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NotesMath AA SLTopic 1.9Pascal's triangle & nCr
Back to Math AA SL Topics
1.9.11 min read

Pascal's triangle & nCr

IB Mathematics: Analysis and Approaches • Unit 1

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Contents

  • Pascal's triangle
  • The ⁿCᵣ formula & the GDC
  • The binomial theorem
The big idea: The coefficients of (a + b)ⁿ are row n of Pascal's triangle — start each row with 1, and every inside number is the sum of the two above it.

Building the triangle

  • Row 0: 1
  • Row 1: 1 1
  • Row 2: 1 2 1 (2 = 1 + 1)
  • Row 3: 1 3 3 1 (3 = 1 + 2)
  • Row 4: 1 4 6 4 1 (6 = 3 + 3)

IB-style question — expand with the triangle

Use Pascal's triangle to expand (a + b)³.

Step by step

  1. Row 3 of the triangle gives the coefficients.
  2. Powers of a fall 3 → 0; powers of b rise 0 → 3.

Final answer

a³ + 3a²b + 3ab² + b³.

Small n only: Pascal's triangle is quickest for small n (up to about 6). For a bigger power, use ⁿCᵣ (next section) instead of writing out every row.
Coefficients without the triangle: For bigger powers, a coefficient is a combination ⁿCᵣ — compute it with the formula, or with the GDC's nCr button.
ⁿCᵣ — the (r + 1)th entry of row n of Pascal's triangle.

IB-style question — by hand

Find ⁵C₂.

Step by step

  1. Substitute n = 5, r = 2.
  2. Cancel the 3! and compute.

Final answer

⁵C₂ = 10 (matches row 5: 1, 5, 10, 10, 5, 1).

IB-style question — a bigger one

Find ¹⁰C₄.

Step by step

  1. Substitute n = 10, r = 4; keep four factors on top.
  2. Compute.

Final answer

¹⁰C₄ = 210.

IB-style question — read a coefficient straight off

Find the coefficient of x³ in the expansion of (1 + x)⁴.

Step by step

  1. Every term of (1 + x)⁴ is a ⁿCᵣ times a power of x, so the coefficient of xʳ is just ⁴Cᵣ. The x³ coefficient is therefore ⁴C₃.
  2. Work out ⁴C₃ with the formula (or the GDC's nCr).

Final answer

The coefficient of x³ is ⁴C₃ = 4 — you never have to expand the whole bracket.

GDC tip (Paper 2): On the GDC: type n, then MATH → ▶ (PRB) → 3: nCr, then r.

So 10 nCr 4 = 210 — much faster than the triangle for large n.

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Putting it together: The binomial theorem writes (a + b)ⁿ as a sum of terms ⁿCᵣ aⁿ⁻ʳ bʳ — with exactly n + 1 terms.
Coefficients ⁿCᵣ; powers of a fall, powers of b rise, and in every term they sum to n.

IB-style question — read the structure

Write out the structure of (a + b)⁵.

Step by step

  1. Row 5 coefficients (six of them).
  2. Powers of a fall 5 → 0; powers of b rise 0 → 5.

Final answer

Six terms (n + 1 = 5 + 1).

How many terms?: (a + b)ⁿ always has n + 1 terms.

The power of a counts down from n to 0; the power of b counts up from 0 to n; in every term the two powers sum to n.

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Write down the number of terms in the expansion of (3x − 2)¹⁵. [1 mark]

Related Math AA SL Topics

Continue learning with these related topics from the same unit:

1.1.1Writing standard form
1.1.2Standard form by hand
1.2.1nth term
1.2.2Sum of n terms
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