<p>Three a's, three b's and three c's are placed randomly in a \(3 \times 3\) matrix. The probability that no row or column contain two identical letters can be expressed as \(\frac{p}{q}\), where p and q are coprime then (p + q) equals to:</p>
Step-by-Step Solution
Key Concept: This is a Latin square problem where each symbol appears exactly once in each row and column. Count valid arrangements and divide by total arrangements.
<p>Total ways to arrange 3 a's, 3 b's, and 3 c's in a $3 \times 3$ matrix = $\frac{9!}{3! \cdot 3! \cdot 3!} = 1680$</p><p>For no row or column to contain two identical letters, each row and each column must contain exactly one a, one b, and one c.</p><p>This is equivalent to finding the number of $3 \times 3$ Latin rectangles using symbols {a, b, c}.</p><p>Number of such valid arrangements = 12</p><p>Probability = $\frac{12}{1680} = \frac{1}{140}$</p><p>Since p = 1 and q = 140 are coprime, p + q = 141</p>
Correct Answer: b