In a circle there are $m$ chords and no $3$ are concurrent, $n$ intersections in the interior. Show there are $m+n+1$ regions dividied by the chords. 
In a circle there are $m$ chords such that no $3$ are concurrent and
  there are $n$ intersections of these chords in the interior of the
  circle. Prove that the number of regions divided by the chords is
  given by $m+n+1$

For example in the above diagram there are $3$ chords, $3$ points of intersection and $7$ regions.
Tried both counting the regions and establishing some sort of a bijection with a set of $m+n+1$ elements but didnt get very far. Any suggestions?
 A: Hint:
Try to prove the following statements:


*

*There is one initial region.

*Every chord with no intersection adds one region.

*Every chord with $k$ intersections adds $k+1$ regions.

*Each intersection is determined by two chords.
A: Alteratively you can consider the whole thing as a planar graph where:


*

*The vertices are the $n$ interior intersections and the $2m$ intersections between the circle and the chords.

*The edges are the segments of chord between two consecutive intersections, the arcs between two consecutive chords and the segments of chord bewteen a chord and the circle (with no intersections in between).
Then you want to use Euler's characteristic to find the number of faces:
$$F + |V| = |E| + 2$$
We can calculate $|E|$ because we know the degree of each vertex:


*

*There are $2m$ intersections between chords and the circle: these verices have degree $3$.

*There are $n$ interior intersections (between two chords): these verices have degree $4$.
So using:
$$\sum_{v\in V} \deg v = 2|E|$$
So $2|E| = 3\cdot 2m + 4\cdot n \Rightarrow |E| = 3m+2n$.
We also know $|V| = 2m + n$ as stated above.
Therefore:
$$F + 2m + n = 3m + 2n + 2 \Rightarrow F = m + n + 2$$
Note that the regions we want are the ones inside the circle, and there is one region outside of it, so the amount of regions is $F-1$:
$m + n + 1$
