3. Let the sequence be composed of ordered pairs of positive integers $(x, y)$: $(1,1),(1,2),(2,1),(1,3),(2,2)$, $(3,1),(1,4),(2,3),(3,2),(4,1), \cdots$, arranged in ascending order of $x+y$, and when $x+y$ is equal, in ascending order of $x$. Then the position of the ordered pair $(m, n)(m, n$ both being positive integers $)$ in this sequence is (.
(A) the $2 m+n-1$ th position
(B) the $2 m+n-2$ th position
(C) the $\frac{(m+n-1)(m+n)}{2}+m$ th position
(D) the $\frac{(m+n-2)(m+n-1)}{2}+m$ th position
3. D.

Group by the value of $x+y$.
When $x+y=2$, it is the first group;
When $x+y=3$, it is the second group;
.....
When $x+y=m+n$, it is the $(m+n-1)$-th group.
Therefore, the first $(m+n-2)$ groups of the sequence have a total of ordered pairs
$$
\frac{(m+n-2)(m+n-1)}{2} \text { (pairs). }
$$

For the ordered pair $(m, n)$, when $x=m$, it is the $m$-th position in the $(m+n-1)$-th group, so the ordered pair $(m, n)$ is in the $\frac{(m+n-2)(m+n-1)}{2}+m$-th position of the sequence.