142. On a plane, there are $n$ points ( $n \geq 3$ ), no three of which lie on the same line. How many lines can be drawn, each passing through two of the given points?
$\triangle$ By analogy with the solution of the previous problem, we get $n(n-1)$ lines. However, the point is that a line, unlike a vector, can be considered as an unordered pair of points - in the sense that line $A B$ coincides with line $B A$. Therefore, in such a calculation, each line is counted twice. Consequently, the number of lines will be half of what is indicated above, which means it is $\frac{n(n-1)}{2}$.

Answer: $\frac{n(n-1)}{2}$. $\mathbf{A}$

Problems that are essentially the same as Problem 142 occur quite frequently. For example: in a round-robin chess tournament, $n$ players play; how many games do they play in total? Or: $n$ friends met and shook hands; how many handshakes were there in total? They have also appeared in this book (see, for example, problems $35-39$).