How many pairs?

We consider a general solution to the problem of how many pairs $(x,y)$ there are with $x \in \{0,1,2, \ldots, n\},$ $y \in \{0,2,4,\ldots, 2n\}$ and $x < y.$ We consider two cases. Either $n = 2k$ or $n = 2k+1.$

When $n = 2k,$ we count the number of possible $x$ values for each $y$ value. For $0 \leq y \leq 2k$ there are $y$ values of $x$ that will give $x < y.$ When $y > 2k,$ all $n+1$ values of $x$ will give $x < y.$ In the first instance, there are $2 + 4 + \cdots + 2k = \frac{n(n+2)}{4}$ pairs $(x,y)$ which will satisfy the conditions and in the second there are $\frac{n(n+1)}{2}$ pairs. In total this gives $\frac{3n^2 + 4n}{4}$ pairs when $n = 2k.$

When $n = 2k+1$ we count in a similar manner. When $0 < y \leq 2k$ there are $y$ values of $x$ with $x < y.$ When $y > 2k$ there are $n + 1$ values of $x$ with $x < y.$ This gives $2 + 4 + \cdots + n-1 = \frac{(n+1)(n-1)}{4}$ pairs for the first instance and $\frac{(n+1)(n+1)}{2}$ pairs for he second instance, for a total of $\frac{3n^2 + 4n + 1}{4}$ pairs when $n$ is odd.

We can combine the two formulae to get $\left\lfloor\frac{3n^2 + 4n + 1}{4}\right\rfloor$ as the general value for each $n.$ When $n = 25$ this gives $494.$