Question

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

• A. $f\left(20\right)=382$
• B. $f\left(n\right)$ is always an even number
• C. $f^{-1}\left(92\right)=10$
• D. $f\left(n\right)$ can be odd

Answer 1

Answer: (A), (B), (C)

The correct options are
A $f\left(20\right)=382$
B $f\left(n\right)$ is always an even number
C $f^{-1}\left(92\right)=10$

The number of regions for ${}^{\prime }{n}^{\prime }$ circles be $f\left(n\right)$. Clearly, $f\left(1\right)=2$. Now,
$f\left(n\right)=f(n-1)+2(n-1),\forall n\ge 2$
$\Rightarrow f\left(n\right)-f(n-1)=2(n-1)$
Putting $n=2,3,...,n$, we get
$f\left(n\right)-f\left(1\right)=2(1+2+3+...n-1)=(n-1)n$
$\Rightarrow f\left(n\right)=n(n-1)+2=(n^2-n+2)$ (which is always even)
$\Rightarrow f\left(20\right)={20}^2-20+2=382$
Also,
$n^2-n+2=92$
$\Rightarrow n^2-n-90=0\Rightarrow n=10$.