Binomial theorem & Pascal’s triangle
(x + y)ⁿ = Σ C(n,k)x^(n−k)y^k — the coefficients are exactly row n of Pascal’s triangle, because choosing which k of the n brackets contribute a y is a combination. Pascal’s identity C(n,k) = C(n−1,k−1) + C(n−1,k) is what builds each row from the one above, and it has a one-line counting proof.
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Method: how to approach it
The order below is what examiners expect to see, and each step carries its own marks.
- Write the general termThe term in y^k is C(n,k)x^(n−k)y^k. For a specific power, solve for k first.
- Handle coefficients and signs inside the binomialFor (2x − 3)ⁿ the general term is C(n,k)(2x)^(n−k)(−3)^k — raise the whole bracket contents to the power.
- Use Pascal’s triangle for small nEach entry is the sum of the two above it. Row 5 is 1, 5, 10, 10, 5, 1.
- Exploit the row identitiesThe row sums to 2ⁿ (set x = y = 1), and the alternating sum is 0 (set x = 1, y = −1).
Worked example
Find the coefficient of x³ in (2x + 1)⁵.
- General term: C(5,k)(2x)^(5−k)(1)^k, so the power of x is 5 − k.
- For x³ we need 5 − k = 3, so k = 2.
- Term: C(5,2)(2x)³ = 10 × 8x³.
- = 80x³.
Answer. The coefficient of x³ is 80.
Where marks get dropped
These are the specific errors that cost credit on binomial theorem & pascal’s triangle questions — QED's rubric penalises each of them separately.
- Forgetting to raise the numerical coefficient. In (2x)³ the 2 is cubed to 8, and missing this is the most common error.
- Getting the index wrong: the term with y^k has x^(n−k), not x^k.
- Dropping the sign in (a − b)ⁿ. The signs alternate because (−b)^k carries (−1)^k.
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Binomial theorem & Pascal’s triangle — frequently asked questions
Why do the rows sum to 2ⁿ?
Setting x = y = 1 gives (1+1)ⁿ = ΣC(n,k). Combinatorially, the total number of subsets of an n-set is 2ⁿ.
How is Pascal’s identity proved?
Count the k-subsets of an n-set by whether they contain a fixed element: those that do number C(n−1,k−1), those that do not number C(n−1,k).
Does the binomial theorem work for negative or fractional powers?
Yes, as an infinite series — Newton’s generalised binomial theorem — valid for |x| < 1. That is where the expansion of (1+x)^(1/2) comes from.
The rest of Combinatorics
Counting principles, permutations, combinations. Each subtopic below has its own method, worked example and mark-losing traps.
- 1Sum & product rules
- 2Permutations & combinations
- 3Binomial theorem & Pascal’s triangle
- 4The pigeonhole principle
- 5Inclusion–exclusion
- 6Counting with repetition (stars & bars)
- 7Derangements & counting surjections
- 8Double counting & bijective proofs
- 9Setting up & solving counting recurrences
- 10Generating functions — an introduction
- 11Multinomial coefficients & repeated items
- 12Hockey stick & Vandermonde identities
- 13Catalan numbers & lattice-path counting
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