The greatest number of points of intersection of n circles and m straight lines is- The question is about combinatorics. I have no idea on how to start solving the problem. Please guide me.

$(a) 2mn+ {m \choose 2}$


$(b) \frac{1}{2}m(m-1)+n(2m+n-1)$

$(c) {m \choose 2}+2({n \choose 2})$

(d) none of these

I tried solving this question, but I couldn't even find a starting point??
 A: Here are some examples of maximal number of points of intersection for $n,m\in\{1,2,3,4,5\}$:



*

*The number of ways you can choose a pair from $n$ circles is $\binom n2$. Each such pair intersect in at most two points.

*The number of ways you can choose a pair from $m$ lines is $\binom m2$. Each such pair intersect in at most one point.

*The number of ways you can choose one circle and one line from $n$ circles and $m$ lines is $n\cdot m$. Each such pair intersect in at most two points.


So as a total we have:
$$
\begin{align}
2\binom n2+\binom m2+2n\cdot m&=n(n-1)+\frac{m(m-1)}{2}+2n\cdot m\\
&=\tfrac 12m(m-1)+n(2m+n-1)
\end{align}
$$
which leads directly to the correct answer.
A: Solution 1:
Make a diagram. Two geometric figures intersect if they have one or more points in common. Draw two circles which intersect in $2$ points. Draw a line which intersects the two circles in $4$ points. Draw another line which intersects the two circles in $4$ points and also intersects the first line. There are $\boxed{11}$ points of intersection.
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Solution 2:
Make a table of the maximum number of points of intersection.$$
\begin{array}{|c|c|}
\hline
\text{Geometric Figures} & \text{Number of Common Points} \\
\hline
\text{2 Circles} & 2 \\
\hline
\text{Line (1) and 2 circles} & 4 \\
\hline
\text{Line (2) and 2 circles} & 4 \\
\hline
\text{2 lines} & 1 \\
\hline
& \text{Total = \boxed{11}} \\
\hline
\end{array}
$$
