Today: Multiplication & Addition Principles
The two foundational rules of counting — and the prism topology bonus that earns AIMO Q2 marks.
What you will learn today
How this lesson is structured
- Phase 0 (Steps 2–3): Prerequisites — multiplication, factorial intuition, "and" vs "or" in English.
- Phase 1 (Steps 4–5): Visual intuition. Decision trees, Venn diagrams for mutually exclusive sets, an animated tree diagram, and the n-prism topology picture.
- Phase 1.5 (Step 6): Formula handbook — the four atomic skills laid out as one cheat-sheet.
- Phase 2 (Steps 7–9): Three guided derivations — multiplication, addition, prism Euler.
- Phase 3 (Steps 10–14): Five Worked Examples (⭐ → ⭐⭐⭐⭐⭐), each fully written out.
- Phase 4 (Step 15): Eight independent practice problems with hints + solutions.
- Phase 5 (Steps 16–18): Three real AIMO past papers in exam format with the 5-step Observe template.
- Phase 5.5 (Steps 19–20): Two synthesis problems combining ≥2 atomic skills each.
- Step 21: Atomic skill matrix + 8 micro-validations (2 per skill) summary.
- Step 22: Mini Mock Test (3 in-Part originals, auto-graded with full-solution reveal).
- Step 23: Summary, cheat sheet, ⭐ self-rating, and error-book preview.
Phase 0 — what you need to bring
Three small ideas. If any of these feel shaky, fix them now before continuing.
1. Multiplication is repeated independent choice
If you have \(a\) ways to do step 1 AND \(b\) ways to do step 2, with the choices not affecting each other, you have \(a \cdot b\) total outcomes. This is just multiplication, but the framing is what matters.
2. Addition is "split into non-overlapping cases"
If a question can be answered by EITHER doing it in case A (\(a\) ways) OR in case B (\(b\) ways), and the cases share no outcomes, the total is \(a + b\). The word here is "or", and the cases must be disjoint.
3. English keywords
- "and" → multiply (independent steps within one outcome)
- "or" → add (different mutually-exclusive cases)
- "at least one" → complement: \(\text{total} - \text{(none)}\)
- "no two adjacent equal" → multiply with one fewer choice for each later step
Phase 0 verification — five quick checks
Five 30-second problems. If you get all five, you are ready. Otherwise re-read Step 2.
A meal is one starter and one main. There are 3 starters and 4 mains. How many meals?
Answer
"and" → multiply: \(3 \cdot 4 = \mathbf{12}\).
A snack is one piece of fruit or one biscuit. There are 4 fruits and 5 biscuits. How many choices?
Answer
"or" with disjoint sets → add: \(4 + 5 = \mathbf{9}\).
Outfit = 1 shirt and 1 pant and 1 shoe. There are 3 shirts, 4 pants, 2 shoes. How many outfits?
Answer
\(3 \cdot 4 \cdot 2 = \mathbf{24}\).
A coin is flipped 3 times. There are 8 sequences total. How many sequences have at least one heads? (Hint: count "no heads" first.)
Answer
"No heads" = TTT only, that is 1 sequence. So "at least one heads" = \(8 - 1 = \mathbf{7}\).
A 4-digit binary string (each digit is 0 or 1). How many strings?
Answer
\(2 \cdot 2 \cdot 2 \cdot 2 = 2^4 = \mathbf{16}\).
The decision tree picture
Forget formulas. Just count leaves on a tree.
You own 3 shirts and 2 pairs of pants. An outfit is one shirt and one pair of pants. How many outfits?
Draw it as a tree. The trunk splits into 3 branches (one per shirt). Each branch then splits into 2 sub-branches (one per pant). Count the leaves at the bottom.
SVG 1 — animated decision tree. Branches grow in via stroke-dasharray; leaves fade in last.
Now the addition picture — Venn for mutually exclusive sets
What if instead the problem said "today's outfit is either a shirt-and-pants combo (6 ways) or a single dress (4 ways)"? You don't multiply 6 and 4. You add them, because the two cases share no outcomes.
