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

Question

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 (A) ((n-1) n/2) (B) ((n-1)(n-2)/2) (C) (n(n+1)/2) (D) ((n+1)(n+2)/2)

Tardigrade Admin · Accepted Answer

The number of triangles = number of selections of 2 lines from the (n-1) lines which are cut by the last line = n-1 C2=((n-1) !/2 !(n-3) !)=((n-1)(n-2)/2)

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

$\frac{( n - 1 ) n}{2}$

$\frac{( n - 1 ) ( n - 2 )}{2}$

$\frac{n ( n + 1 )}{2}$

$\frac{( n + 1 ) ( n + 2 )}{2}$

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

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

2(n−1)n​

2(n−1)(n−2)​

2n(n+1)​

2(n+1)(n+2)​