My notes state the Counting process for knowing no. of regions a plane is divided into by $n$ lines in general position :=
Let $h_1(n)=$ No. of parts a line is divided by $n$ distinct points=$n+1$
$h_2(n)=$ No. of regions a plane is divided into by $n$ lines in general position(i.e. no two of them are parallel and no three of them are concurrent).
To find $h_2(n)$ we proceed as follows:
Suppose the plane has already been divided into parts by $n$ straight lines in general positions;let these lines be $L_1,L_2,\ldots,L_n$. Let $L_{n+1}$ be $(n+1)^{\text{th}}$ line which is in general position with $L_1,\ldots L_n$.
Let $P_i=L_i\cap L_{n+1}$ ,where $i=1,2,\ldots,n$ and $P_i$'$s$ are all distinct .We have $P_1,P_2\ldots P_n$ $n$ distinct points lying on $L_{n+1}$ .These points divided $L_{n+1}$ in $h_1(n)$ parts ,with each part lying exactly in one of the regions determined by $n$ lines and it divides that region into $2$ parts .
$\therefore$ we obtain :
$h_2(n+1)=h_2(n)+h_1(n)$ or $h_2(n+1)-h_2(n)=h_1(n)$
$\implies \triangle h_2(n)=h_1(n)$
$\implies h_2(n)=\triangle^{-1} h_1(n)$
$~~~~~~~~~~~~~~~~~~~~=\triangle^{-1} \Big(\binom{n}{0}+\binom{n}{1}\Big)$
$~~~~~~~~~~~~~~~~~~~~= \Big(\binom{n}{1}+\binom{n}{2}+c\Big)$
Now, $h_2(1)=n$ .Also $h_2(1)=\binom{1}{1}+\binom{1}{2}+c$=$2=0+1+c$$\implies c=1=\binom{n}{0}$
$\therefore h_2(n)=\binom{n}{0}+\binom{n}{1}+\binom{n}{2}$.
I can't understand this method as I missed my class and now I'm $\mathbf STUCK$ because I've my exam day after.Please if anyone can help me understanding logic behind this .......