Let A be a set containing 10 elements. Suppose A is partitioned...

The relation $R$ is defined such that $(x, y) \in R \iff x, y \in A_i$.

This means $R = (A_1 \times A_1) \cup (A_2 \times A_2) \cup (A_3 \times A_3)$.

Since $A_1, A_2, A_3$ are disjoint, the number of ordered pairs in $R$ is given by $n(R) = |A_1|^2 + |A_2|^2 + |A_3|^2$.

Let $x, y, z$ be the number of elements in $A_1, A_2, A_3$ respectively. We are given $x + y + z = 10$, where $x, y, z \ge 1$ are integers.

To minimize $x^2 + y^2 + z^2$, the values of $x, y, z$ should be as close to each other as possible.

Thus, we choose $x = 3, y = 3, z = 4$.

The minimum number of ordered pairs is $3^2 + 3^2 + 4^2 = 9 + 9 + 16 = 34$.

Answer: $34$