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Module 4 · L15 of 15 ~40 min ⚡ +95 XP available

Mixed Combinatorics & Exam Technique

You've mastered permutations, combinations, the Binomial Theorem, inclusion-exclusion, pigeonhole, and combinatorial identities. Now it's time to synthesise everything. In this final lesson you'll build a decision framework for approaching any combinatorics problem cold, and lock in the exam habits that separate Band 5 from Band 6.

Today's hook, A committee of 5 is chosen from 6 men and 4 women. How many committees contain at least 2 women? Most students reach for their calculator before they've decided what method to use. By the end of this lesson you'll have a five-question decision framework that makes the method obvious every time.

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Recall, your gut answer first

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A committee of 5 is chosen from 6 men and 4 women. Without calculating roughly how many committees do you think contain at least 2 women? And how would you approach this type of problem in an exam? Jot your instinct before reading on.

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Orientation and key terms

The five-question decision framework

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Every combinatorics problem can be decoded by answering these five questions in order. Practise applying them until it becomes automatic.

Does order matter? Yes → permutation ($^nP_r$); No → combination ($^nC_r$).
Are there restrictions? (fixed positions, items that must stay together, items that cannot be adjacent)
Is the arrangement linear or circular? Circular → fix one element, arrange the rest: $(n-1)!$
Are there overlapping conditions? Use inclusion-exclusion: $|A \cup B| = |A| + |B| - |A \cap B|$
Does the problem require proving existence? Yes → Pigeonhole Principle.

A decision tree that routes a counting problem by whether order matters to nPr or nCr, then to the relevant restriction technique before solving and verifying.

$^nC_r = \dfrac{n!}{r!\,(n-r)!}$

Order is the key gate

Ask: does swapping two elements create a different outcome? If yes, it's a permutation; if no, it's a combination.

Complement for "at least"

For "at least $k$" problems, check whether it's easier to subtract the complement (fewer cases) than to enumerate all qualifying cases.

Write your reasoning

Show clearly why you multiply or add steps. Define what $n$ and $r$ represent. Examiners award method marks even when arithmetic slips.

What you'll master

Know

Key facts

The five-question decision framework for combinatorics
When to use complement vs. direct case analysis for "at least" problems
Exam writing conventions: define variables, justify multiply/add steps

Understand

Concepts

How all Module 4 techniques connect: permutations, combinations, circular, PIE, pigeonhole, Binomial Theorem
Why impossible cases (e.g., choosing 4 from 3) must be checked before summing
How exam mark allocations signal the expected method and length of response

Can do

Skills

Apply the decision framework to decode unseen problems
Correctly split "at least" problems into cases or use complement
Identify and discard impossible cases before calculating

Key terms

Permutation $^nP_r$An ordered arrangement of $r$ objects chosen from $n$ distinct objects: $^nP_r = \dfrac{n!}{(n-r)!}$.

Combination $^nC_r$An unordered selection of $r$ objects from $n$: $^nC_r = \dfrac{n!}{r!\,(n-r)!}$.

Circular permutationArrangements in a circle; one element is fixed as a reference point, giving $(n-1)!$ distinct arrangements.

Inclusion-exclusion (PIE)$|A \cup B| = |A| + |B| - |A \cap B|$. Extends to three or more sets.

Pigeonhole PrincipleIf $n$ items are placed in fewer than $n$ containers, at least one container holds more than one item.

General term (Binomial)The $(r+1)$th term of $(a+b)^n$ is $T_{r+1} = \binom{n}{r} a^{n-r} b^r$.

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Exam technique

core concept

Strong exam technique in combinatorics is as important as knowing the formulas. Follow these four principles in every response:

Write your reasoning: Show clearly why you multiply (independent steps) or add (mutually exclusive cases). Never write a string of numbers without explanation.
Define variables: State explicitly what $n$ and $r$ represent in context. E.g., "choosing $r = 3$ items from $n = 8$ distinct books (order doesn't matter)".
Simplify factorials: When exact values are large, leave answers as products of factorials or simplified fractions. $\frac{8!}{3!\cdot 5!}$ is a perfectly valid final form.
Check plausibility: Does your answer make sense? The number of committees can't exceed the total number of ways to choose 5 from 10 people ($^{10}C_5 = 252$).

Mark-allocation signal. A 1-mark question wants a single numerical answer. A 2-mark question expects at least one line of working. A 3-mark question expects a structured method, working, and final answer, usually with at least two distinct reasoning steps visible.

