, without reusing a person. How many assignments?
7. A library offers two digital tutorials with two language options each, or three printed guides. How many single learning-resource options are there if each digital language version counts separately and no guide is also a digital version?
8. For case study codes Q, E, A, list the ordered pairs of distinct codes and group those with the same underlying two-feature selection. Explain why counting all ranked triples from eight codes does not directly count three-feature sets.

14 · Common Mistakes

15 · Summary

• Add the counts of separate cases; multiply across stages when each first-stage choice has the stated number of continuations.
• For two overlapping sets, subtract their intersection once: |A∪B|=|A|+|B|−|A∩B|.
• More objects than boxes force some box to share; small capacity arguments give stronger guarantees.
• An ordered assignment changes when roles or positions are exchanged. Distinct items cannot be reused if the task forbids it.
• Counting candidate feature choices does not identify good features or evaluate any predictive model.

16 · Visual Concept Map

17 · Looking Ahead

Week 12 completes the feature-set case: count selections without order, compare them with rankings, and handle simple restrictions. We will introduce compact notation only after we have checked small examples by listing and grouping.