Counting Principles is one of the highest-leverage, lowest-rote-memory topics in IB Mathematics Analysis & Approaches HL. The formulas are short — factorials, $^nP_r$, $^nC_r$ and Pascal's identity — but the questions reward students who can choose between three or four valid approaches under exam pressure: direct counting, complementary counting, the block method, or residue classes. Almost all marks are lost not on arithmetic but on misreading the question (order matters or not?), forgetting the leading-zero rule, or failing to divide by $k!$ for unlabelled groups.
This cheatsheet condenses every rule, trigger pattern, trick and trap from SL 1.9 and AHL 1.10–1.11 into one revision page. Scroll to the bottom for the printable PDF, the full Notes, the worked tutorials and the marked-up solutions in the gated student library.
§1 — Quick Reference: Core Counting Formulas SL 1.9, AHL 1.10–1.11
| Formula | Notation | Use when… |
|---|---|---|
| $n_1 \times n_2 \times \cdots \times n_k$ | FCP | Sequential independent choices |
| $n! = n(n-1) \cdots 2 \cdot 1$ | Factorial | Arrange $n$ distinct objects in a line |
| $\dfrac{n!}{(n-r)!}$ | ${}^nP_r$ | Select and arrange $r$ from $n$ (roles) |
| $\dfrac{n!}{r!(n-r)!}$ | ${}^nC_r = \binom{n}{r}$ | Choose $r$ from $n$ (no roles) |
| $\binom{n}{r} + \binom{n}{r-1} = \binom{n+1}{r}$ | Pascal's identity | Inductive step in proofs |
| $\dfrac{n!}{(r!)^k \cdot k!}$ | Equal unlabelled groups | Divide into $k$ groups of $r$, no labels |
Key values: $0! = 1$, $1! = 1$, $2! = 2$, $3! = 6$, $4! = 24$, $5! = 120$, $6! = 720$, $7! = 5040$, $8! = 40320$.
Symmetry: $\binom{n}{r} = \binom{n}{n-r}$. Boundaries: $\binom{n}{0} = 1$, $\binom{n}{1} = n$, $\binom{n}{n} = 1$.
§2 — FCP: Independent vs Dependent Slots SL 1.9, AHL 1.10
Pattern: "How many codes / numbers / sequences…" with or without repetition.
The trick: Count the choices at each stage and multiply. If digits/objects are reusable: $n^k$. If no repetition: $n(n-1)(n-2)\cdots$
- With repetition (3 digits from $\{1,2,3,4,5\}$): $5 \times 5 \times 5 = 125 = 5^3$.
- Without repetition (same): $5 \times 4 \times 3 = 60 = {}^5P_3$.
§3 — Permutations: Order Matters AHL 1.10
Pattern: "Assign roles", "fill ranked positions", "arrange in a line", "how many $r$-digit PINs".
The trick: Use ${}^nP_r = \dfrac{n!}{(n-r)!}$. Ask: "Does swapping two chosen items give a different outcome?" If yes, it's a permutation. If no, it's a combination.
- Elect chair, VP, secretary from 10: ${}^{10}P_3 = 720$.
- First, second, third in a race of 8: ${}^8P_3 = 8 \times 7 \times 6 = 336$.
§4 — Block Method: "Must Be Adjacent" AHL 1.10
Pattern: "Group $X$ must sit/stand together", "must be adjacent", "immediately after".
The trick: Glue the $r$ objects into one super-object. Arrange $(n - r + 1)$ objects. Multiply by $r!$ for internal arrangements. If order within the block is fixed (e.g. "A immediately after B"), do not multiply by $r!$.
- 3 girls together in a 9-person line: $7! \times 3! = 5040 \times 6 = 30240$.
- Jack immediately after Andrea (8 runners): $7! = 5040$ (block is fixed-order, no $r!$ multiplier).
§5 — Complementary: "Must NOT Be Adjacent" AHL 1.10
Pattern: "Must not be together", "not in the same group", "at least one seat apart".
The trick: Almost always fastest: Total (no restriction) $-$ Bad (they ARE together). Use the block method to count the bad cases.
- 3 friends not all together (row of 12): ${}^{12}P_3 - 3! \times 10 = 1320 - 60 = 1260$.
