# Counting of number of Triangles

378 views
Given a grid of $4\times4$ points,how many triangles with their vertices on the grid can be drawn?
• 🚩 Duplicate | | 💬 “merge it here please "https://gateoverflow.in/159677/p-and-c-doubt"”
1 flag
0
Can you tell the answer? I am getting a huge value :P
0
$516$ is the correct answer.

.....

selected
0
wow awesome concept !
0
great job brother thanks a lot
0
thanks @utkarsh
0
:)
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 – (40+4) = 516

## Related questions

1
260 views
How many triangles are formed by selecting points from a set of 15 points out of which 8 are collinear? (a) 800 (b) 824 (c) 844 (d) 854
How many pairs $(x,y)$ such that $x+y <= k$, where x y and k are integers and $x,y>=0, k > 0$. Solve by summation rules. Solve by combinatorial argument.
Let $A = \left \{1, 2, 3, 4 \right \}$. Number of functions possible on $A$ which are neither $1-1$ nor on-to is _________.