Worked Solutions
Combinatorics — Worked Solutions (HSC Maths Extension 1)
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Worked examples for HSC 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.
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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 — Term in a binomial expansion
Question
Find the coefficient of $x^6$ in the expansion of $\left(2x^2 + \dfrac{1}{x}\right)^9$.
Solution
General term: $T_{k+1} = \dbinom{9}{k}(2x^2)^{9-k}\left(\dfrac{1}{x}\right)^k$.
Power of $x$: $x^{2(9-k)}\cdot x^{-k} = x^{18 - 3k}$. Set $18 - 3k = 6 \Rightarrow k = 4$.
Coefficient: $\dbinom{9}{4} \cdot 2^{5} = 126 \cdot 32 = 4032$.
Find $k$ from the power of $x$ first — then the coefficient is just the binomial number times the power of 2.
The binomial theorem gives every term as $T_{k+1} = \dbinom{9}{k}(2x^2)^{9-k}\left(\dfrac{1}{x}\right)^k$. We don't want the whole expansion — just the one term with $x^6$.
So track the power of $x$. From $(2x^2)^{9-k}$ we get $x^{2(9-k)}$, and from $\left(\tfrac1x\right)^k = x^{-k}$ we get $x^{-k}$. Together that's $x^{18 - 3k}$.
Set the exponent equal to 6: $18 - 3k = 6$, so $k = 4$. Now build that term's coefficient — the $x$ part is done, we just need the constants:
$$\binom{9}{4}\cdot 2^{9-4} = 126 \cdot 2^5 = 126 \cdot 32 = 4032.$$
We solve for $k$ from the exponent because the coefficient only makes sense once we know which term carries $x^6$.
- $T_{k+1} = \binom{9}{k}(2x^2)^{9-k}(x^{-1})^k$
- $x$-power: $18 - 3k = 6$ ⇒ $k = 4$
- Coefficient $= \binom{9}{4}2^{5} = 126\cdot32 = 4032$
Where the marks go
- 1 mark: Writes the general term $\binom{9}{k}(2x^2)^{9-k}(x^{-1})^k$
- 1 mark: Solves $18 - 3k = 6$ to get $k = 4$
- 1 mark: Correct coefficient $4032$
Key idea
Use the general term $\binom{n}{k}a^{n-k}b^k$, find $k$ from the required power of $x$, then evaluate the constant factor.
Example 2 — Arrangements with a restriction
Question
Seven different books, including a dictionary and a thesaurus, are arranged in a row on a shelf. In how many arrangements are the dictionary and thesaurus next to each other?
Solution
Treat the dictionary and thesaurus as one block. That leaves $6$ items to arrange: $6! = 720$.
Inside the block, the two books can swap: $2! = 2$.
Total $= 6! \times 2! = 720 \times 2 = 1440$.
The "items together" trick is always block them as one, arrange, then multiply by the internal orderings.
When two items must be together, the neat idea is to glue them into a single block so they can't be separated. Now instead of 7 separate books we have 6 things to arrange in a row: the block plus the other 5 books.
Those 6 items arrange in $6! = 720$ ways.
But the block itself has an internal order — the dictionary can be on the left or the thesaurus can be, so there are $2! = 2$ ways to arrange inside it.
By the multiplication principle, total $= 6! \times 2! = 720 \times 2 = 1440$.
We multiply because each external arrangement pairs with each internal one — that's why "tie them together, then multiply by $2!$" works.
- Block the 2 together ⇒ 6 items: $6! = 720$
- Internal order of block: $2! = 2$
- Total $= 720 \times 2 = 1440$
Where the marks go
- 1 mark: Treats the two books as a single block
- 1 mark: Arranges the 6 items: $6! = 720$
- 1 mark: Accounts for the internal order: $2! = 2$
- 1 mark: Correct total $1440$
Key idea
For items that must be together, bundle them into one block, arrange the resulting items, then multiply by the arrangements within the block.