What is the binomial theorem in HSC Maths Extension 1?

The binomial theorem expands $(a + b)^n$ as a sum $\\sum_{k = 0}^{n} \\binom{n}{k} a^{n - k} b^k$. Each term has a binomial coefficient $\\binom{n}{k}$ and powers of $a$ and $b$ that sum to $n$. The coefficients are the entries of row $n$ of Pascal's triangle.

What is the general term in a binomial expansion?

The $(k + 1)$-th term in the expansion of $(a + b)^n$ is $T_{k + 1} = \\binom{n}{k} a^{n - k} b^k$. This formula is the workhorse for finding specific terms, coefficients of given powers, or the term independent of the variable.

How is combinatorics examined in HSC Maths Extension 1?

Combinatorics typically appears in Section II as a counting problem (how many arrangements or selections satisfy certain conditions) or as a problem using the binomial theorem (find a coefficient, find the term independent of $x$). Sum identities involving binomial coefficients also show up.

What is the difference between a permutation and a combination?

A permutation is an ordered arrangement (order matters); a combination is an unordered selection (order does not matter). The formulas differ by a factor of $r!$ - the permutation is the combination multiplied by $r!$ to account for the orderings.

How do you find the term independent of $x$ in a binomial expansion?

Write the general term $T_{k + 1}$, find the power of $x$ in it, set that power to $0$, and solve for $k$. Substitute back to get the value of the independent term.

What is Pascal's triangle, and why does it matter?

Pascal's triangle is the array of binomial coefficients arranged in rows. Row $n$ contains $\\binom{n}{0}, \\binom{n}{1}, \\dots, \\binom{n}{n}$, and each entry is the sum of the two above it (Pascal's rule). It is a quick way to find coefficients for small $n$ and reveals the structural identities used in many HSC proofs.

§-Study guide

NSWMaths Extension 1

HSC Maths Extension 1 binomial theorem and combinatorics: deep dive (2026 guide)

A complete deep dive into the binomial theorem and combinatorics for HSC Mathematics Extension 1. Counting principles, permutations, combinations, the binomial theorem in full, Pascal's triangle, standard exam techniques (general term, independent term, sum identities), and an embedded practice question set with full solutions.

Generated by Claude Opus 4.822 min readNESA Mathematics Extension 1 Stage 6 Syllabus (2017), Working with Combinatorics (ME-A1, Year 11) and the binomial theoremUpdated 2026-07-02

Reviewed by: AI editorial process; not yet individually human-reviewed

Jump to a section

What this guide covers
Counting principles
Permutations
Combinations
The binomial theorem
Standard exam techniques
Common exam traps
Practice patterns
Exam strategy
Check your knowledge
Linked dot points

What this guide covers

The binomial theorem and the underlying combinatorics are bread-and-butter Extension 1 topics. They appear in Section II most years, often as a 4-mark coefficient question or a 3-mark identity proof.

This guide covers:

The fundamental counting principles (multiplication, addition, complementary).
Permutations and combinations with worked examples.
The binomial theorem in full, including the general term.
Standard exam techniques: coefficient of $x^k$ , term independent of $x$ , identity proofs.
Sum identities like $\sum \binom{n}{k} = 2^n$ .
Ratio of consecutive terms and the greatest binomial coefficient.

Counting principles

The multiplication principle

If a procedure has steps $1, 2, \dots, k$ and step $i$ can be done in $n_i$ ways (independent of the previous steps), the total number of ways is $n_1 \cdot n_2 \cdots n_k$ .

The addition principle

If a procedure can be completed in one of several disjoint ways, the total is the sum of the counts for each way.

Complementary counting

$(\text{at least 1}) = (\text{total}) - (\text{none}).$

For "at least" and "at most" problems, the complement is often easier.

Permutations

The number of ways to arrange $n$ distinct objects in a row is $n!$ . To arrange $r$ out of $n$ : ${}^n P_r = \frac{n!}{(n - r)!}$ .

For objects with repeats (say $n_1$ of one kind, $n_2$ of another), the number of distinct arrangements is

\frac{n!}{n_1! \, n_2! \cdots n_k!}.

Circular arrangements of $n$ distinct objects: $(n - 1)!$ (fix one to break rotational symmetry).

