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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, and standard exam techniques (general term, independent term, sum identities).

Generated by Claude OpusReviewed by Better Tuition Academy12 min readNESA-MATH-EXT1-A1

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 xkx^k, term independent of xx, identity proofs.
  • Sum identities like βˆ‘(nk)=2n\sum \binom{n}{k} = 2^n.

Counting principles

The multiplication principle

If a procedure has steps 1,2,…,k1, 2, \dots, k and step ii can be done in nin_i ways (independent of the previous steps), the total number of ways is n1β‹…n2β‹―nkn_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

(atΒ leastΒ 1)=(total)βˆ’(none).(\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 nn distinct objects in a row is n!n!. To arrange rr out of nn: nPr=n!(nβˆ’r)!{}^n P_r = \frac{n!}{(n - r)!}.

For objects with repeats (say n1n_1 of one kind, n2n_2 of another), the number of distinct arrangements is

n!n1! n2!β‹―nk!.\frac{n!}{n_1! \, n_2! \cdots n_k!}.

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

Combinations

The number of unordered selections of rr from nn distinct objects is

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

Key identities:

  • Symmetry: (nr)=(nnβˆ’r)\binom{n}{r} = \binom{n}{n - r}.
  • Pascal's rule: (nr)=(nβˆ’1rβˆ’1)+(nβˆ’1r)\binom{n}{r} = \binom{n - 1}{r - 1} + \binom{n - 1}{r}.
  • Sum: βˆ‘r=0n(nr)=2n\sum_{r = 0}^{n} \binom{n}{r} = 2^n.

The binomial theorem

For any non-negative integer nn,

(a+b)n=βˆ‘k=0n(nk)anβˆ’kbk.(a + b)^n = \sum_{k = 0}^{n} \binom{n}{k} a^{n - k} b^k.

The general term is

Tk+1=(nk)anβˆ’kbk.T_{k + 1} = \binom{n}{k} a^{n - k} b^k.

So T1=anT_1 = a^n (when k=0k = 0) and Tn+1=bnT_{n + 1} = b^n (when k=nk = n).

Pascal's triangle

The first few rows:

111121133114641151010511615201561\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 nn gives the coefficients of (a+b)n(a + b)^n.

Standard exam techniques

Find the coefficient of xkx^k in IMATH_35

Use the general term: Tk+1=(nk)(cx)nβˆ’kdkT_{k + 1} = \binom{n}{k} (c x)^{n - k} d^k. Wait, this gives xnβˆ’kx^{n - k}, not xkx^k. Adjust: for the power of xx to be jj in (cx+d)n(c x + d)^n, we need nβˆ’k=jn - k = j, so k=nβˆ’jk = n - j. Coefficient: (nnβˆ’j)cjdnβˆ’j=(nj)cjdnβˆ’j\binom{n}{n - j} c^{j} d^{n - j} = \binom{n}{j} c^{j} d^{n - j}.

Concrete example: coefficient of x3x^3 in (2x+5)7(2 x + 5)^7.

Tk+1=(7k)(2x)7βˆ’k(5)kT_{k + 1} = \binom{7}{k} (2 x)^{7 - k} (5)^k. For x3x^3: 7βˆ’k=37 - k = 3, k=4k = 4. Coefficient: (74)(2)3(5)4=35β‹…8β‹…625=175 000\binom{7}{4} (2)^3 (5)^4 = 35 \cdot 8 \cdot 625 = 175 \, 000.

Find the term independent of IMATH_52

The general term Tk+1T_{k + 1} has some power of xx. Set the power to 00 and solve for kk.

Example: find the term independent of xx in (x2+3x)6\left( x^2 + \frac{3}{x} \right)^6.

Tk+1=(6k)(x2)6βˆ’k(3x)k=(6k)3kx12βˆ’2kβˆ’k=(6k)3kx12βˆ’3kT_{k + 1} = \binom{6}{k} (x^2)^{6 - k} \left( \frac{3}{x} \right)^k = \binom{6}{k} 3^k x^{12 - 2 k - k} = \binom{6}{k} 3^k x^{12 - 3 k}.

Independent: 12βˆ’3k=012 - 3 k = 0, so k=4k = 4. T5=(64)34=15β‹…81=1215T_5 = \binom{6}{4} 3^4 = 15 \cdot 81 = 1215.

Sum identities

Several sum identities follow from the binomial theorem by substitution.

Identity
IMATH_63 . (Set a=b=1a = b = 1.)
Identity
IMATH_65 for nβ‰₯1n \ge 1. (Set a=1a = 1, b=βˆ’1b = -1.)
Identity
IMATH_69 . (Differentiate (1+x)n(1 + x)^n and set x=1x = 1.)

These are standard HSC identity-proof items. A typical question:

"Prove that (n0)+(n2)+(n4)+β‹―=(n1)+(n3)+(n5)+…\binom{n}{0} + \binom{n}{2} + \binom{n}{4} + \dots = \binom{n}{1} + \binom{n}{3} + \binom{n}{5} + \dots."

Subtract βˆ‘k=0n(βˆ’1)k(nk)=0\sum_{k = 0}^{n} (-1)^k \binom{n}{k} = 0 from βˆ‘k=0n(nk)=2n\sum_{k = 0}^{n} \binom{n}{k} = 2^n to get that the alternating sum is 00, then the even-indexed sum equals the odd-indexed sum, each being 2nβˆ’12^{n - 1}.

Common exam traps

Trap: Off-by-one in IMATH_77

T1T_1 has k=0k = 0 (ana^n), and Tn+1T_{n + 1} has k=nk = n (bnb^n). Many students get the indexing wrong, costing marks.

Trap: Wrong power-balance equation

For (axp+bxq)n\left( a x^p + b x^q \right)^n, the power of xx in Tk+1T_{k + 1} is p(nβˆ’k)+qkp(n - k) + q k. Not pk+q(nβˆ’k)p k + q(n - k).

Trap: Forgetting coefficients

In (2x+3)n\left( 2 x + 3 \right)^n, the general term includes powers of both 22 and 33. Easy to forget.

Trap: Combinations vs permutations

Read carefully whether order matters. "A group of 55" is unordered (combination); "a sequence of 55" or "a queue" is ordered (permutation).

Trap: Repeated elements in permutations

Forgetting to divide by r1!r2!β‹―r_1! r_2! \cdots for repeats massively over-counts.

Practice patterns

Pattern A: Coefficient of a specific power

Find the coefficient of x5x^5 in (2+3x)8(2 + 3 x)^8. Find the coefficient of x10x^{10} in (x2+1x)12\left( x^2 + \frac{1}{x} \right)^{12}.

Pattern B: Term independent of IMATH_99

Find the term independent of xx in (2xβˆ’1x2)9\left( 2 x - \frac{1}{x^2} \right)^9 or (x+2x3)12\left( x + \frac{2}{x^3} \right)^{12}.

Pattern C: Identity proof using IMATH_103

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

Pattern D: Counting with constraints

How many 44-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 55 books be chosen from 1010 and arranged on a shelf? (105)β‹…5!\binom{10}{5} \cdot 5!, equivalently 10P5{}^{10} P_5.

Exam strategy

Binomial theorem questions are typically 2-4 marks each. Allocate 4βˆ’64-6 minutes per question. Steps:

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

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

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