Combinatorics — Worked Solutions (Preliminary Maths Extension 1)
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Worked examples for Preliminary Maths Extension 1 combinatorics. Each shows where the marks are awarded, the key idea, and the full solution explained by your choice of tutor — Stella, Ella or Cassie.
In short: full step-by-step worked solutions for Preliminary Maths Extension 1 — Combinatorics. Every question is worked through with the method and reasoning shown, so you can check how to get the answer, not just the final result.
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Try each question first, then check your working. Use the tutor tabs to read the full solution in the style that suits you: Stella is direct and challenging, Ella is warm and explains the why, and Cassie is concise and analytical.
Example 1 — Permutations and combinations
Question
A committee of $5$ is to be chosen from $7$ women and $4$ men.
(a) In how many ways can the committee be chosen if there are no restrictions?
(b) In how many ways can it be chosen if it must contain exactly $2$ men?
Solution
Choosing a committee is unordered, so use combinations $\binom{n}{r}$.
(a) Choose any $5$ from $11$: $\binom{11}{5} = \dfrac{11!}{5!\,6!} = 462$.
(b) Exactly $2$ men means $2$ from $4$ men and $3$ from $7$ women — multiply because the choices are independent.
$\binom{4}{2}\binom{7}{3} = 6 \times 35 = 210$.
Order never matters for a committee, so it's $\binom{n}{r}$, not $^nP_r$. Mixing those up is the classic slip.
A committee has no ranking — being chosen first or last doesn't matter — so we count with combinations rather than permutations.
(a) With no restrictions we simply pick $5$ people from all $11$: $\binom{11}{5} = 462$.
(b) "Exactly $2$ men" splits the job into two independent choices: pick the men, then pick the women. Two men from the four available is $\binom{4}{2} = 6$, and since the committee has $5$ seats the remaining $3$ are women, chosen from seven: $\binom{7}{3} = 35$.
We multiply these because every way of choosing the men can pair with every way of choosing the women: $6 \times 35 = 210$.
The multiplication principle is the heart of it — independent stages of a choice multiply together.
Unordered → combinations.
(a) $\binom{11}{5} = 462$
(b) Exactly $2$ men, so $3$ women:
- Men: $\binom{4}{2} = 6$
- Women: $\binom{7}{3} = 35$
- Multiply: $6 \times 35 = 210$
Where the marks go
- 1 mark: Recognises combinations and computes $\binom{11}{5} = 462$ for part (a)
- 1 mark: Sets up part (b) as $\binom{4}{2}\binom{7}{3}$
- 1 mark: Correct individual values $\binom{4}{2} = 6$ and $\binom{7}{3} = 35$
- 1 mark: Correct final answer $210$
Key idea
Committees are unordered, so use $\binom{n}{r}$; independent choices multiply via the multiplication principle.
Example 2 — Pigeonhole principle
Question
Show that if $5$ integers are chosen, at least two of them must leave the same remainder when divided by $4$.
Solution
This is the pigeonhole principle. The remainders on division by $4$ are $0, 1, 2, 3$ — only $4$ possible values.
Treat the remainders as $4$ "boxes" and the $5$ integers as the "objects" to place in them.
With $5$ objects and only $4$ boxes, at least one box holds two or more objects.
Therefore at least two of the integers share the same remainder mod $4$. The reasoning is airtight: $5 > 4$, so they can't all be different.
The pigeonhole principle says that if you put more objects than boxes into the boxes, some box must hold at least two. Here we just need to spot the boxes.
When you divide any integer by $4$, the remainder is one of only four values: $0, 1, 2$ or $3$. Those four remainders are our boxes.
We're placing $5$ integers (the pigeons) into these $4$ boxes (the remainders). Since $5 > 4$, there aren't enough boxes for each integer to land in a different one.
So at least one remainder must be shared — meaning two of the chosen integers leave the same remainder when divided by $4$. The principle works because you simply can't fit five things into four slots without doubling up.
Pigeonhole principle.
- Possible remainders mod $4$: $\{0, 1, 2, 3\}$ → $4$ boxes
- $5$ integers → $5$ objects
- $5 > 4 \Rightarrow$ some box holds $\geq 2$
Hence two integers share a remainder mod $4$.
Where the marks go
- 1 mark: Identifies the $4$ possible remainders $0, 1, 2, 3$ as the pigeonholes
- 1 mark: Treats the $5$ integers as objects placed into the $4$ remainder classes
- 1 mark: Concludes via $5 > 4$ that two share a remainder
Key idea
Pigeonhole principle: with more objects than boxes, some box holds at least two — division by $4$ gives only $4$ remainder boxes.
Frequently asked questions
Step-by-step solutions to Combinatorics questions in Preliminary Maths Extension 1, with the full method shown for each — so you can follow the reasoning, not just the final answer.
Attempt each question yourself first, then compare your method — not just your answer — against the worked solution. The aim is to learn the approach so you can handle unfamiliar questions, rather than memorising answers.
Yes — they follow the NESA syllabus for Preliminary Maths Extension 1, using the methods and notation expected in exams and assessments.