Combinations

The number of unordered selections of $r$ from $n$ distinct objects is

\binom{n}{r} = \frac{n!}{r! (n - r)!}.

Key identities:

Symmetry: $\binom{n}{r} = \binom{n}{n - r}$ .
Pascal's rule: $\binom{n}{r} = \binom{n - 1}{r - 1} + \binom{n - 1}{r}$ .
Sum: $\sum_{r = 0}^{n} \binom{n}{r} = 2^n$ .

The binomial theorem

For any non-negative integer $n$ ,

(a + b)^n = \sum_{k = 0}^{n} \binom{n}{k} a^{n - k} b^k.

The general term is

T_{k + 1} = \binom{n}{k} a^{n - k} b^k.

So $T_1 = a^n$ (when $k = 0$ ) and $T_{n + 1} = b^n$ (when $k = n$ ).

Pascal's triangle

The first few rows:

\begin{array}{c} 1 \\ 1 \quad 1 \\ 1 \quad 2 \quad 1 \\ 1 \quad 3 \quad 3 \quad 1 \\ 1 \quad 4 \quad 6 \quad 4 \quad 1 \\ 1 \quad 5 \quad 10 \quad 10 \quad 5 \quad 1 \\ 1 \quad 6 \quad 15 \quad 20 \quad 15 \quad 6 \quad 1 \end{array}

Each row $n$ gives the coefficients of $(a + b)^n$ .

Standard exam techniques

Find the coefficient of a specific power

The general term is $T_{k + 1} = \binom{n}{k} a^{n - k} b^k$ . Identify the power of the variable in $T_{k + 1}$ , set it equal to the target, and solve for $k$ .

Find the coefficient of

x^3

in the expansion of

(2x + 5)^7

Step 1. Write the general term. With $a = 2x$ , $b = 5$ , $n = 7$ :

T_{k+1} = \binom{7}{k} (2x)^{7-k} (5)^k = \binom{7}{k} 2^{7-k} 5^k \, x^{7-k}.

Step 2. Match the target power. The power of $x$ is $7 - k$ . Set $7 - k = 3$ , so $k = 4$ .

Step 3. Substitute $k = 4$ . Pick off the coefficient at this value of $k$ .

T_5 = \binom{7}{4} 2^{3} 5^{4} \, x^3.

Step 4. Evaluate the numerical factors. Compute $\binom{7}{4} = 35$ , $2^3 = 8$ , $5^4 = 625$ .

\text{Coefficient} = 35 \cdot 8 \cdot 625 = 175\,000.

The coefficient of $x^3$ in $(2x + 5)^7$ is $175\,000$ .

Find the term independent of the variable

The general term $T_{k + 1}$ has some power of $x$ formed from the two factors. Set the power to $0$ and solve for $k$ .

Find the term independent of

x

in the expansion of

\left( x^2 + \dfrac{3}{x} \right)^6

Step 1. Write the general term. With $a = x^2$ , $b = 3/x$ , $n = 6$ :

T_{k+1} = \binom{6}{k} (x^2)^{6-k} \left( \frac{3}{x} \right)^k = \binom{6}{k} 3^k \, x^{2(6-k) - k} = \binom{6}{k} 3^k \, x^{12 - 3k}.

Step 2. Set the power of $x$ to zero. For the term to be independent of $x$ , $12 - 3k = 0$ , so $k = 4$ .

Step 3. Substitute $k = 4$ .

T_5 = \binom{6}{4} 3^4.

Step 4. Evaluate. $\binom{6}{4} = 15$ and $3^4 = 81$ , so the term is $15 \cdot 81 = 1215$ .

The term independent of $x$ is $1215$ .

Coefficient of $x^k$ in a product of two binomial expansions

For an expression like $(a + bx)^n (c + dx)^m$ , the coefficient of $x^k$ comes from summing all ways to pick a power $x^j$ from the first factor and $x^{k-j}$ from the second.

