There are n concurrent lines and another line parallel to one of them. The number of…
Mathematics · JEE Advanced · NTA Exams — Permutations and Combinations
There are n concurrent lines and another line parallel to one of them. The number of different triangles that will be formed by the (n + 1) lines, is
- \(\frac{(\mathrm{n}-1) \mathrm{n}}{2}\)
- \(\frac{(\mathrm{n}-1)(\mathrm{n}-2)}{2}\)
- \(\frac{\mathrm{n}(\mathrm{n}+1)}{2}\)
- \(\frac{(\mathrm{n}+1)(\mathrm{n}+2)}{2}\)
Answer
(B) ( n -1)( n -2) 2
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