SVG 2 — Venn diagram for mutually exclusive sets. Circles do not overlap, so the addition principle applies cleanly.
The n-prism — counting V, E, F by picture
A surprise topology bonus that earns AIMO 2004 Q2 marks. One picture, three formulas.
An n-sided prism is the 3D shape you get by taking an n-sided polygon and stretching it straight up. Two parallel n-gons (top and bottom) plus n rectangular sides connecting them.
The four atomic skills — at a glance
Memorise this table. Every problem in this Part is one of these four, or a combination.
| ID | Skill | Trigger words | Formula / pattern |
|---|---|---|---|
| C1-A1 | Multiplication principle | "and", consecutive steps | \(n_1 \cdot n_2 \cdots n_k\) |
| C1-A2 | Addition principle (mutually exclusive) | "or", disjoint cases | \(|A| + |B|\) when \(A \cap B = \emptyset\) |
| C1-A3 | Subtraction / complement counting | "at least one", "not all", "no" | \(\text{total} - \text{(complement)}\) |
| C1-A4 | n-prism topology | "n-sided prism", "edges/faces of a prism" | \(V = 2n,\ E = 3n,\ F = n + 2\) |
Derivation 1 — Multiplication principle
Why \(n_1 \cdot n_2\) really is the right count. From scratch.
Setup. Suppose step 1 has \(n_1\) outcomes \(\{a_1, a_2, \ldots, a_{n_1}\}\). After fixing one of these, step 2 has \(n_2\) outcomes (the same number for each step-1 outcome — that is the "independent" condition).
List the pairs. Pair each step-1 outcome with each step-2 outcome:
Total. \(n_1\) groups, each of size \(n_2\), no two sharing a pair (because step-1 labels differ). Total = \(n_1 \cdot n_2\).
Extension to \(k\) steps: apply the same argument repeatedly. The product is \(n_1 \cdot n_2 \cdots n_k\).
Derivation 2 — Addition principle
Why splitting a problem into disjoint cases just adds the counts.
Setup. Suppose the problem allows you to answer it in case A (with \(|A|\) outcomes) or case B (with \(|B|\) outcomes), and the cases are mutually exclusive: no outcome appears in both \(A\) and \(B\).
Then the total number of outcomes is \(|A \cup B| = |A| + |B|\).
Why? When sets are disjoint, listing all of \(A\) first then all of \(B\) gives a list of length \(|A| + |B|\) with no repeats. That list is exactly \(A \cup B\).
What if the cases overlap? Then we use \(|A \cup B| = |A| + |B| - |A \cap B|\) — the inclusion-exclusion principle (Week 8 · Part 2). For now, keep cases strictly disjoint.
Combined with multiplication. A typical problem has structure: "(do X with \(a\) ways AND Y with \(b\) ways) OR (do Z alone with \(c\) ways)" → answer is \(a \cdot b + c\). This is exactly WE2.
Derivation 3 — n-prism V, E, F formulas
Use the multiplication and addition principles directly to count vertices, edges, and faces.
Vertices: \(V = 2n\)
The prism has 2 polygons (top + bottom). Each polygon has \(n\) vertices. Top vertices and bottom vertices are disjoint sets, so by addition: \(n + n = 2n\).
Edges: \(E = 3n\)
Three disjoint kinds of edges:
- Top polygon edges: \(n\)
- Bottom polygon edges: \(n\)
- Vertical edges connecting top to bottom: \(n\)
Disjoint cases → addition: \(n + n + n = 3n\).
Faces: \(F = n + 2\)
Two disjoint kinds of faces:
- Side faces (rectangles): \(n\)
- Top + bottom polygon faces: \(2\)
Disjoint cases → addition: \(n + 2\).
Sanity check — Euler's formula
For any convex polyhedron: \(V - E + F = 2\). Plug in: \((2n) - (3n) + (n+2) = 2\). ✓ for every n.
Worked Example 1 — outfit count ⭐
The cleanest possible application of the multiplication principle.
A wardrobe has 3 shirts, 4 pants, and 2 pairs of shoes. An outfit is one shirt and one pant and one pair of shoes. How many distinct outfits are possible?