Find the coefficient of

x^4

in the expansion of

(1 + 2x)^5 (1 - x)^4

Step 1. Identify the general terms: From $(1 + 2x)^5$ the term in $x^j$ is $\binom{5}{j} 2^j x^j$ . From $(1 - x)^4$ the term in $x^i$ is $\binom{4}{i} (-1)^i x^i$ .
Step 2. Set up the convolution: To get $x^4$ , choose $j + i = 4$ with $0 \leq j \leq 5$ and $0 \leq i \leq 4$ . The valid pairs $(j, i)$ are $(0, 4)$ , $(1, 3)$ , $(2, 2)$ , $(3, 1)$ , $(4, 0)$ .
Step 3. Compute each contribution: Multiply the two coefficients for each pair.

$(0, 4)$ : $\binom{5}{0} 2^0 \cdot \binom{4}{4} (-1)^4 = 1 \cdot 1 \cdot 1 \cdot 1 = 1$ .
$(1, 3)$ : $\binom{5}{1} 2^1 \cdot \binom{4}{3} (-1)^3 = 5 \cdot 2 \cdot 4 \cdot (-1) = -40$ .
$(2, 2)$ : $\binom{5}{2} 2^2 \cdot \binom{4}{2} (-1)^2 = 10 \cdot 4 \cdot 6 \cdot 1 = 240$ .
$(3, 1)$ : $\binom{5}{3} 2^3 \cdot \binom{4}{1} (-1)^1 = 10 \cdot 8 \cdot 4 \cdot (-1) = -320$ .
$(4, 0)$ : $\binom{5}{4} 2^4 \cdot \binom{4}{0} (-1)^0 = 5 \cdot 16 \cdot 1 \cdot 1 = 80$ .

Step 4. Sum the contributions.

1 - 40 + 240 - 320 + 80 = -39.

The coefficient of $x^4$ in $(1 + 2x)^5 (1 - x)^4$ is $-39$ .

Ratio of consecutive terms and the greatest binomial coefficient

For an expansion such as $(1 + x)^n$ at a fixed value of $x$ , the ratio $T_{k+2} / T_{k+1}$ tells you whether successive terms grow or shrink. The largest term occurs where the ratio crosses $1$ .

In the expansion of

(1 + x)^{20}

x = 2

, find the value of

k

for which the term

T_{k+1}

is greatest

Step 1. Write the general term. $T_{k+1} = \binom{20}{k} 2^k$ for $k = 0, 1, \dots, 20$ .

Step 2. Form the ratio of consecutive terms.

\frac{T_{k+2}}{T_{k+1}} = \frac{\binom{20}{k+1} 2^{k+1}}{\binom{20}{k} 2^k} = \frac{20 - k}{k + 1} \cdot 2 = \frac{2(20 - k)}{k + 1}.

Step 3. Find where the ratio crosses $1$ . Terms grow while the ratio exceeds $1$ and shrink once it drops below $1$ . Solve $\frac{2(20 - k)}{k+1} > 1$ :

2(20 - k) > k + 1 \implies 40 - 2k > k + 1 \implies 39 > 3k \implies k < 13.

So $T_{k+2} > T_{k+1}$ for $k < 13$ and $T_{k+2} < T_{k+1}$ for $k > 13$ .

Step 4. Identify the largest term. The growth stops after $k = 13$ . So the largest term is $T_{14}$ , corresponding to $k = 13$ .

Step 5. Verify at the boundary. At $k = 13$ : ratio $= \frac{2 \cdot 7}{14} = 1$ , meaning $T_{15} = T_{14}$ . Both $T_{14}$ and $T_{15}$ are equal and are the largest terms. (If a single answer is required, give $k = 13$ , so the greatest term is $T_{14} = \binom{20}{13} 2^{13}$ .)

Find the value of

r

for which the binomial coefficient

\binom{n}{r}

is greatest, for

(1 + x)^n

x = 1

and

n = 12

Step 1. Restate the question: At $x = 1$ the term in $(1+x)^{12}$ corresponding to power $r$ is $\binom{12}{r}$ . We want the value of $r$ that maximises $\binom{12}{r}$ for $r \in \{0, 1, \dots, 12\}$ .
Step 2. Use symmetry: $\binom{12}{r} = \binom{12}{12 - r}$ , so the sequence of coefficients is symmetric about $r = 6$ .
Step 3. Form the ratio of consecutive coefficients

\frac{\binom{12}{r+1}}{\binom{12}{r}} = \frac{12 - r}{r + 1}.

Step 4. Find where the ratio crosses $1$ . Solve $\frac{12 - r}{r + 1} > 1$ :

12 - r > r + 1 \implies 11 > 2r \implies r < 5.5.

