Let f(n) be the number of regions in which n coplanar circles can divide the plane. If it is known that each pair of circles intersect in two different points and no three of them have common point of intersection, then

Let the number of regions for n circles be f(n). 
Clearly, f(1) =2. Now,
$\begin{array}{r}\mathrm{f}\left(\mathrm{n}\right)=\mathrm{f}(\mathrm{n}-1)+2(\mathrm{n}-1),\mathrm{\forall }\mathrm{n}\ge 2\\ \Rightarrow \mathrm{f}\left(\mathrm{n}\right)-\mathrm{f}(\mathrm{n}-1)=2(\mathrm{n}-1)\end{array}$
Putting n = 2, 3, . . ., n,we get
$\begin{array}{l}\mathrm{f}\left(\mathrm{n}\right)-\mathrm{f}\left(1\right)=2(1+2+3+\cdots \mathrm{n}-1)=(\mathrm{n}-1)\mathrm{n}\\ \Rightarrow \mathrm{f}\left(\mathrm{n}\right)=\mathrm{n}(\mathrm{n}-1)+2=\left({\mathrm{n}}^2-\mathrm{n}+2\right)\left(\text{ which is always even }\right)\\ \Rightarrow \mathrm{f}\left(20\right)={20}^2-20+2=382\end{array}$

Also,

$\begin{array}{l}{\mathrm{n}}^2-\mathrm{n}+2=92\\ \mathrm{or} {\mathrm{n}}^2-\mathrm{n}-90=0\text{ or }\mathrm{n}=10\end{array}$