<p>Choosing <i>{x, y, z} ⊂ S</i>, such that <i>x, y, z</i> are in AP, is</p>
Step-by-Step Solution
Key Concept: For three numbers in AP, the sum of first and third must be even, which happens when both are odd or both are even.
<p><strong>Solution:</strong> Given, <i>x, y, z</i> are in AP, so <i>2y = x + z</i>.</p><p>It is clear that sum of <i>x</i> and <i>z</i> is even. This means <i>x</i> and <i>z</i> both are even or odd out of set <i>S</i>.</p><p>Set <i>S</i> contains 11 odd numbers (1, 3, 5, ..., 21) and 10 even numbers (2, 4, 6, ..., 20).</p><p>Number of favourable cases = $\binom{11}{2} + \binom{10}{2} = 55 + 45 = 100$</p><p>Total number of ways choosing <i>x, y, z</i> is $\binom{21}{3} = \frac{21 \times 20 \times 19}{1 \times 2 \times 3} = 1330$</p><p>Required probability = $\frac{100}{1330} = \frac{10}{133}$</p>
Correct Answer: B