So $\binom{12}{r+1} > \binom{12}{r}$ for $r < 5.5$ , and the coefficients keep increasing until $r = 6$ .

Step 5. Conclude. The greatest binomial coefficient is $\binom{12}{6} = 924$ at $r = 6$ .

Sum identities

Several sum identities follow from the binomial theorem by substitution.

Prove that

\binom{n}{0} + \binom{n}{2} + \binom{n}{4} + \dots = \binom{n}{1} + \binom{n}{3} + \binom{n}{5} + \dots = 2^{n-1}

for all integers

n \geq 1

Step 1. Substitute $a = 1$ , $b = 1$ into the binomial theorem.

2^n = \binom{n}{0} + \binom{n}{1} + \binom{n}{2} + \dots + \binom{n}{n}. \tag{i}

Step 2. Substitute $a = 1$ , $b = -1$ into the binomial theorem. For $n \geq 1$ :

0 = \binom{n}{0} - \binom{n}{1} + \binom{n}{2} - \dots + (-1)^n \binom{n}{n}. \tag{ii}

Step 3. Add (i) and (ii). The odd-indexed terms cancel; the even-indexed ones double.

2^n + 0 = 2 \left[ \binom{n}{0} + \binom{n}{2} + \binom{n}{4} + \dots \right].

So $\binom{n}{0} + \binom{n}{2} + \binom{n}{4} + \dots = 2^{n-1}$ .

Step 4. Subtract (ii) from (i). The even-indexed terms cancel; the odd-indexed ones double.

2^n - 0 = 2 \left[ \binom{n}{1} + \binom{n}{3} + \binom{n}{5} + \dots \right].

So $\binom{n}{1} + \binom{n}{3} + \binom{n}{5} + \dots = 2^{n-1}$ .

Conclusion. Both the even-indexed and odd-indexed sums equal $2^{n-1}$ , and so they equal each other.

Common exam traps

Trap: Off-by-one in $T_{k + 1}$

$T_1$ has $k = 0$ ( $a^n$ ), and $T_{n + 1}$ has $k = n$ ( $b^n$ ). Many students get the indexing wrong, costing marks.

Trap: Wrong power-balance equation

For $\left( a x^p + b x^q \right)^n$ , the power of $x$ in $T_{k + 1}$ is $p(n - k) + q k$ . Not $p k + q(n - k)$ .

Trap: Forgetting coefficients

In $\left( 2 x + 3 \right)^n$ , the general term includes powers of both $2$ and $3$ . Easy to forget.

Trap: Combinations vs permutations

Read carefully whether order matters. "A group of $5$ " is unordered (combination); "a sequence of $5$ " or "a queue" is ordered (permutation).

Trap: Repeated elements in permutations

Forgetting to divide by $r_1! r_2! \cdots$ for repeats massively over-counts.

Practice patterns

Pattern A: Coefficient of a specific power

Find the coefficient of $x^5$ in $(2 + 3 x)^8$ . Find the coefficient of $x^{10}$ in $\left( x^2 + \frac{1}{x} \right)^{12}$ .

Pattern B: Term independent of $x$

Find the term independent of $x$ in $\left( 2 x - \frac{1}{x^2} \right)^9$ or $\left( x + \frac{2}{x^3} \right)^{12}$ .

Pattern C: Identity proof using $(1 + x)^n$

Show that $\sum_{k = 0}^{n} \binom{n}{k} 2^k = 3^n$ . (Set $x = 2$ in $(1 + x)^n = \sum \binom{n}{k} x^k$ .)

Pattern D: Counting with constraints

How many $4$ -letter words from MATHEMATICS contain at least one vowel? Use complementary counting from no-vowels.

Pattern E: Combined permutation/combination

In how many ways can $5$ books be chosen from $10$ and arranged on a shelf? $\binom{10}{5} \cdot 5!$ , equivalently ${}^{10} P_5$ .

Exam strategy

Binomial theorem questions are typically 2-4 marks each. Allocate $4-6$ minutes per question. Steps:

Write the general term $T_{k + 1}$ for the given expression.
Identify what is being asked: coefficient, specific term, independent term.
Form the right equation (matching power of $x$ , etc.) and solve for $k$ .
Substitute back to compute the answer.
State your answer clearly.

