Permutations & Combinations
Grade 11
Question:
<p>Triplet <span class="math-tex">\((x, y, z)\)</span> is chosen from the set <span class="math-tex">\(\{1,2,3, \ldots, n\}\)</span>, such that <span class="math-tex">\(x \leq y \lt z\)</span>. The number of such triplets is</p>
<p style="display:inline"><span class="math-tex">\(\mathrm{n}^3\)</span></p>
<p style="display:inline"><span class="math-tex">\({ }^n \mathrm{C}_2\)</span></p>
<p style="display:inline">none of these</p>
<p style="display:inline"><span class="math-tex">\({ }^n C_3\)</span></p>
Step-by-Step Solution
Key Concept: The constraint is x ≤ y < z (allowing x = y), which differs from the standard combination formula ⁿC₃ that counts strictly increasing selections (x < y < z). This requires counting cases where x = y separately or recognizing the constraint allows multiset-like behavior in the first two positions.
<p>Any three numbers <span class="math-tex">\(x, y, z\)</span> from <span class="math-tex">\(\{1,2,3, \ldots\}\)</span> can be chosen in <span class="math-tex">\({ }^n C_3\)</span> ways and we get unique triplet ( <span class="math-tex">\(x, y, z\)</span> ), <span class="math-tex">\(x<y<z\)</span>. Again any two numbers <span class="math-tex">\(x, z\)</span> can be chosen from <span class="math-tex">\(\{1,2,3, \ldots, n\}\)</span> in <span class="math-tex">\({ }^n C_2\)</span> ways and we get the triplet ( <span class="math-tex">\(x, x, z\)</span> ), <span class="math-tex">\(x<z\)</span>. Hence total number of required triplets is <span class="math-tex">\({ }^{{n}} {C}_2+{ }^{{n}} {C}_3\)</span>.</p>
Correct Answer: C