§6 — At Least / At Most AHL 1.10
Pattern: "At least $k$…", "at most $k$…", "select at least two girls".
The trick:
- At least 1: use complementary — Total $-$ (none).
- At least $k \geq 2$: list valid cases and sum.
- At most $k$: list valid cases or use complementary — Total $-$ (more than $k$).
Example: Select 5 from 6 boys + 3 girls, at least 2 girls:
$\binom{3}{2}\binom{6}{3} + \binom{3}{3}\binom{6}{2} = 60 + 15 = 75$.
Complementary check: $\binom{9}{5} - \binom{3}{1}\binom{6}{4} - \binom{6}{5} = 126 - 45 - 6 = 75$. ✓
§7 — Grid / Table Adjacency AHL 1.10
Pattern: Seating in an $r \times c$ grid, desk arrangement, with forbidden adjacent pairs.
The trick: Count adjacent pairs first, before any calculation:
$$r(c-1) + c(r-1) = 2rc - r - c \;\text{ total adjacent pairs in an }r \times c\text{ grid.}$$
Then: Bad $=$ (adjacent pairs) $\times$ (orderings of forbidden pair) $\times$ (arrangements of rest).
Example: $3\times 2$ grid: $3(1) + 2(2) = 7$ adjacent pairs. 5 sheep, 1 per pen, A and B not adjacent: Bad $= 7 \times 2 \times 4! = 336$. Valid $= 6! - 336 = 720 - 336 = 384$.
§8 — Equal Unlabelled Groups AHL 1.10
Pattern: "Divide $n$ people into $k$ equal teams", "split into equal groups with no names/labels".
The trick: Divide by $k!$ to remove overcounting from the $k!$ orderings of identical groups:
$$\dfrac{\binom{n}{r}\binom{n-r}{r}\cdots}{k!} = \dfrac{n!}{(r!)^k \cdot k!}$$
Example: 15 students, 3 unlabelled teams of 5: $\dfrac{\binom{15}{5}\binom{10}{5}\binom{5}{5}}{3!} = \dfrac{3003 \times 252 \times 1}{6} = 126\,126$.
If teams are labelled (Red/Blue/Green): $6 \times 126\,126 = 756\,756$.
§9 — Counting Equations: Find $n$ AHL 1.10
Pattern: "Find $n$ such that the number of arrangements is halved / doubled / equals $X$", "every student plays every other student… find class size".
The trick: Express both counts as polynomials in $n$. Form an equation, expand factorials, solve the polynomial. Always check domain restrictions (e.g. group sizes must be $\geq 0$) and reject extraneous roots.
- Two students refuse same group; count halved. $2 \cdot {}^{n-2}C_2 = \tfrac{1}{2} {}^n C_3$ gives $n^2 - 13n + 36 = 0$, so $n = 4$ or $n = 9$. Group 2 needs $\geq 3$, so reject $n = 4$. $n = 9$.
- Every pair plays twice, 513 total games, one player left early after 7 games: $(n-1)(n-2) + 7 = 513 \Rightarrow n = 24$.
§10 — Divisibility via Residue Classes AHL 1.10
Pattern: "Sum divisible by $d$", "product/sum with a specific remainder".
The trick: Classify all numbers by residue mod $d$. For $d = 3$: three groups R0, R1, R2. Sum $\equiv 0$ iff residues are all-same or one-of-each.
Example: From $\{1, \ldots, 30\}$ (10 each of R0, R1, R2), choose 3 with sum $\equiv 0 \pmod 3$:
All same: $3 \times \binom{10}{3} = 360$. One of each: $10^3 = 1000$. Total: 1360.
§11 — Proof by Induction with Combinations AHL 1.10–1.11
Pattern: "Prove $\displaystyle\sum_{r=1}^{n} {}^r C_k = {}^{n+1} C_{k+1}$" or similar combinatorial sum by induction.
The 4-step script (never deviate):
- Base case ($n = 1$): evaluate both sides numerically. "It works" without values $\Rightarrow$ R0.
- Assume true for $n = k$: write the $k$-statement explicitly. "Let $n = k$" $\Rightarrow$ M0.
- Inductive step: write LHS for $n = k + 1$, substitute the assumption, apply Pascal's identity.