For identity proofs, the standard moves are: set specific values into $(1 + x)^n$ , add or subtract two such substitutions, or differentiate and substitute. The library of identities you need is small.

Check your knowledge

Try these questions before reading the solutions. They are roughly arranged from easier to harder.

Find the coefficient of $x^5$ in the expansion of $(2 + 3x)^8$ .
Find the term independent of $x$ in the expansion of $\left( 2x - \dfrac{1}{x^2} \right)^9$ .
Show that $\sum_{k=0}^{n} \binom{n}{k} 2^k = 3^n$ for all integers $n \geq 0$ .
Find the coefficient of $x^2$ in the expansion of $(1 + x)^5 (1 + 2x)^4$ .
Find the value of $r$ for which $\binom{15}{r}$ is greatest.
In how many ways can $4$ letters be selected from the word MATHEMATICS so that the selection includes at least one vowel? Use complementary counting.
In the expansion of $(1 + 3x)^{18}$ , find the value of $k$ for which the term $T_{k+1}$ is greatest when $x = 1$ .
Show that $\sum_{k=0}^{n} k \binom{n}{k} = n \cdot 2^{n-1}$ for all integers $n \geq 1$ by differentiating $(1 + x)^n$ .

Solutions

Q1
Step 1. Write the general term: $T_{k+1} = \binom{8}{k} 2^{8-k} (3x)^{k} = \binom{8}{k} 2^{8-k} 3^k x^k$ .
Step 2. Match the target power: Set $k = 5$ .
Step 3. Substitute: $T_6 = \binom{8}{5} 2^3 3^5 x^5 = 56 \cdot 8 \cdot 243 \, x^5$ .
Step 4. Evaluate: $56 \cdot 8 = 448$ and $448 \cdot 243 = 108\,864$ .

The coefficient of $x^5$ is $108\,864$ .

Q2
Step 1. Write the general term: $T_{k+1} = \binom{9}{k} (2x)^{9-k} \left( -\dfrac{1}{x^2} \right)^k = \binom{9}{k} 2^{9-k} (-1)^k \, x^{9-k} \cdot x^{-2k} = \binom{9}{k} 2^{9-k} (-1)^k \, x^{9 - 3k}$ .
Step 2. Set the power of $x$ to zero: $9 - 3k = 0$ , so $k = 3$ .
Step 3. Substitute $k = 3$: $T_4 = \binom{9}{3} 2^{6} (-1)^3 = 84 \cdot 64 \cdot (-1)$ .
Step 4. Evaluate: $84 \cdot 64 = 5376$ , so the term is $-5376$ .

The term independent of $x$ is $-5376$ .

Q3
Step 1. Start from the binomial theorem: $(1 + x)^n = \sum_{k=0}^{n} \binom{n}{k} x^k$ .
Step 2. Substitute $x = 2$: $(1 + 2)^n = \sum_{k=0}^{n} \binom{n}{k} 2^k$ .
Step 3. Simplify: LHS $= 3^n$ .
Conclusion: $\sum_{k=0}^{n} \binom{n}{k} 2^k = 3^n$ for all integers $n \geq 0$ .
Q4
Step 1. Identify the general terms: From $(1+x)^5$ , the term in $x^j$ is $\binom{5}{j} x^j$ . From $(1+2x)^4$ , the term in $x^i$ is $\binom{4}{i} 2^i x^i$ .
Step 2. List the pairs giving $x^2$: $(j, i) \in \{(0, 2), (1, 1), (2, 0)\}$ .
Step 3. Compute contributions

$(0, 2)$ : $\binom{5}{0} \cdot \binom{4}{2} 2^2 = 1 \cdot 6 \cdot 4 = 24$ .
$(1, 1)$ : $\binom{5}{1} \cdot \binom{4}{1} 2^1 = 5 \cdot 4 \cdot 2 = 40$ .
$(2, 0)$ : $\binom{5}{2} \cdot \binom{4}{0} = 10 \cdot 1 = 10$ .

Step 4. Sum. $24 + 40 + 10 = 74$ .

The coefficient of $x^2$ is $74$ .

