First slide

Combinations

Question

There are n straight lines in a plane, no two are parallel and no three are concurrent. If the points of intersection of these lines are joined, then the number of new lines is

Easy

A

$3 (\begin{matrix} n \\ 3 \end{matrix})$
B

$2 (\begin{matrix} n \\ 4 \end{matrix})$
C

$3 (\begin{matrix} n \\ 4 \end{matrix})$
D

$(\begin{matrix} n \\ 4 \end{matrix})$

Solution

The n lines intersect in $(\begin{matrix} n \\ 2 \end{matrix})$ points.
The number of lines formed by joining these points is $\frac{1}{2} (\begin{matrix} n \\ 2 \end{matrix}) ((\begin{matrix} n \\ 2 \end{matrix}) - 1)$

= $\frac{1}{2} . \frac{n (n - 1)}{2} . (\frac{n (n - 1)}{2} - 1) = \frac{1}{8} n (n - 1) (n + 1) (n - 2)$

On each line there are $n - 1$ points

The number of lines formed by these points is $n (\begin{matrix} n - 1 \\ 2 \end{matrix}) = \frac{n}{2} (n - 1) (n - 2)$

The number of new lines is $\frac{1}{8} n (n - 1) (n + 1) (n - 2) - \frac{n}{2} (n - 1) (n - 2)$

$= \frac{1}{8} n (n - 1) (n - 2) (n - 3) = 3 (\begin{matrix} n \\ 4 \end{matrix})$

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