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  4. 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

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Q. 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

Permutations and Combinations

Solution:

The number of triangles $=$ number of selections of $2$ lines from the $(n-1)$ lines which are cut by the last line
$={ }^{n-1} C_{2}=\frac{(n-1) !}{2 !(n-3) !}=\frac{(n-1)(n-2)}{2}$