Strategy
Three independent steps joined by "and": pick a shirt (3 ways), pick a pant (4 ways), pick shoes (2 ways). Multiply.
Calculation
Worked Example 2 — breakfast options ⭐⭐
Combining multiplication and addition in one problem.
Breakfast is either (one of 3 hot dishes and one of 2 drinks) or (one of 4 cold dishes alone, no drink). How many breakfast options are there?
Strategy
Two disjoint cases. Case A = hot dish AND drink (multiply). Case B = cold dish alone (count directly). The cases are mutually exclusive (a hot+drink breakfast is not a cold-only breakfast). Add the two counts.
Calculation
Worked Example 3 — n-sided prism with 2004 edges ⭐⭐⭐
The exact AIMO 2004 Q2 setup. Apply the prism formula and solve a one-step equation.
An n-sided prism has 2004 edges. Find n and the number of faces.
Strategy
Apply the prism edge formula \(E = 3n\). Solve for n. Then plug into \(F = n + 2\).
Calculation
Worked Example 4 — 3-digit numbers, no two adjacent equal ⭐⭐⭐⭐
A constrained multiplication. Each step has a different available count.
Count the 3-digit numbers in which no two adjacent digits are equal. (Leading digit must be non-zero, as always for 3-digit numbers.)
Strategy
Three independent steps — but each later step's count depends on the previous digit (must differ from it). The available count is constant per step, so multiplication still applies.
Calculation
Worked Example 5 — school code, no repeats ⭐⭐⭐⭐⭐
A 5-step multiplication with two no-repeat constraints. The decreasing-pool pattern.
A school code consists of 2 uppercase letters followed by 3 digits, with NO repeated letter and NO repeated digit. How many such codes exist?
Strategy
Two independent halves: letter half (2 positions, no repeat) AND digit half (3 positions, no repeat). Each half is a "decreasing pool" multiplication.
Calculation
Phase 4 — 8 independent practice problems
Try each cold. Hint and Show Solution buttons are progressive — open only when you genuinely cannot proceed.
A restaurant offers 4 starters, 5 mains, and 3 desserts. A meal is one of each. How many distinct meals?
A path from A to B has 5 routes via X and 3 routes via Y, with X and Y disjoint. How many routes total?
A licence plate has 3 letters followed by 3 digits, no constraints (letters or digits can repeat). How many possible plates?
A 5-sided pyramid (apex + 5-gon base) — how many edges does it have?
How many 4-digit numbers are odd?
A 4-character string from {A, B, C}, with no two adjacent characters equal. How many strings?
An n-sided prism has F = 14 faces. Find n and the number of edges E.
How many 3-digit numbers have all three digits distinct AND are divisible by 5? (A number is divisible by 5 iff its last digit is 0 or 5.)
Case B (last = 5): leading from {1..9}\\{5} = 8 choices, middle from {0..9}\\{5, leading} = 8 choices = \(8 \cdot 8 = 64\).
Total: \(72 + 64 = \mathbf{136}\).
AIMO 2004 · Q2
2 marks. Try cold first. Use Hints on the right only if stuck.
(1) Observe — what to notice
1. Keywords: "n-sided prism", "edges"
2. Known: total edges = 2004
3. Unknown: n; also V and F
4. Intermediate: the formula \(E = 3n\)
5. Hidden constraints: n must be a positive integer ≥ 3 (a prism needs at least a triangle base)
(2) Strategy — how to think about it
Why this technique works in general: Any "given X count, find n" prism question reduces to one of three linear equations \(V = 2n\), \(E = 3n\), \(F = n + 2\). The pattern is universal — recognise the polyhedron, write the formula, solve.
(3) Step 1 — first action
(4) Step 2 — second action
(5) Step 3 — Euler check
Strategy
Direct application of \(E = 3n\) for an n-prism, then substitute back.
Solution
Euler check: \(1336 - 2004 + 670 = 2\) ✓.
AIMO 2005 · Q2
2 marks. Counting integer solutions by enumeration.
