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

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The number of diagonal paths from $(0, a)$ to $(n, b)$ is $(\binom{n}{u})$ for the natural number $u$ with $2 u = n + b - a$ , and $0$ when no such $u$ exists

Statement

Let $a, b \in Z$ and $n \in N$ , and let $W ((0, a), (n, b))$ be the set of diagonal lattice paths of length $n$ from $(0, a)$ whose height function ends at $b$ (Diagonal lattice paths with steps $U = (1, 1)$ and $D = (1, - 1)$ , and the height function).

Suppose $2$ divides $n + b - a$ and $- n \leq b - a \leq n$ , and let $u \in N$ be the natural number with $2 u = n + b - a$ ; then $0 \leq u \leq n$ and

$∣ W ((0, a), (n, b)) ∣ = (\binom{n}{u}) .$
If either condition fails then $W ((0, a), (n, b)) = \emptyset$ , so its cardinality is $0$ .

In both cases the set is finite, and the count depends on $a$ and $b$ only through the difference $b - a$ .

Facts & Assumptions

Given: integers $a$ and $b$ and a natural number $n$ .

[F1]

A diagonal path of length $n$ from $(0, a)$ has $v (i) = (i, h (i))$ with $h (0) = a$ and $h (i) - h (i - 1) \in {1, - 1}$ ; with $μ (n)$ the number of up-steps its endpoint height is $h (n) = a + 2 μ (n) - n$ ; and $W ((0, a), (n, b))$ is nonempty exactly when $2$ divides $n + b - a$ and $- n \leq b - a \leq n$ (Diagonal lattice paths with steps $U = (1, 1)$ and $D = (1, - 1)$ , and the height function).

[L1]

For $u, d \in N$ and $n = u + d$ , replacing $U$ by $N$ and $D$ by $E$ is a bijection $W ((0, 0), (n, u - d)) \to M ((0, 0), (d, u))$ (The two step sets describe the same objects: $U \mapsto N$ , $D \mapsto E$ is a bijection matching the diagonal $y = x$ with the level $0$ ).

[L2]

For $m, n^{'} \in N$ the set $M ((0, 0), (m, n^{'}))$ is finite with $∣ M ((0, 0), (m, n^{'})) ∣ = (\binom{m + n^{'}}{n^{'}})$ ( $∣ M ((0, 0), (m, n)) ∣ = (\binom{m + n}{n})$ ).

[L3]

For $f : A \to B$ : $f$ is a bijection if and only if there is a function $g : B \to A$ with $g \circ f = Δ_{A}$ and $f \circ g = Δ_{B}$ ( $f : A \to B$ is a bijection if and only if there is a function $g : B \to A$ with $g \circ f = Δ_{A}$ and $f \circ g = Δ_{B}$ ; such a $g$ is unique, equals the inverse relation $f^{- 1}$ , and is itself a bijection).

[L4]

If $A$ is finite and $f : A \to B$ is a bijection then $B$ is finite and $∣ B ∣ = ∣ A ∣$ ; and a finite set has cardinality $0$ exactly when it is empty (The cardinality $∣ A ∣$ of a finite set).

[L5]

$(\binom{n}{k})$ is the number of $k$ -element subsets of the natural number $n$ , and $(\binom{n}{0}) = (\binom{n}{n}) = 1$ (The set $[A]^{k}$ of $k$ -element subsets and the binomial coefficient $(\binom{n}{k}) : = ∣ [n]^{k} ∣$ ).

Proof

technique · direct

1.1

Subtracting the constant $(0, a)$ leaves every difference $v (i) - v (i - 1)$ unchanged and sends $(0, a)$ to $(0, 0)$ and $(n, b)$ to $(n, b - a)$ , and adding it back inverts the operation; so it is a bijection $W ((0, a), (n, b)) \to W ((0, 0), (n, b - a))$ and the two sets have the same cardinality.

F1L3L4

2.1

If $2$ does not divide $n + b - a$ , or if $b - a$ lies outside the range from $- n$ to $n$ , then $W ((0, a), (n, b))$ is empty and its cardinality is $0$ , which is clause 2.

F1L4step 1.1

2.2

If the two conditions hold, let $u$ be the natural number with $2 u = n + b - a$ and put $d : = n - u$ . From $2 u = n + b - a$ and $u + d = n$ one gets $u - d = b - a$ , and $0 \leq u \leq n$ because $0 \leq n + b - a \leq 2 n$ .

F1step 1.1algebra

3.1

By step 2.2 and [L1] the set $W ((0, 0), (n, b - a)) = W ((0, 0), (n, u - d))$ is in bijection with $M ((0, 0), (d, u))$ , which by [L2] is finite with $(\binom{d + u}{u}) = (\binom{n}{u})$ elements; transporting along that bijection and along the translation of step 1.1 gives clause 1. At $n = 0$ the conditions force $b = a$ and $u = 0$ , and the one empty path is counted by $(\binom{0}{0}) = 1$ ; at $b - a = n$ they force $u = n$ , and the one all-up path is counted by $(\binom{n}{n}) = 1$ .

L1L2L4L5step 1.1step 2.1step 2.2∎

Remarks

The vanishing clause is used, not decorative. The reflection principle below subtracts one of these counts from another, and both the parity and the range conditions can fail for the reflected endpoint while holding for the original; the difference is correct only because the count is then $0$ rather than undefined.
Why the answer is stated through $u$ rather than as a quotient. The natural number $u$ with $2 u = n + b - a$ exists exactly under the stated hypotheses, and writing $\frac{1}{2} (n + b - a)$ would name an element of a field where the hypothesis of the statement is that the halving is exact in $Z$ .

Depends on

Used by

The reflection principle: paths from (0,a) to (n,b) staying strictly above level c are counted by a difference of two binomial coefficients Theorem

Dependency tree · two levels

30 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, §10.3 (standard reference, not scraped)

The number of diagonal paths from (0,a)(0,a) to (n,b)(n,b) is (nu)(un​) for the natural number uu with 2u=n+b−a2u=n+b−a, and 00 when no such uu exists