Discrete Mathematics for BSAI

01 · Opening Challenge: Same Features, Different Order?

A learning-support team has eight candidate measurements. It can select three for a dashboard, or rank the top three in first, second and third place. Are those two tasks asking for the same number?

Predict: The selections {Q,E,A} and {A,Q,E} contain the same measurements. The rankings (Q,E,A) and (A,Q,E) assign different roles. Week 11 counted 8×7×6=336 rankings. We now find how many distinct selected sets those rankings represent.

02 · Learning Objectives and Three-Hour Guide

By the end of Week 12, you should be able to count ordered assignments with simple restrictions; group ordered listings that give one unordered selection; explain factorial, permutation and combination notation; choose an appropriate counting method; and complete the eight-feature case with justified counts and limits on interpretation.

Classroom essentials Sections 01–11 and Practice 1–6. Optional self-study Colab and Practice 7–8. List tiny outcomes before relying on a formula.

03 · First Count Ordered Positions

Four distinct applicants fill the roles of chairperson and deputy, with no person in both roles. There are four choices for chairperson. Once chosen, three remain for deputy: 4×3=12 assignments. Swapping the two people changes who has which role.

With n distinct candidates and r ordered positions, filled without reuse, the number of choices decreases one by one:

04 · Apply a Restriction Before Multiplying

With five distinct applicants, assign a chairperson and deputy. If applicant A must be chairperson, the first position has one choice and the deputy has four: 1×4=4. If A cannot be chairperson, the first position has four choices, and each leaves four people for deputy: 4×4=16.

The final check 4+16=20 works because “A first” and “A not first” are separate cases covering every unrestricted assignment. The repeated number 4 in 4×4 has two different meanings: four eligible first choices, then four remaining people for the second role.

05 · Group Orders That Make the Same Selection

From three distinct codes Q, E, A, Week 11 listed six ordered pairs: QE, QA, EQ, EA, AQ, AE. But choosing two codes for a set has only three outcomes:

Each two-element set was counted twice in the ordered list. Therefore 3×2÷2=3 selections. More generally, for three selected distinct objects, there are 3×2×1=6 internal orders of those same objects. Dividing an ordered count by six gives the number of three-object selections when every selected triple has all six possible orders and there are no restrictions favoring one order.

06 · Factorial Is Shorthand for a Falling Product

For a positive whole number n, factorial is the product n!=n×(n−1)×···×2×1. For example, 4!=4×3×2×1=24. We set 0!=1; this makes the counting formulas work when no choices remain to be arranged.

Factorial is a compact way to write a product, not a new kind of outcome. To count only the first three ordered positions from five, use 5×4×3=60, not all 5!=120 full arrangements.

07 · Compact Notation After the Reasoning

Let 0≤r≤n. We use P(n,r), also written ₙPᵣ, for ordered arrangements of r distinct choices from n without reuse. Its formula shortens the falling product:

We use C(n,r), also written (ⁿᵣ) or “n choose r,” for unordered selections. Each selected set of r distinct objects appears in r! different orders, so:

These formulas require distinct objects, no repetition, and the stated order interpretation. For a small task, a list or stage-by-stage count is often clearer than substituting immediately.

08 · Choose the Counting Method

Problems may combine methods. For example, a restriction can be applied first, followed by a product count; disjoint restricted cases can then be added. Give a reason for each step.

09 · Choice or Rank?

Predict → Test → Explain Each card asks for a count. Identify whether roles matter, select an answer, and read the correction for repeated orders or restrictions.

10 · Major Case Study 3 Concludes: Exploring Possible Feature Sets

Scenario → Understand → Represent → Reason → Check → Interpret. We continue with the same eight distinct candidate measurements from Week 11. No measurement may be repeated within a set or ranking.

Understand and represent

A two-feature set treats {Q,E} and {E,Q} as the same outcome. An ordered pair treats (Q,E) and (E,Q) as different. A ranking of the top three also uses distinct positions.

