Catalan Number
Equivalent Definitions of the Catalan Number
The Catalan number $C_n$ can be defined in several equivalent ways:
- Parentheses Matching: $C_n$ is the number of valid ways to form $n$ pairs of parentheses such that they are properly matched (e.g.,
(()())or(())()for $n = 3$). - Dyck Paths: $C_n$ is the number of lattice paths from $(0,0)$ to $(2n,0)$ in the integer plane that:
- Use $n$ steps of $(1,1)$ (up) and $n$ steps of $(1,-1)$ (down),
- Never dip below the $x$-axis at any point.
- Triangulations of a Polygon: $C_n$ is the number of ways to triangulate a convex polygon with $n+2$ vertices.
- Binary Trees: $C_n$ is the number of distinct binary trees with $n$ nodes.
Calculation of $C_n$
The $n$-th Catalan number is given by:
\[C_n = \frac{1}{n+1} \binom{2n}{n}.\]Proof Using Reflection Principle
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Consider a lattice path from $(0,0)$ to $(2n,0)$. The total number of such paths, without any restrictions, is $\binom{2n}{n}$, as we need to choose $n$ upward steps (or equivalently, downward steps) from $2n$ total steps.
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The invalid paths dip below the $x$-axis. Reflect any such path at the first point where it touches $y = -1$. The reflection creates a path from $(0,0)$ to $(2n,-2)$ with $n-1$ up steps and $n+1$ down steps.
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The reflection construction is a bijection between the invalid paths and the lattice path from $(0,0)$ to $(2n, -2)$.
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The number of such invalid paths is $\binom{2n}{n-1}$, as we are choosing $n-1$ upward steps from $2n$ total steps.
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Subtracting these invalid paths from the total gives:
Using the identity $\binom{2n}{n-1} = \frac{n}{n+1} \binom{2n}{n}$, we simplify:
\[C_n = \binom{2n}{n} - \frac{n}{n+1} \binom{2n}{n} = \frac{1}{n+1} \binom{2n}{n}.\]Example Applications
Ticket Line Problem
Scenario: At a theater, $2n$ people are waiting to buy tickets. $n$ of them have only \$5 bills, and the other $n$ have only \$10 bills. Each ticket costs \$5, and the ticket seller has no change initially. What is the probability that all people can buy their tickets without anyone needing to change their position in line?
Solution:
- Model this situation as a lattice path problem where:
- A person with a \$5 bill corresponds to an upward step $(1,1)$,
- A person with a \$10 bill corresponds to a downward step $(1,-1)$.
- The requirement that no one needs to change positions translates to ensuring that the lattice path never dips below the $x$-axis.
The number of valid paths is $C_n = \frac{1}{n+1} \binom{2n}{n}$, and the total number of paths (without restrictions) is $\binom{2n}{n}$. Thus, the probability is:
\[p = \frac{\text{Valid paths}}{\text{Total paths}} = \frac{C_n}{\binom{2n}{n}} = \frac{1}{n+1}.\]