Framework: Order? → Restrictions? → Linear/Circular? → Overlapping? → Existence?; Multiply steps when they are independent; add when cases are mutually exclusive

Pause, copy the five-question framework into your book: (1) order? → P or C; (2) restrictions? → fix/glue/gap; (3) linear/circular?; (4) overlapping cases?; (5) existence? → Pigeonhole; multiply steps when independent, add when mutually exclusive.

Quick check: A password must use 3 different letters from the alphabet (26 letters, order matters). Which formula applies?

"At least" problems, two methods

core concept

We just saw the five-question framework: (1) order? (2) restrictions? (3) linear or circular? (4) overlapping? (5) existence?, working through this systematically identifies which formula and method to apply. That raises a question: "at least $k$" problems seem to require summing many cases, is there a shortcut? This card answers it → yes: use the complement, total outcomes minus (fewer than $k$ outcomes), which usually has fewer cases.

Problems using "at least" language can be solved two ways. Choose whichever has fewer cases:

Direct case analysis: List every qualifying case and add. Best when there are only 2–3 cases.
Complement method: Total $-$ (cases that fail the condition). Best when failure is a single case or very few cases.

$$\text{at least } k = \text{Total} - \text{(fewer than } k\text{)}$$

Always check whether all listed cases are actually possible. For example, "choose 4 consonants from EQUATION" has only 3 consonants (Q, T, N), that case is impossible and contributes 0 to the count.

Impossible case trap. Before summing cases, verify that $r \leq n$ for every case. If $r > n$, that case contributes zero, but writing "$^3C_4 = 0$" (rather than skipping it silently) earns communication marks.

"At least k": try complement first, total - (fewer than k); Direct case method: list all qualifying cases, compute each, then add (use when complement is complex)

Pause, copy the "at least" strategy into your book: try complement first, total $-$ (fewer than $k$); use direct case method only when the complement is more complex than listing the qualifying cases.

Did you get this? True or false: the complement method for "at least 1 woman" in a committee problem gives Total $-$ (0 women).

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Worked examples

Worked examples · 3 in a row, reveal as you go

PROBLEM 1 · AT-LEAST (CASE ANALYSIS)

A committee of 5 is to be chosen from 6 men and 4 women. How many committees contain at least 2 women?

Cases with at least $2$ women: exactly $2W + 3M$, exactly $3W + 2M$, exactly $4W + 1M$.

List all qualifying cases. Check all are possible: max women available = 4, which is $\leq$ 5. All three cases have $r \leq n$.

PROBLEM 2 · IMPOSSIBLE CASE

How many 4-letter arrangements from the word EQUATION contain at least one vowel?

EQUATION has $8$ distinct letters: E, Q, U, A, T, I, O, N. Identify which are vowels and which are consonants.

Vowels: E, U, A, I, O (5 vowels). Consonants: Q, T, N (3 consonants).

PROBLEM 3 · BINOMIAL GENERAL TERM

Find the coefficient of $x^3$ in the expansion of $(1 + 2x)^5$.

$T_{r+1} = \binom{5}{r}(1)^{5-r}(2x)^r = \binom{5}{r} 2^r x^r$

Write the general term formula. Here $a = 1$, $b = 2x$, $n = 5$.

Fill the gap: In the committee example (6 men, 4 women, choose 5), the number of committees with exactly 3 women is $^4C_3 \times \,{}^6C_2 = $ .

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Misconceptions to fix

Misconceptions to fix · the 3 traps that cost marks

Trap 01

Treating "at least" as a single formula

Students often write a single expression for "at least 2 women" instead of splitting into cases (2W, 3W, 4W). There is no shortcut formula, you must enumerate cases or use the complement. Skipping the case split always costs marks.

Trap 02

Forgetting to check for impossible cases

In the EQUATION example, attempting $^3P_4$ without recognising $r > n$ is impossible leads to a formula error. Always verify $r \leq n$ before evaluating each case, impossible cases contribute 0, not an error.

Trap 03

Overcomplicating by ignoring simpler total counts

Sometimes the simplest observation (e.g., "all arrangements contain at least one vowel because there aren't enough consonants") solves the problem immediately. Apply the decision framework and look for obvious simplifications before calculating.

Did you get this? True or false: $^3C_4 = 0$ because you cannot choose 4 objects from only 3.

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Activities

Activities · practice with the ideas

A team of 5 is chosen from 7 boys and 5 girls. How many teams have more boys than girls? (Hint: list cases where boys > girls.)

Find the number of ways to arrange the letters of BANANA.

Find the coefficient of $x^3$ in the expansion of $(1 + 2x)^5$. Show full working using the general term formula.

