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



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.

  • $\begingroup$ Can i know why in first case every polygon is counted twice $\endgroup$ – 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 $126$.


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