Reason: base counts

Reason: a required measurement

Suppose every selected pair must contain Q. Fix Q and choose its partner from the seven other codes: 7 sets. Do not multiply by two: {Q,E} and {E,Q} are one set. If every selected triple must contain Q, choose two from the other seven: C(7,2)=7×6÷2=21 triples.

Reason: an incompatible pair

For a hypothetical design constraint, say Q and E may not appear together in a selected pair. Among 28 two-feature sets, exactly one is the forbidden set {Q,E}. Therefore 28−1=27 valid two-feature sets. If Q is also required, its allowed partner is one of the other six codes, so 6 valid pairs. The incompatibility is an exercise condition, not a finding about the actual measurements.

Check and submit

Check the ordered-pair count against the set count: 28×2=56. Check the triple count: 56×6=336. A case submission should state (1) what one outcome is, (2) whether order matters, (3) the count and reason for each row above, (4) the two restrictions, and (5) one sentence about what these counts cannot tell us.

11 · What the Counts Mean

The result 28 counts possible two-feature sets under the stated distinctness rule. It does not tell us whether any pair is useful. The result 27 depends on the additional prohibition of {Q,E}; removing that policy returns 28. The result 56 for three-feature sets is an inventory of possible selections, not a score for the best dashboard.

12 · Optional Colab: Check a Hand Count

Self-study · mathematics first First list the six two-feature sets from four codes Q,E,A,T: QE, QA, QT, EA, ET, AT. Python can verify the count after you have reasoned that C(4,2)=4×3÷2=6.

1 · Reuse loops and indices

range(4) supplies positions 0,1,2,3. For a chosen first position i, range(i + 1, 4) supplies only positions after it, so the pair is printed once and no code is paired with itself.

features = ["Q", "E", "A", "T"]
for i in range(4):
    for j in range(i + 1, 4):
        print(features[i], features[j])

Predict the six printed pairs before running. This small list checks your hand enumeration; it does not prove a general formula on its own.

2 · Optional ready-made enumeration

The standard-library function combinations lists selections without order. import makes that function available. With the same four-code list, the printed result should again contain six pairs.

from itertools import combinations
features = ["Q", "E", "A", "T"]
print(list(combinations(features, 2)))

To check the eight-feature case, extend the list to all eight codes and change nothing else. The output should contain C(8,2)=28 pairs; explain why it prints (Q,E) only once and no (E,Q) separately.

13 · Practice and Case Submission

Decide first whether order matters, then justify the count. Selected solutions are at the end.

1. From five distinct books, select two to borrow. Count selections. Separately, how many first-and-second reading orders can be formed from the five books?
2. From six distinct applicants, rank first, second and third. Count rankings. How many three-applicant sets?
3. Four distinct tasks are arranged in a row. How many orders? Expand the factorial rather than giving only a symbol.
4. From five distinct applicants, assign chairperson and deputy, but applicant A must be chairperson. Count. Repeat when A cannot be chairperson.
5. For the eight case-study features, count selected pairs containing Q. How many selected pairs remain if {Q,E} is forbidden? How many remain if both restrictions apply?
6. For eight features, find the number of three-feature sets by grouping the 336 rankings. Explain exactly why you divide by 3!.
7. At a student event, 12 joined coding, 10 joined design, and 3 joined both. How many joined at least one? State the counting method.
8. Seven papers are distributed among three folders. What minimum folder size is guaranteed? Give a short capacity argument.

14 · Common Mistakes

15 · Summary

• For ordered positions without reuse, multiply the number of remaining choices at each stage: P(n,r)=n!/(n−r)!.
• For selections without order, each selected set has r! orders: C(n,r)=P(n,r)/r!.
• Factorial is a product shorthand, with 0!=1.
• Handle a restriction by defining allowed outcomes first; add separate cases or subtract forbidden outcomes only when they have been counted correctly.
• In the eight-feature case, there are 56 ordered pairs, 28 two-feature sets, 336 ranked triples, and 56 three-feature sets under the stated assumptions.

16 · Visual Concept Map

17 · Looking Ahead

Week 13 begins graph theory. We will model buildings and walkways as vertices and edges, then examine routes that inspect each walkway. The skill of stating exactly what counts as a distinct object or route continues to matter.