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Given a 4 *4 grid  points , how many Triangles with vertices on the grid can be formed?
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Can you tell the answer? I am getting a huge value :P
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$516$ is the correct answer.
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3 Answers

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I have calculated it as 516.

The selection of 3 points out of 16 can be done in 16C3 = 560.

This include points that are collinear. If you calculate the number of collinear lines with 3 grid points then you will get 44 such lines. 

So, final number of possible triangles are 560 -44 = 516

This stackexchange Q&A helped me to solve this: https://math.stackexchange.com/a/634131

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4 Comments

16C3 is total 3 points selected. -560

10* 4C3 - collinear lines 4 vertical 4 horizontal 2 diagonal -40

I m unable to visualise the other 4 collinear lines

Can u plzz help me out

I think I m missing something...
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Please consider counting these 4 lines too

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Thanku....:)
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For 5* 5 grid

Will it be like this

25C3  all 3 points selected

12* 5C3 = 120   collinear lines 5 each horizontal vertical and 2 diagonal

4 non diagonal collinear lines with 3  grid points

4* 4C3 non diagonal collinear lines with 4 points

I m getting 2160...

But there how they got 12 collinear lines with 3 grid points....:(
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7 votes
7 votes

.....

4 Comments

I'm not getting diagonal cases??
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thanks, brother

now I'm get the solution and concept also
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nicely explained @Utkarsh Joshi!

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1 vote
1 vote

Total possible triangles = $\binom{16}{3}$ = 560

Now we know that if all points are collinear then the triangle is not possible.

So we will simply deduct the cases wherein we are selecting 3 points which are collinear,

How many groups of 4 collinear points do we have?

4 horizontal lines + 4 vertical lines + 2 diagonals = 10

So $10*\binom{4}{3}=40$ invalid counts

How many groups of 3 collinear points do we have?

Only 4.

So $4*\binom{3}{3}=4$ invalid counts

Therefore the total number of triangles =

560 – (40+4) = 516

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