Six straight lines are drawn in a plane with no two parallel and no three concurrent. The number of regions into which they divide the plane is:
$\textbf{(A)}\ 16 \qquad \textbf{(B)}\ 20\qquad \textbf{(C)}\ 22 \qquad \textbf{(D)}\ 24 \qquad \textbf{(E)}\ 26$
Solution 1
The first line divides the plane into two regions.  The second line intersects one line, creating two regions.  The third line intersects two lines, creating three regions.  Similarly, the fourth line intersects three lines and creates four regions, the fifth line intersects four lines and creates five regions, and the sixth line intersects five lines and creates six regions.
Totaling the regions created results in $2 + 2 + 3 + 4 + 5 + 6 = 22$ regions, which is answer choice $\boxed{\textbf{(C)}}$.

Solution 2
[1963 AHSME Problem 27.png](https://artofproblemsolving.com/wiki/index.php/File:1963_AHSME_Problem_27.png)
With careful drawing, one can draw all six lines and count the regions.  There are $22$ regions in total, which is answer choice $\boxed{\textbf{(C)}}$.

Solution 3
We can use the fact that the number of regions that $n$ lines divide a plane is given by the equation $L_n = \frac{n^2 + n +2}{2}$, and in this problems, $n=6$, from which the answer is $\boxed{22}$. 
The formula is proved in Engel's Extremal Principles chapter or [https://www.cut-the-knot.org/proofs/LinesDividePlane.shtml](https://artofproblemsolving.comhttps://www.cut-the-knot.org/proofs/LinesDividePlane.shtml) gives two proofs for it.
(OmicronGamma 2019)