Problem Solving

kiskopata wrote:For A:
need 3 points for a triangle from 6points
choice for 1st point: 6
for 2nd: 5
for 3rd: 4
Total 6x5x4=120

This is not correct. If you count this way, you are assuming the order of the points matters- you'd be counting triangle ABF as though it were different from triangle AFB.

pranaysad wrote: B) Now If the same question was : There is a regular pentagon (ABCDE) how many ways can you join the sides of the pentagon.
You would do 5C2 = 10 ways but since 2 points are adjacent you - 5 from it to get 5 ways.

I assume you're asking how many 'diagonals' can be formed in the interior of a regular pentagon by connecting the vertices. Your solution is perfect: you can connect the vertices in 5C2 ways, but you don't want to count the 5 edges.

You can also look at the problem this way: from point A, you can make 5-3 = 2 diagonals; you can't connect A to itself or to either of the two points with which it forms an edge. For a regular n-sided shape, you can make n-3 diagonals. This is true for all five points: you can make 5*2 = 10 diagonals... but you've counted every diagonal exactly twice in this way (you've counted AC and CA as though they were different, but they're the same diagonal). So you must divide by 2, and the answer is 5.

pranaysad wrote: Question is there something I have missed in B above? Is there a general formula to use when dealing with geometric figures and joining the sides or creating triangles?

I have been able to figure out via trial and error that when you are asked to figure out how many lines can be made by joining the side of a geometric figure like a triangle, square etc, the answer is

nC2 - n
where n is the number of sides.

Can someone confirm this? Are they any other such formulas someone can share in geometry?

For diagonals inside a regular n-sided shape, your formula is perfectly good:

diagonals = nC2 - n

Or, looking at it the way I did above, from each point you can make (n-3) diagonals; there are n points; we must divide by two to avoid counting each diagonal twice:

diagonals = n*(n-3)/2

If you expand the formula you found above (nC2 - n), you'll see that both of these are equal.