# no. of quadrilateral in 12 sided polygon

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

In the first case there are 12 different sides, so there are 12 different pairs of vertices to count. Your reasoning to leave vertices $A_3$ and $A_{12}$ is OK, but you need to make another restriction after the selection of the third vertex, because if we chose the vertices $A_1, A_2, A_5, A_6$, then we'll have two common sides, right?

Now we have to distinct cases. The first is when our third vertex is $A_{11}$ or $A_4$. Then we will have 6 other vertices to chose those are $A_4, A_5, A_6, A_7, A_8, A_9$ for $A_{11}$.

The number of combinations is $2 \times 6 = 12$.

In the second case the third vertex is one of the rest. So for example if we choose $A_5$ as our third vertex we'll have 5 other options for the fourth vertex. Those are: $A_7, A_8, A_9, A_{10}, A_{11}$. So there are $6 \times 5 = 30$ combinations.

Now add those together and divide by 2, because every polygon is counted twice. Then we'll have:

$$\frac{12 + 30}{2} = \frac{42}{2} = 21 = \binom {7}{2} \text{ quadrilaterial with one common side}$$

Multiply by 12, because we said there are 12 different ways to chose a pair of consecutive vertices and you'll get $252$ distinct quadrilaterials.

For the second problem you should look in two cases. The one is when the two common sides are consecutive (we use 3 consecutive vertices). Again there are 12 such triples.

Note that if we select 3 consecutive vertices then there are 7 available vertices to choose.

There are total of $12 \times 7 = 84$ quadrilaterials.

For the second cas you need again to chose pair of adjacent vertices and anothe pair of adjacent vertices. Because for every pair of vertices there are 8 available vertices, there are 7 other pairs so for selected pair $A_1, A_2$ we can chose the other pair as $(A_4,A_5)$, $(A_5,A_6)$, $(A_6,A_7)$, $(A_7,A_8)$, $(A_8,A_9)$, $(A_9,A_{10})$, $(A_{10},A_{11})$

So there are total of $\frac{12 \times 7}{2} = 42$ quadrilaterial.

We are dividing by two, because the same quadrilaterial can be reached starting from $A_1, A_2$ and $A_5, A_6$, so every quadrilaterial is calcualted twice.

Add those two numbers and you'll end up with: $84 + 42 = 126$ quadrilaterials with two common sides.

• Can i know why in first case every polygon is counted twice – Umesh shankar Apr 4 '17 at 2:29

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 $21\times 12=252$.

The polygon cannot be arbitrary. Odd things can happen, for example with a $12$-sided cross. Let our polygon be regular. By quadrilateral we mean convex quadrilateral.

(1) As in your solution, there are $12$ ways to choose the side in common with the $12$-gon. The "opposite" side's vertices are chosen from the $8$ remaining candidate vertices. There are $\binom{8}{2}$ ways to choose $2$ vertices. But $7$ of these pairs are adjacent, leaving $21$ choices, for a total of $(12)(21)$.

(2) We can choose $3$ consecutive vertices in $12$ ways, and for each way choose a non-consecutive in $7$ ways. Thus far we have a total of $84$.

Now we count the cases where the edges shared with the $12$-gon are opposite. Choose an edge of the $12$-gon, and colour it blue. There are $12$ ways to do this. Now choose a non-adjacent edge and colour it red. There are $7$ ways to do this. The product $84$ double-counts our quadrilaterals. So there are $42$ of this type, for a total of $126$.