User:Aspaine/sandbox

Proof 1
This proof involves extension of the Andres Reflection method as used in the second proof for the Catalan number. The following shows how every path from the bottom left $$(0,0)$$ to the top right $$(k, n)$$ of the diagram that crosses the constraint $$n-k+m-1 = 0$$ can also be reflected to the end point $$(n + m, k - m)$$.

We consider three cases to determine the number of paths from $$(0,0)$$ to $$(k, n)$$ that do not cross the constraint:

(1) when $$m > k$$ the constraint cannot be crossed, so all paths from $$(0,0)$$ to $$(k, n)$$ are valid, i.e. $$C_{m}(n,k)=\left(\begin{array}{c} n+k\\ k \end{array}\right)$$.

(2) when $$k - m + 1 > n$$ it is impossible to form a path that does not cross the constraint, i.e. $$C_{m}(n,k)= 0 $$.

(3) when $$ m\leq k\leq n+m-1 $$, then $$C_{m}(n,k)$$ is the number of 'red' paths $$\left(\begin{array}{c} n+k\\ k \end{array}\right)$$ minus the number of 'yellow' paths that cross the constraint, i.e. $$\left(\begin{array}{c} (n+m)+(k-m)\\ k-m \end{array}\right) = \left(\begin{array}{c} n+k\\ k-m \end{array}\right)$$.

Thus the number of paths from $$(0,0)$$ to $$(k, n)$$ that do not cross the constraint $$n - k + m - 1 = 0$$ is as indicated in the formula in the previous section "Generalization".

Proof 2
Firstly, we confirm the validity of the recurrence relation $$C_{m}(n,k)=C_{m}(n-1,k)+C_{m}(n,k-1)$$ by breaking down $$C_{m}(n,k)$$ into two parts, the first for XY combinations ending in X and the second for those ending in Y. The first group therefore has $$C_{m}(n-1,k)$$ valid combinations and the second has $$C_{m}(n,k-1)$$. Proof 2 is completed by verifying the solution satisfies the recurrence relation and obeys initial conditions for $$C_{m}(n,1)$$ and $$C_{m}(1,k)$$.