number of quadrilaterals that can be made by using the vertices of a polygon of 10 sides as their vertices and having exactly two sides a common with polygon

i have got the answer for 3 sides common with the polygon as 10

2. If you take only quadrilaterals with two common adjacent sides, you have 5x10 = 50 quadrilaterals. (5 possible quadrilaterals for each of the 10 pair of adjacent sides)

Now, taking non adjacent sides, it will be a problem similar to:
How many different pair of letters you can make from a through j, without repeat and order is not important, ie ab and ba are the same, provided two adjacent letters cannot be paired, including a and j.

You have the sides:
ae
af
ag
ah
be
bf
bg
bh
bi
cf
cg
ch
ci
cj
dg
dh
di
da (but you already have ad!, so this is rejected. I won't put other repeats)
eh
ei
ej
fi
fj
gj

For a total of (24 + 50) possible quadrilaterals.

3. Originally Posted by Unknown008
If you take only quadrilaterals with two common adjacent sides, you have 5x10 = 50 quadrilaterals. (5 possible quadrilaterals for each of the 10 pair of adjacent sides)

Now, taking non adjacent sides, it will be a problem similar to:
How many different pair of letters you can make from a through j, without repeat and order is not important, ie ab and ba are the same, provided two adjacent letters cannot be paired, including a and j.

You have the sides:
ae
af
ag
ah
be
bf
bg
bh
bi
cf
cg
ch
ci
cj
dg
dh
di
da (but you already have ad!, so this is rejected. I won't put other repeats)
eh
ei
ej
fi
fj
gj

For a total of (24 + 50) possible quadrilaterals.
thanks

4. Hello, prasum!

Number of quadrilaterals that can be made by using the vertices of a polygon
of 10 sides as their vertices and having exactly two sides in common with polygon.

From the available 10 vertices, choose a pair that is adjacent.
. . There are 10 such pairs.

For the second pair of adjacent vertices,
. . we must not use the vertices adjacent to the first two. .**

From the available 6 vertices, choose a pair that is adjacent.
. . There are 5 such pairs.

Therefore, there are: .$\displaystyle 10 \times 5 \:=\:50$ such quadrilaterals.

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

**

Label the vertices: .$\displaystyle A,B,C,D,E,F,G,H,I,J$ in that order.

Suppose we select $\displaystyle \,AB$ for the first pair.

Then we must not select $\displaystyle C\!D$ or $\displaystyle I\!J$ for the second pair,
. . or we have a quadrilateral with three sides in common.