Q5
Step 1. Use the ratio of consecutive coefficients: $\dfrac{\binom{15}{r+1}}{\binom{15}{r}} = \dfrac{15 - r}{r + 1}$ .
Step 2. Solve ratio $> 1$: $15 - r > r + 1 \implies r < 7$ .
Step 3. Interpret: Coefficients increase while $r < 7$ , so the largest is at $r = 7$ or $r = 8$ . Check: $\binom{15}{7} = 6435 = \binom{15}{8}$ . They are equal.

The greatest binomial coefficient is $\binom{15}{7} = \binom{15}{8} = 6435$ .

Q6
Step 1. Identify the distinct letters available: MATHEMATICS contains the distinct letters $\{M, A, T, H, E, I, C, S\}$ , that is $8$ different letters. The vowels are $A, E, I$ ( $3$ vowels) and the consonants are $M, T, H, C, S$ ( $5$ consonants).
Step 2. Count total selections of $4$ distinct letters: $\binom{8}{4} = 70$ .
Step 3. Count selections with no vowels (complement): Choose all $4$ from the $5$ consonants: $\binom{5}{4} = 5$ .
Step 4. Apply complementary counting: Selections with at least one vowel $= 70 - 5 = 65$ .

The number of selections is $65$ .

Q7
Step 1. Write the general term at $x = 1$: $T_{k+1} = \binom{18}{k} 3^k$ for $k = 0, 1, \dots, 18$ .
Step 2. Form the ratio of consecutive terms

\frac{T_{k+2}}{T_{k+1}} = \frac{\binom{18}{k+1} 3^{k+1}}{\binom{18}{k} 3^k} = \frac{18 - k}{k + 1} \cdot 3 = \frac{3(18 - k)}{k + 1}.

Step 3. Solve ratio $> 1$ . $3(18 - k) > k + 1 \implies 54 - 3k > k + 1 \implies 53 > 4k \implies k < 13.25$ .

Step 4. Interpret. Terms increase while $k < 13.25$ , so the largest is at $k = 13$ (giving $T_{14}$ ).

The greatest term is $T_{14} = \binom{18}{13} 3^{13}$ , corresponding to $k = 13$ .

Q8
Step 1. Write the binomial expansion: $(1 + x)^n = \sum_{k=0}^{n} \binom{n}{k} x^k$ .
Step 2. Differentiate both sides with respect to $x$: Use the chain rule on the LHS and term-by-term differentiation on the RHS.

n (1 + x)^{n - 1} = \sum_{k=0}^{n} k \binom{n}{k} x^{k-1}.

Step 3. Substitute $x = 1$ . LHS $= n \cdot 2^{n-1}$ . RHS $= \sum_{k=0}^{n} k \binom{n}{k}$ .

Conclusion. $\sum_{k=0}^{n} k \binom{n}{k} = n \cdot 2^{n-1}$ for all integers $n \geq 1$ .

Linked dot points

For more detail, see the dot point pages at permutations, combinations, binomial theorem and Pascal's triangle, and pigeonhole principle.

For NESA past papers and marking guidelines, refer to educationstandards.nsw.edu.au.

binomial-theorem
combinatorics
hsc-maths-extension-1
pascal
2026

What this guide covers

Counting principles

The multiplication principle

The addition principle

Complementary counting

Permutations

Combinations

The binomial theorem

Pascal's triangle

Standard exam techniques

Find the coefficient of a specific power

Find the term independent of the variable

Coefficient of xkx^kxk in a product of two binomial expansions

Ratio of consecutive terms and the greatest binomial coefficient

Sum identities

Common exam traps

Trap: Off-by-one in Tk+1T_{k + 1}Tk+1​

Trap: Wrong power-balance equation

Trap: Forgetting coefficients

Trap: Combinations vs permutations

Trap: Repeated elements in permutations

Practice patterns

Pattern A: Coefficient of a specific power

Pattern B: Term independent of xxx

Pattern C: Identity proof using (1+x)n(1 + x)^n(1+x)n

Pattern D: Counting with constraints

Pattern E: Combined permutation/combination

Exam strategy

Check your knowledge

Linked dot points

Coefficient of $x^k$ in a product of two binomial expansions

Trap: Off-by-one in $T_{k + 1}$

Pattern B: Term independent of $x$

Pattern C: Identity proof using $(1 + x)^n$