(1) Observe — what to notice
1. Keywords: "different ways to allocate", "exactly full"
2. Known: 105 passengers; taxi sizes 6, 10, 15
3. Unknown: number of solutions \((a, b, c)\) to \(6a + 10b + 15c = 105\)
4. Intermediate: non-negative integers a, b, c
5. Hidden constraints: order of taxi departures doesn't matter (so we count multisets, i.e. unordered triples \((a, b, c)\))
(2) Strategy — case-split + enumerate
Why this technique works in general: Any "count integer solutions to a sum equation" reduces to disjoint cases on the largest coefficient. Within a case, the residual is a 2-variable equation that can be enumerated with simple divisibility and bounding.
(3) Step 1 — bound c
(4) Step 2 — solve each c case
c = 7: \(6a + 10b = 0\) → (0, 0). 1 solution.
c = 5: \(6a + 10b = 30\) → (a,b) with \(3a + 5b = 15\): (0,3), (5,0). 2 solutions.
c = 3: \(6a + 10b = 60\) → \(3a + 5b = 30\): (0,6),(5,3),(10,0). 3 solutions.
c = 1: \(6a + 10b = 90\) → \(3a + 5b = 45\): (0,9),(5,6),(10,3),(15,0). 4 solutions.
Even c gives no solutions (LHS even, RHS odd).
(5) Step 3 — sum the case counts
Strategy
Count non-negative integer solutions \((a, b, c)\) to \(6a + 10b + 15c = 105\). Disjoint case-split on c (number of 15-seat taxis), then enumerate a, b for each case.
Solution
For each value of c, the residual is \(6a + 10b = 105 - 15c\), or dividing by 2: \(3a + 5b = (105 - 15c)/2\). Need RHS to be a non-negative integer, so \(105 - 15c\) must be even ⇒ c odd. Try c ∈ {1, 3, 5, 7} (c = 9 gives \(15 \cdot 9 = 135 > 105\) — invalid):
AIMO 2001 · Q3
3 marks. A 5-step decreasing-pool multiplication.
(1) Observe — what to notice
1. Keywords: "5-digit", "all distinct", "non-zero leading"
2. Known: digits drawn from {0, 1, ..., 9}
3. Unknown: count of valid 5-digit numbers
4. Intermediate: digit-by-digit choice counts (decreasing pool)
5. Hidden constraints: first digit ∈ {1..9}, others ∈ {0..9} \ {previously used digits}
(2) Strategy — left to right
Why this technique works in general: Any "k-digit number with constraints" problem can be modelled as a sequence of independent positional choices, where each position's pool is determined by the constraint structure. Multiply the per-position counts. This is the universal model for permutation counting.
(3) Step 1 — leading digit
(4) Step 2 — second digit
(5) Step 3 — remaining digits and multiply
Strategy
Build left to right. First digit avoids 0. Each subsequent digit avoids the previously used digits.
Solution
Synthesis 1 — 5-digit password, no two adjacent equal
Combines C1-A1 (multiplication) with the adjacency constraint from WE4.
(1) Observe — what to notice
1. Keywords: "no two adjacent equal", "non-zero leading"
2. Known: 5 positions, digit pool {0..9}
3. Unknown: count of valid passwords
4. Intermediate: per-position choice count under adjacency constraint
5. Hidden constraints: first digit ≠ 0; later digits avoid only the immediately previous digit (NOT all earlier digits — repetitions further back are fine)
(2) Strategy
Why this technique works in general: "No two adjacent equal" is a local constraint, not a global one. Each later step independently has \(b - 1\) choices when the alphabet is size \(b\). The total is \(b \cdot (b-1)^{k-1}\) for k positions (with leading-zero adjustment if needed).
(3) Step 1
(4) Step 2
(5) Step 3
Strategy
Multiplication principle position by position. The "no two adjacent equal" rule reduces each later position's pool by exactly 1 (from 10 to 9). The leading-zero rule already gives 9 for position 1. So all five positions independently have 9 choices.
Solution
Synthesis 2 — pyramid + Euler verification
Combines C1-A4 (prism topology) with the Euler formula extended to pyramids.
