VeritasPrepKarishma wrote:
dimri10 wrote:
if anyone can help please to clarify the methos:
let's say that the Q was:
How many triangles with positive area can be drawn on the coordinate plane such that the vertices have integer coordinates (x,y) satisfying 1≤x≤6 and 1≤y≤6?
how will it be solved:
will 3C36 minus 6 vertical and 6 horizontal minus 2 diagonals will be the answer or will the answer be different.
thank's in advance
I think your question is quite similar to yogesh1984's question above. I missed answering his question (thought of doing it later due to the diagram involved but it skipped my mind).
Anyway, let me show you how I would solve such a question. Both the questions can be easily answered using this method.
How many triangles with positive area can be drawn on the coordinate plane such that the vertices have integer coordinates (x,y) satisfying 1≤x≤6 and 1≤y≤6?
Ok, so we have a total of 36 co-ordinates (as shown below by the red and black dots). We need to make triangles so we need to select a triplet of co-ordinates out of these 36 which can be done in 36C3 ways. Out of these, we need to get rid of those triplets where the points are collinear. How many such triplets are there?
Look at the diagram:
Attachment:
Ques2.jpg
The Black dots are the outermost points. Red dots are the inside points. Now each of these red dots is the center point for 4 sets of collinear points (as shown by the red arrows). Hence the 4*4 = 16 red dots will make 16*4 = 64 triplets of collinear points.
These 64 triplets account for all collinear triplets except those lying on the edges. Each of the 4 edges will account for 4 triplets of collinear points shown by the black arrows. Hence, there will be another 4*4 = 16 triplets of collinear points.
Total triplets of collinear points = 64 + 16 = 80
Therefore, total number of triangles you can make = 36C3 - 80
Similarly you can work with 1<=x<=5 and -9<=y<=3.
The number of red dots in this case = 11*3 = 33
So number of collinear triplets represented by red arrows will be = 33*4 = 132
Number of black arrows will be 3 + 11 + 3 + 11 = 28
Total triplets of collinear points = 132 + 28 = 160
Total triangles in this case = 65C3 - 160
Ma'am,
It would like to point out tht the resoning given is wrong. the triplets need not necessarily be adjacent. tht's the flaw.
my way:
no: of collinear points=?
horizontal and vertical lines both give the same no: and each line of 6 points gives 6C3 possibs.
hence horz and vert. lines give a total of 2*6*6C3.
next 2 diagonals give same no: of such possibs.
consider any diagonal direction. it gives 3,4,5,6,5,4,3 collinear points along 6 parallel lines corresponding to any diagonalic direction and each of the points gives us their corresponding triples-3C3+4C3+5C3+6C3+5C3+4C3+3C3.
along 2 such dirs. this adds up to 2*(2*(3C3+4C3+5C3)+6C3).
total no: of line forming selections= 2*6*6C3+ 2*(2*(3C3+4C3+5C3)+6C3).