CorollaryStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-26

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For $n \geq 1$ the pairs of non-intersecting monotone paths $(1, 0) \to (n + 1, n)$ and $(0, 1) \to (n, n + 1)$ number ${(\binom{2 n}{n})}^{2} - (\binom{2 n}{n + 1}) (\binom{2 n}{n - 1})$

Statement

Let $n \geq 1$ . Then the number of pairs $(P, Q)$ such that

$P$ is a monotone lattice path from $(1, 0)$ to $(n + 1, n)$ ,
$Q$ is a monotone lattice path from $(0, 1)$ to $(n, n + 1)$ , and
$P$ and $Q$ do not intersect,

${(\binom{2 n}{n})}^{2} - (\binom{2 n}{n + 1}) (\binom{2 n}{n - 1}) .$

Facts & Assumptions

Given: a natural number $n \geq 1$ .

[L1]

In the compatible case, the determinant of the matrix of path counts equals the number of non-intersecting systems joining the starts to the corresponding ends ( $\det (∣ M (A_{i}, E_{j}) ∣)_{i, j} = \sum_{π \in S_{r}} sgn (π) \cdot # {non-intersecting π -systems}$ ).

[L2]

If $a \leq c$ and $b \leq d$ , and $m = c - a$ , $n^{'} = d - b$ are the corresponding natural numbers, then the number of monotone paths from $(a, b)$ to $(c, d)$ is $(\binom{m + n^{'}}{n^{'}})$ ( $∣ M ((0, 0), (m, n)) ∣ = (\binom{m + n}{n})$ ).

Proof

technique · direct

1.1

With $A_{0} = (1, 0)$ , $A_{1} = (0, 1)$ , $E_{0} = (n + 1, n)$ and $E_{1} = (n, n + 1)$ , the four path counts are $m_{0, 0} = m_{1, 1} = (\binom{2 n}{n}), m_{0, 1} = (\binom{2 n}{n + 1}), m_{1, 0} = (\binom{2 n}{n - 1})$ by [L2].

1.2

Every monotone path $P : (1, 0) \to (n, n + 1)$ meets every monotone path $Q : (0, 1) \to (n + 1, n)$ . After $t$ steps, both paths lie on the line $x + y = t + 1$ ; writing their $x$ -coordinates as $x_{P} (t)$ and $x_{Q} (t)$ , the difference $x_{P} (t) - x_{Q} (t)$ starts at $1$ and ends at $- 1$ , and each step changes it by at most $1$ . So some index $t$ has $x_{P} (t) = x_{Q} (t)$ , and then the common value of $x + y$ forces the same $y$ -coordinate as well.

givenalgebra

2.1

Step 1.2 is exactly the compatibility condition for these two pairs of endpoints, so [L1] applies and gives the count as the determinant $\det (\begin{array}{cc} (\binom{2 n}{n}) & (\binom{2 n}{n + 1}) \\ (\binom{2 n}{n - 1}) & (\binom{2 n}{n}) \end{array}) = {(\binom{2 n}{n})}^{2} - (\binom{2 n}{n + 1}) (\binom{2 n}{n - 1}) .$

L1step 1.1step 1.2

3.1

At $n = 1$ this gives $2^{2} - 1 \cdot 1 = 3$ , which matches the direct count: there are four ordered pairs of paths, and exactly one pair meets at the point $(1, 1)$ .

step 2.1∎

Remarks

The determinant is already nontrivial at $n = 1$ : the count is not the product of the two individual path counts because the compatibility condition removes the intersecting pair.

Depends on

Used by

A two-by-two determinant counting non-intersecting path pairs Example

Dependency tree · two levels

22 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

C. Krattenthaler, "Lattice Path Enumeration", ch. 10 of the Handbook of Enumerative Combinatorics, Corollary 10.13.2 (standard reference, not scraped)

For n≥1n≥1 the pairs of non-intersecting monotone paths (1,0)→(n+1,n)(1,0)→(n+1,n) and (0,1)→(n,n+1)(0,1)→(n,n+1) number (2nn)2−(2nn+1)(2nn−1)(n2n​)2−(n+12n​)(n−12n​)