QED
Combinatorics · step 13 of 13

Catalan numbers & lattice-path counting

Monotone lattice paths from (0,0) to (m,n) number C(m+n, n), since a path is determined by which of its steps go up. Constraining a path to stay below the diagonal gives the Catalan numbers Cₙ = C(2n,n)/(n+1), counted by the reflection principle. They also count balanced bracket strings, binary tree shapes and triangulations of a polygon.

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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.

  1. Count unconstrained paths with a binomialA path with m rights and n ups is a word in R and U, so there are C(m+n, n) of them.
  2. Apply reflection for the constrained countReflect the offending prefix of a bad path to build a bijection with paths to a shifted endpoint, then subtract.
  3. Use the Catalan formulaCₙ = C(2n,n)/(n+1), giving 1, 1, 2, 5, 14, 42, 132.
  4. Recognise Catalan structuresBalanced brackets, binary trees, non-crossing chords, and polygon triangulations all reduce to the same count.

Worked example

How many monotone lattice paths from (0,0) to (4,3), and how many balanced bracket strings with 4 pairs?

  1. Paths: 7 steps of which 3 are up, so C(7,3).
  2. = 35.
  3. Balanced brackets with 4 pairs: the Catalan number C₄ = C(8,4)/5.
  4. = 70/5.

Answer. 35 lattice paths; 14 balanced bracket strings with 4 pairs.

Where marks get dropped

These are the specific errors that cost credit on catalan numbers & lattice-path counting questions — QED's rubric penalises each of them separately.

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Catalan numbers & lattice-path counting — frequently asked questions

What is the reflection principle?

Bad paths crossing a boundary are put in bijection with all paths to a reflected endpoint, so the bad count is a binomial and can be subtracted from the total.

What do Catalan numbers count?

Balanced brackets, binary tree shapes, triangulations of an (n+2)-gon, non-crossing chord diagrams, and Dyck paths — dozens of families, all bijective with each other.

What is the Catalan recurrence?

Cₙ₊₁ = Σ_{k=0}^{n} CₖCₙ₋ₖ, obtained by splitting a structure at its first return to the axis, or at the root of a binary tree.

The rest of Combinatorics

Counting principles, permutations, combinations. Each subtopic below has its own method, worked example and mark-losing traps.

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