Permutations & Combinations
Selection from Polygon Vertices
Grade 11
Question:
<p>Let <strong>T<sub>n</sub></strong> denote the number of triangles which can be formed using the vertices of a regular polygon of '<strong>n</strong>' sides. If T<sub>n+1</sub> − T<sub>n</sub> = 21, then '<strong>n</strong>' equals:</p>
<p>(A) 5</p>
<p>(B) 7</p>
<p>(C) 6</p>
<p>(D) 4</p>
Step-by-Step Solution
Key Concept: The number of triangles from n vertices is \(\binom{n}{3}\). The difference \(T_{n+1} - T_n\) simplifies to \(\frac{n(n-1)}{2}\).
<p><strong>Solution:</strong> The number of triangles that can be formed using vertices of an n-sided polygon is \(T_n = \binom{n}{3}\).</p><p>Thus, \(T_{n+1} - T_n = \binom{n+1}{3} - \binom{n}{3} = 21\)</p><p>\(\binom{n+1}{3} - \binom{n}{3} = \frac{(n+1)n(n-1)}{6} - \frac{n(n-1)(n-2)}{6}\)</p><p>\(= \frac{n(n-1)[(n+1) - (n-2)]}{6} = \frac{n(n-1) \cdot 3}{6} = \frac{n(n-1)}{2} = 21\)</p><p>\(n(n-1) = 42\)</p><p>\(n^2 - n - 42 = 0\)</p><p>\((n-7)(n+6) = 0\)</p><p>Since n > 0, we have <strong>n = 7</strong>.</p>
Correct Answer: B