- Conclusion: "Since true for $n = 1$, and true for $n = k$ implies true for $n = k + 1$, by PMI true for all $n \in \mathbb{Z}^+$." This sentence earns a dedicated R1 — never skip it.
Example: Prove $\displaystyle\sum_{r=1}^{n} {}^r C_1 = {}^{n+1} C_2$. Inductive step: ${}^{k+1}C_2 + {}^{k+1}C_1 = {}^{k+2}C_2$ by Pascal. Done.
§12 — GDC Use on Paper 2 AHL 1.10
Counting questions are largely Paper 1 (no GDC) but some appear on Paper 2. The GDC is useful for:
| Task | TI-84 / Nspire syntax |
|---|---|
| $\binom{n}{r}$ | MATH → PRB → nCr: type n nCr r |
| ${}^nP_r$ | MATH → PRB → nPr: type n nPr r |
| $n!$ | MATH → PRB → !: type n! |
| Solve polynomial | solve(n^2 - 13n + 36 = 0, n) or table of values |
| Verify large factorials | Store intermediately; avoid overflow |
solve may give non-integer or negative roots — always verify $n \in \mathbb{Z}^+$ and check the domain.§13 — Exam Attack Plan SL 1.9, AHL 1.10–1.11
| If the question says… | Reach for… |
|---|---|
| "$k$-digit integer from $\{0, \ldots, 9\}$" | FCP; first digit: 9 choices |
| "Distinct digits in increasing order" | $\binom{n}{k}$ — one arrangement per set |
| "Assign roles / ranked positions" | ${}^nP_r$ |
| "Must be adjacent / together" | Block method $\times r!$ |
| "Immediately after / before" | Block (fixed order, no $r!$) |
| "Must not be adjacent" | Total $-$ block method |
| "Anywhere after / before" | Symmetry: $n! / 2$ |
| "At least 1 [condition]" | Total $-$ none |
| "At least $k \geq 2$" | Sum of cases |
| Grid / table seating problem | Count $r(c-1) + c(r-1)$ pairs first |
| "Divide into equal unlabelled teams" | $\div\, k!$ for identical groups |
| "Find $n$; count is halved/doubled" | Algebra of combinations |
| "Sum divisible by $d$" | Residue classes mod $d$ |
| "Prove $\sum \binom{r}{k} = \binom{n}{m}$" | Induction + Pascal; conclude formally |
Worked Example — IB-Style Committee Selection
Question (HL Paper 1 style — 7 marks)
A committee of 6 students is to be chosen from a class of 7 boys and 5 girls.
- How many committees have exactly 2 girls?
- How many committees have at least 2 girls?
- If two specific boys, Alex and Brian, refuse to serve together, how many committees with exactly 3 boys and 3 girls are now possible?
Solution
- Exactly 2 girls: choose 2 girls from 5 and 4 boys from 7. $\binom{5}{2} \cdot \binom{7}{4} = 10 \cdot 35 = 350$. (M1)(A1)
- At least 2 girls: list cases — 2G+4B, 3G+3B, 4G+2B, 5G+1B.
$\binom{5}{2}\binom{7}{4} + \binom{5}{3}\binom{7}{3} + \binom{5}{4}\binom{7}{2} + \binom{5}{5}\binom{7}{1}$
$= 350 + 350 + 105 + 7 = 812$. (M1)(A1) - 3B + 3G with Alex and Brian not both serving: Total 3B+3G unrestricted $= \binom{7}{3}\binom{5}{3} = 35 \cdot 10 = 350$. (M1)
- Bad cases (Alex AND Brian both in): pick the third boy from the remaining 5, and 3 girls from 5: $\binom{5}{1}\binom{5}{3} = 5 \cdot 10 = 50$. (A1)
- Valid $= 350 - 50 = 300$. (A1)
Examiner's note: Part (c) tests complementary counting under a constraint. The most common error is to compute "Alex in OR Brian in" instead of "both in" — these are different. Always identify the exact bad event before subtracting from the total.
Common Student Questions
When do I use a permutation versus a combination?
How does the block method work for "must be adjacent"?
How do I count $k$-digit numbers when 0 is one of the digits?
Why do I divide by $k!$ for equal unlabelled groups?
How do I do a "sum divisible by $d$" counting question?
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