# Number of quadrilatera in polygon

#### jacks

##### Well-known member
Number of Quadrilateral that can be made using the vertex of a polygon of 10 sides as

there vertices and having Exactly $2$ sides common with the polygon

My solution:

first we will take $2$ sides common is $=10$ ways now if we take two sides as $A_{1}A_{2}$ and $A_{1}A_{3}$ then we

will not take vertex $A_{4}$ and $A_{10}$ So we will choose $1$ vertes from $5$ vertices which can be done by

$\displaystyle \binom{5}{1}$ ways So Total No. of ways in which $2$ sides common is $=10 \times \displaystyle\binom{5}{1} =50$ but answer given

is $=75$.

But I did not understand it. Would anyone like to explain me where i am wrong

Thanks

#### caffeinemachine

##### Well-known member
MHB Math Scholar
Re: number of quadrilatera in polygon

Number of Quadrilateral that can be made using the vertex of a polygon of 10 sides as

there vertices and having Exactly $2$ sides common with the polygon

My solution:

first we will take $2$ sides common is $=10$ ways now if we take two sides as $A_{1}A_{2}$ and $A_{1}A_{3}$ then we

will not take vertex $A_{4}$ and $A_{10}$ So we will choose $1$ vertes from $5$ vertices which can be done by

$\displaystyle \binom{5}{1}$ ways So Total No. of ways in which $2$ sides common is $=10 \times \displaystyle\binom{5}{1} =50$ but answer given

is $=75$.

But I did not understand it. Would anyone like to explain me where i am wrong

Thanks
You can also have a quadrilateral $A_1A_2A_4A_5$. It too has exactly $2$ of its sides common with the polygon. There, of course, many more such quadrilaterals.

#### soroban

##### Well-known member
Hello, jacks!

I think I've got it . . .

of a polygon of 10 sides as their vertices and having
exactly 2 sides common with the polygon.

Code:
              A   B

J           C

I               D

H           E

G   F
Case 1: the two sides are not adjacent.

The first side can be any of the 10 adjacent vertex pairs:
. . $$AB, BC, CD, \text{ . . . } JA$$

Suppose the first side is $$AB.$$
Then the second side has 5 choices:
. . $$DE, EF, FG, GH, HI$$

It seems there are $$10\times 5 \,=\,50$$ such quadrilaterals.
But this list includes $$\{AB,FG\}$$ and $$\{FG,AB\}$$

Hence, there are: $$\tfrac{50}{2}\,=\,25$$ such quadrilaterals.

Case 2: The two sides are adjacent.
There are 10 triples of vertices:
. . $$ABC,BCD,CDE,\text{ . . . }JAB$$

Suppose the triple is $$ABC.$$
Then the fourth vertex can be: $$\{E,F,G,H,I\}$$

Hence, there are: $$10\times5\,=\,50$$ such quadrilaterals.

Therefore, there are: $$25 + 50 \,=\,75$$ quadrilaterals.