Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 3.9 AMC 10/12 Find the answer China

Let the 9-element set $A = \{a + bi \mid a, b \in \{1, 2, 3\}\}$ , with $i$ being the imaginary unit. $\alpha = (z_1, z_2, \dots, z_9)$ is a permutation of all the elements in $A$ , satisfying $|z_1| \le |z_2| \le \dots \le |z_9|$ . The number of such permutations $\alpha$ is ______.

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

Since
$\begin{align*} |1+i| < |2+i| = |1+2i| < |2+2i| < |3+i| \\ = |1+3i| < |3+2i| = |2+3i| < |3+3i|, \end{align*}$
it follows that
$\begin{align*} z_1 &= 1 + i, \\ \{z_2, z_3\} = \{2 + i, 1 + 2i\}, \\ z_4 &= 2 + 2i, \\ \{z_5, z_6\} = \{3 + i, 1 + 3i\}, \\ \{z_7, z_8\} = \{3 + 2i, 2 + 3i\}, \\ z_9 = 3 + 3i. \end{align*}$
By the multiplication principle, the number of permutations $\alpha$ satisfying the condition is $2^3 = 8$ . $\square$

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.