1...).there are 3,4 and 5 points marked on the three sides of a triangle.how many triangles can be formed with these  points as vertices if none of the marked points are at a vertex of the triangle.

2. find the number of triangles that can be formed with the vertices of a polygon of 10 sides as thier vertices if

a) the triangle cannot have more than one side comon with polygon.

b) the triangle cannot have any side comon with polygon.

1)I think the answer is

(3c₁)(4c₁)(5c₁)+(3c₂)(9c₁)+(4c₂)(8c₁)+(5c₂)(7c₁)

=60+27+48+70=205.

If correct,I will explain in detail.(If u want)

YUP, ABSOLUTELY CORRECT. NOW SHOW ME YOUR WORKING

Ya..Ryuamakusa explained it.We need to select three non-collinear points.

This can be done by selecting one point each from three sides or two points on one side and remaining on a diff.side.

Hope you got it.

Are the answers to 2nd question

(a)(1/3)(10c₁)(7c₂)+(1/2)(10c₁)(2c₁)(7c₁)=210/3+140/2=70+70=140.

(b)(1/3)(10c₁)(7c₂)=210/3=70.

Edited after thinking again on the problem.Now are they correct?

2

a  10C1 X 9C1 X 8C1 - 10C1 X2 X2

b 10C1 X 9C1 X 8C1 - 10C1 X2 X2 - 10C1 X 2 X 6C1

a total - no with two sides common

b total - no with 2 sides cmmn - no with 1 side common

in the second one

answer to a)110

           b)50

i got the right answer to (a)

see my method  in a)

i have calculated the total no. of triangles =10 C3  and then subtracted the triangle with two side common i.e=10

=10 C3-10

=110

ok i have also got the second one,

just total-2 side comm.-1 side common

=10C3 - 10(for 2 side) - 60(for 1 side)

=50

if u will check in a  five sided - 1 triangle is formed with one side common(from one vertex)

                                   six  "     -  2  "            "                "        "

                                  ..ten sided- 5 traiangle per vertex

therefore total -60 triangles from ten vertex.

also(thanks allamraju and ray amasuka)