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Given a grid of $4\times4$ points,how many triangles with their vertices on the grid can be drawn?
  • ๐Ÿšฉ Duplicate | ๐Ÿ‘ฎ reboot | ๐Ÿ’ฌ โ€œmerge it here please "https://gateoverflow.in/159677/p-and-c-doubt"โ€
in Combinatory 1 flag 378 views
0
Can you tell the answer? I am getting a huge value :P
0
$516$ is the correct answer.

2 Answers

7 votes
 
Best answer

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0
wow awesome concept !
0
great job brother thanks a lot
0
thanks @utkarsh
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:)
0
I'm not getting diagonal cases??
1
thanks, brother

now I'm get the solution and concept also
0

nicely explained @Utkarsh Joshi!

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 โ€“ (40+4) = 516

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