Permutations & Combinations
Counting
Grade 11
Question:
<p>Let \(A\) be a set of \(n\) (\(\geq 3\)) distinct elements. The number of triplets \((x, y, z)\) of the \(A\) elements in which at least two coordinates is equal to</p>
<p>(1) \({}^nP_3\)</p>
<p>(2) \(n^3 - {}^nP_3\)</p>
<p>(3) \(3n^2 - 2n\)</p>
<p>(4) \(3n^2(n-1)\)</p>
Step-by-Step Solution
Key Concept: Use complementary counting: total triplets minus triplets with all distinct elements. For n elements, total ordered triplets = n³, and triplets with all distinct = n(n-1)(n-2).
<p><strong>Step 1:</strong> Find total number of ordered triplets (x, y, z) from set A with n elements.</p><p>Total triplets = n³ (each position can be filled with any of n elements)</p><p><strong>Step 2:</strong> Find number of triplets where all three coordinates are distinct.</p><p>Triplets with all distinct elements = n(n-1)(n-2)</p><p>(First position: n choices, second: n-1 choices, third: n-2 choices)</p><p><strong>Step 3:</strong> Use complementary counting to find triplets with at least two equal coordinates.</p><p>Triplets with at least two equal = Total triplets - Triplets with all distinct</p><p>= n³ - n(n-1)(n-2)</p><p>= n³ - n(n² - 3n + 2)</p><p>= n³ - n³ + 3n² - 2n</p><p>= 3n² - 2n</p><p>= n(3n - 2)</p><p>∴ Answer: <strong>n(3n - 2)</strong> or equivalent form</p>
Correct Answer: B