(1) Observe — what to notice
1. Keywords: "n-sided pyramid", "Euler"
2. Known: apex + n-gon base; Euler holds for any convex polyhedron
3. Unknown: V, E, F formulas in n
4. Intermediate: count vertices, edges, faces by category (apex vs base vs lateral)
5. Hidden constraints: n ≥ 3 (a pyramid needs at least a triangular base)
(2) Strategy
Why this technique works in general: Counting V, E, F by category (categorising each vertex/edge/face into a disjoint type) generalises the prism trick to any polyhedron family — pyramid, antiprism, truncated solid, etc.
(3) Step 1 — Vertices
(4) Step 2 — Edges
(5) Step 3 — Faces and Euler
Strategy
Apply the addition principle to count V, E, F by disjoint categories (apex vs base vs lateral). Then verify with Euler.
Compare: square pyramid vs triangular prism
n-pyramid general formulas
Atomic skill matrix — 4 skills, 2 micro-checks each
Quick fluency check before the Mock Test. Aim for 8/8 here.
| ID | Skill | WEs | Phase 4 problems | AIMO problems |
|---|---|---|---|---|
| C1-A1 | Multiplication | WE1, WE4, WE5 | P1, P3, P5, P6 | 2001 Q3, Synthesis 1 |
| C1-A2 | Addition (mut.exclusive) | WE2 | P2 | 2005 Q2 |
| C1-A3 | Complement | (P0-4) | P8 (cases) | (implicit) |
| C1-A4 | n-prism / pyramid topology | WE3 | P4, P7 | 2004 Q2, Synthesis 2 |
Micro-validation — 2 checks per atomic skill (8 total)
C1-A1 — Multiplication principle
5 colours of paint × 4 brushes × 3 canvases. How many combinations?
Answer
\(5 \cdot 4 \cdot 3 = \mathbf{60}\).
A meal is 1 of 6 starters AND 1 of 8 mains. How many meals?
Answer
\(6 \cdot 8 = \mathbf{48}\).
C1-A2 — Addition principle (mutually exclusive)
Either ferry (3 routes) OR train (4 routes) OR bus (2 routes), with no overlap. How many trip choices?
Answer
\(3 + 4 + 2 = \mathbf{9}\).
A book is on 1 of 5 fiction shelves OR 1 of 7 non-fiction shelves. How many possible locations?
Answer
\(5 + 7 = \mathbf{12}\).
C1-A3 — Subtraction / complement
How many 3-digit numbers contain at least one digit equal to '5'?
Answer
Total 3-digit = \(900\). "No 5" anywhere: leading ∈ {1..9}\\{5} = 8 choices, then each of two later positions ∈ {0..9}\\{5} = 9 choices. Complement = \(8 \cdot 9 \cdot 9 = 648\). Answer = \(900 - 648 = \mathbf{252}\).
How many 4-digit numbers do NOT contain the digit 0?
Answer
Each of 4 digits ∈ {1..9} = 9 choices. \(9^4 = \mathbf{6561}\).
C1-A4 — n-prism topology
An octagonal (8-sided) prism: enter the number of edges E.
Answer
\(V = 16,\ E = 24,\ F = 10\). The asked value is \(E = \mathbf{24}\).
A prism has 30 edges. Find n.
Answer
\(3n = 30 \Rightarrow n = \mathbf{10}\), and \(F = 12\).
Mini Mock Test — 3 in-Part originals · 8 marks
Closed book. Time yourself: aim for 10 minutes total. Then click "Submit Mock" for instant grading.
Mini Mock — Multiplication & Addition
3 questions · 8 marks total · ~10 minutes recommended
A simple lock has 3 dials, each showing one of 6 symbols. How many distinct lock combinations?
A drink order is either (1 of 4 sodas AND 1 of 3 ice flavours) OR (1 of 2 hot teas alone). How many distinct drink orders?
A prism has V + F = 32. Find n (the number of sides of the base polygon).
Summary, cheat sheet, self-assessment
Bank what you learned. Rate yourself. Review the error book.