Find the number of quadrilaterals that can be made using the vertices of a polygon of 12 sides as their vertices and having
(1) exactly 1 sides common with the polygon.
(2) exactly 2 sides common with the polygon.
$\underline{\bf{My \; Try}}::$ Let $A_{1},A_{2},A_{3},…………….A_{12}$ points of a polygon of side $=12$.
(1) part:: Let We Select adjacents pairs $A_{1}A_{2}$, Then other two vertices are from $A_{4},A_{5},………A_{11}$.
Here $A_{12}$ is not included because it is Left consecutive point corrosponding to $A_{1}$
So this can be done by $\displaystyle \binom{7}{2}$ similarly we can take another consecutive pairs $A_{2}A_{3}$.
So there are Total $12$ adjacents pairs in Anticlock-wise sence.
So Total no. of Quadrilateral in which one side common with $12$ sided polygon is
$\displaystyle = \binom{7}{2}\times 12 = 21\times 12 = 252$
(2) part :: If $2$ selected sides are consecutive:
Let we select $A_{1}A_{2}$ and $A_{2}A_{3}$. Then we select one points from the vertices $A_{4},A_{5},A_{6},………A_{12}$
This can be done by $\displaystyle {9}{1}$ ways.
Now we select consecutive adjacents sides in Anti-clockwise sence by $(11)$ ways.
So Total ways in above case(for two adjacents sides) is $\displaystyle = \binom{9}{1}\times 11 = 99$
If $2$ selected sides are not consecutive:
Now I did not understand How can i calculate in that case
Help required
Thanks
Best Answer
Let us count the number of quadrilaterals $A_1A_2A_{2+i}A_{2+i+j}$ where $i >1, j > 1, 2+i+j < 12$. Such vertices count the quadrilaterals with exactly $A_1A_2$ as common side. This is same as the number of solutions to $i+j <10, i >1, j > 1 $. Putting $x = i-1, y = j-1$, we need the solutions $x+y < 8$ where $x >0, y > 0$. By stars and bars method this is $\binom{1}{1}+\binom{2}{1}+\binom{3}{1}+\binom{4}{1}+\binom{5}{1} +\binom{6}{1}= 21$. Thus the answer for part 1 is $21\times 12=252$.