There are 12 points in a plane, no three of which are in the same straight line, except 5 points whi

Q: Wait, why do we subtract $^{5}C_3$? Shouldn't we just ignore those points entirely?

That is a great question! If we simply ignored the 5 collinear points, we would be missing out on the triangles that could be formed by picking 2 points from the collinear set and 1 point from the remaining 7 non-collinear points. By calculating the total combinations $^{12}C_3$ and then subtracting the invalid cases $^{5}C_3$, we correctly account for all valid triangles while removing only the specific combinations that fail to form one.

Q: Does the order in which I select the points matter for forming a triangle?

In geometry, a triangle defined by vertices A, B, and C is the same as a triangle defined by B, C, and A. Because the order of vertices does not change the triangle, we use combinations ($^{n}C_{r}$) rather than permutations ($^{n}P_{r}$). We are simply choosing a set of 3 points, not arranging them.

Q: What if the problem said 'no three points are collinear'? How would the formula change?

If no three points were collinear, the condition that prevents triangle formation would vanish. In that case, every single combination of 3 points would form a valid triangle. The answer would simply be $^{12}C_3$. The collinearity constraint is the only reason we have to perform that extra subtraction step.

Question

There are 12 points in a plane, no three of which are in the same straight line, except 5 points which are collinear. Then the total number of triangles that can be formed with the vertices at any three of these 12 points is

Options:

(A) 230

(B) 220

(C) 200

(D) 210

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before an infinite, pristine white plane. Your task is to place twelve distinct points on this canvas. These are not just dots; they are the potential vertices of geometric shapes.

In the world of JEE Advanced, we often find that the most elegant problems are hidden within simple arrangements. Today, we are going to explore the combinatorial beauty of forming triangles from these twelve points.

### The Combinatorial Landscape

To form a triangle, we need exactly three vertices. If we were to pick any three points from our set of twelve, how many unique triangles could we potentially create?

This is a classic problem of selection. We are choosing $r = 3$ points from a total of $n = 12$ points, where the order of selection does not matter. This is the definition of a combination, denoted as $^{n}C_{r}$.

The formula for this is:

$$^{n}C_{r} = \frac{n!}{r!(n-r)!}$$

Applying this to our twelve points, we calculate the total number of unrestricted selections:

$$^{12}C_3 = \frac{12!}{3!(12-3)!} = \frac{12 	imes 11 	imes 10}{3 	imes 2 	imes 1}$$

When we perform this arithmetic, we find that $12 	imes 11 	imes 10 = 1320$, and $3 	imes 2 	imes 1 = 6$. Dividing 1320 by 6 gives us 220.

So, there are 220 ways to pick three points from the twelve. But here is the catch—are all 220 of these selections actually triangles?

### The Trap of Collinearity

Geometry has a strict rule: a triangle must enclose an area. If you pick three points that lie on the same straight line, you do not get a triangle; you get a line segment.

The problem states that out of our twelve points, five are collinear. This means that any selection of three points made exclusively from these five will fail to form a triangle.

We must identify these 'invalid' selections and remove them from our total count. How many such invalid combinations exist? We are choosing 3 points from the 5 collinear ones.

Using our combination formula again, we calculate:

$$^{5}C_3 = \frac{5!}{3!(5-3)!} = \frac{5 	imes 4}{2 	imes 1} = 10$$

These 10 combinations are the 'ghost' triangles—they exist in our initial count of 220, but they are geometrically impossible.

### The Elegant Resolution

Now, we arrive at the final, satisfying step. We have the total number of ways to pick three points (220) and the number of ways that fail to form a triangle (10).

To find the number of valid triangles, we simply subtract the invalid cases from the total:

$$	ext{Total Triangles} = ^{12}C_3 - ^{5}C_3 = 220 - 10 = 210$$

It is a beautiful moment of clarity. By acknowledging the constraint of collinearity, we have refined our count to exactly **210 valid triangles**.

This problem teaches us that in mathematics, as in life, we must first look at the big picture, identify the constraints that limit our possibilities, and then subtract the invalid paths to find the truth. Keep practicing this logical rigor, and you will find that even the most complex JEE problems become stories waiting to be told.

Animated Solution for Mathematics - Permutations and Combinations: There are 12 points in a plane, no three of which are in the same straight line, except 5 points which are collinear. Then the total number of triangles that can be formed with the vertices at any three of these 12 points is

Select Answer:

Visualized Solution

Visualizing the Points

The Collinear Points

Forming a Triangle

Total Unrestricted Selections

Calculating $^{12} C_{3}$

The Collinear Constraint

Invalid Selections

Calculating $^{5} C_{3}$

Total Valid Triangles

Substituting the Values

Final Computation

Final Answer

The Sigma Insight: Combinations and Selection

Analyzing the Setup

The Combinatorial Landscape

The Trap of Collinearity

The Elegant Resolution

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