How many 5-digit numbers can be formed using the digits 1, 2, 3, 4, 5 (no repetition) if the number must be even?

Apply the decision framework to this problem: "In how many ways can 8 people sit around a circular table?" State your answer to each of the five framework questions.

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Revisit your thinking

Earlier you estimated the number of committees with at least 2 women from 6 men and 4 women (choosing 5).

The exact answer is 186: $6 \times 20 + 4 \times 15 + 1 \times 6 = 120 + 60 + 6$. Out of $^{10}C_5 = 252$ total committees, 186 (about 74%) contain at least 2 women. The reason it's so high: with only 6 men, it's actually hard to get committees that are mostly male.

Odd one out: Which of the following does NOT belong to the Module 4 toolkit?

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Multiple choice

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Pick your answer, then rate your confidence. That tells the system what to drill next. Each retry pulls a fresh mix from the bank.

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Short answer

ApplyBand 42 marks

Q1. A team of 5 is chosen from 7 boys and 5 girls. How many teams have more boys than girls? (2 marks)

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ApplyBand 42 marks

Q2. Find the number of distinct arrangements of the letters in the word BANANA. (2 marks)

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AnalyseBand 53 marks

Q3. (a) Find the coefficient of $x^3$ in $(1 + 2x)^5$ using the general term. (b) Explain why the complement method would be inefficient for finding the number of 4-letter arrangements from EQUATION containing at least one vowel, and state the answer directly. (3 marks)

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Comprehensive answers (click to reveal)

Activity 1:

1. More boys than girls (teams of 5): cases 3B+2G, 4B+1G, 5B+0G.
$= \,{}^7C_3 \cdot \,{}^5C_2 + \,{}^7C_4 \cdot \,{}^5C_1 + \,{}^7C_5 \cdot \,{}^5C_0 = 35 \times 10 + 35 \times 5 + 21 \times 1 = 350 + 175 + 21 = 546$

2. BANANA: 6 letters; A appears 3 times, N appears 2 times.
Distinct arrangements $= \dfrac{6!}{3! \times 2!} = \dfrac{720}{6 \times 2} = 60$

3. Coefficient of $x^3$ in $(1+2x)^5$: $T_4 = \binom{5}{3} \cdot 2^3 \cdot x^3 = 10 \times 8 \cdot x^3$. Coefficient $= 80$.

4. Even 5-digit number from {1,2,3,4,5} (no repetition): last digit must be 2 or 4 (2 choices). Remaining 4 positions: $4! = 24$ ways. Total $= 2 \times 24 = 48$.

5. Framework for 8 people in a circle: (1) Order matters, yes (different seating arrangements are distinct); (2) No restrictions; (3) Circular; (4) No overlapping conditions; (5) No existence question. Answer: $(8-1)! = 7! = 5040$.

Q1 (2 marks): Cases: 3B+2G, 4B+1G, 5B+0G [0.5 for identifying all cases].
$\,{}^7C_3 \cdot \,{}^5C_2 + \,{}^7C_4 \cdot \,{}^5C_1 + \,{}^7C_5 \cdot \,{}^5C_0 = 350 + 175 + 21 = \mathbf{546}$ [1.5].

Q2 (2 marks): 6 letters total; A $\times 3$, N $\times 2$ [0.5].
$\dfrac{6!}{3! \times 2!} = \dfrac{720}{12} = \mathbf{60}$ distinct arrangements [1.5].

Q3 (3 marks): (a) $T_{r+1} = \binom{5}{r}(2x)^r$; set $r=3$: $T_4 = 10 \times 8 \cdot x^3$. Coefficient $= \mathbf{80}$ [1.5]. (b) The complement requires 4 consonants from only 3 available (Q, T, N), which is impossible, $^3P_4 = 0$. Therefore all $^8P_4 = 1680$ arrangements contain at least one vowel [1.5].

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Review and finish

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Module 4 summary

Always decide if order matters first: permutation ($^nP_r$) vs. combination ($^nC_r$).
Check for restrictions, circular arrangements, and overlapping conditions before calculating.
For "at least" problems: use case analysis (2–3 cases) or complement (1 failing case). Verify all cases are possible ($r \leq n$).
Show clear working: define $n$ and $r$, justify why you multiply or add, and check plausibility.
Before the Module 4 checkpoint or exam: review any lesson where you felt uncertain. The five-question framework is your starting point for every problem.

Mark lesson as complete

Tick when you've finished the practice and review.

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Module 4 overview · Extension 1 · Checkpoint 3 (L11–15)