If the sides AB, BC and CA of a triangle ABC have 3,5 and 6 interior points respectively, then the t

Q: Wait, why do we subtract the collinear points? Can't we just count the valid ones directly?

You absolutely could, but it would be a combinatorial nightmare! To count directly, you would need to sum up cases like (1 point from AB, 1 from BC, 1 from CA) + (2 from AB, 1 from BC) + (2 from AB, 1 from CA), and so on. Complementary counting—subtracting the 'bad' cases from the 'total universe'—is the elegant, 'JEE-smart' way to solve this. It saves time and minimizes calculation errors.

Q: Does the order of selection matter here? Why are we using combinations instead of permutations?

That is a fantastic question. In geometry, a triangle defined by vertices {P1, P2, P3} is exactly the same as a triangle defined by {P2, P1, P3}. The order in which you pick the vertices does not change the shape. Since order is irrelevant, we use combinations (${}^{n}C_r$) rather than permutations (${}^{n}P_r$).

Q: What if the vertices A, B, and C were included in the count of points? Would that change the logic?

It would change the numbers, but not the logic! If A, B, and C were included, you would simply increase your total $N$. However, you would have to be very careful about collinearity. For example, if A is included on side AB and side CA, it would be part of two different collinear sets. Always define your 'universe' of points clearly before you start counting.

Question

If the sides AB, BC and CA of a triangle ABC have 3,5 and 6 interior points respectively, then the total number of triangles that can be constructed using these points as vertices, is equal to :

Options:

(A) 364

(B) 240

(C) 333

(D) 360

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineer. Today, we are not just solving a problem; we are exploring the architecture of space. We have a triangle, $ABC$, and scattered along its sides are points—like beads on a string.

We have $3$ points on side $AB$, $5$ on side $BC$, and $6$ on side $CA$. Our mission is to find out how many distinct triangles we can construct using these points as vertices.

This is a classic problem that tests your ability to distinguish between the 'Universe of Possibilities' and the 'Reality of Constraints.'

### The Universe of Possibilities

First, let us define our total set of points. If we ignore the geometry for a moment and just look at the raw data, we have a collection of points.

Let $N$ be the total number of points. We simply sum them up: $N = 3 + 5 + 6 = 14$. We have $14$ points in total.

Now, imagine you are a creator. You want to build a triangle. To build a triangle, you need exactly $3$ points.

If we were to pick any $3$ points from our set of $14$ without any restrictions, how many ways could we do that? This is a fundamental combinatorics question. We use the combination formula, ${}^{n}C_{r} = \frac{n!}{r!(n-r)!}$.

Here, $n=14$ and $r=3$. So, we calculate ${}^{14}C_{3}$:

$$ {}^{14}C_{3} = \frac{14 	imes 13 	imes 12}{3 	imes 2 	imes 1} $$

Let us simplify this. Since $3 	imes 2 	imes 1 = 6$ and $12 / 6 = 2$, we are left with $14 	imes 13 	imes 2$. That gives us $364$.

This number, $364$, represents every possible triplet of points we can choose. But here is the catch: not every triplet makes a triangle.

### The Collinear Trap

In the world of geometry, three points form a triangle if and only if they are non-collinear. If you pick three points that lie on the same straight line, you do not get a triangle; you get a line segment. This is the 'Collinear Trap.'

Look at our triangle $ABC$. The points on side $AB$ are all collinear. The points on side $BC$ are all collinear. The points on side $CA$ are all collinear.

If we accidentally pick $3$ points from side $AB$, we have failed to create a triangle. If we pick $3$ points from side $BC$, we have failed. If we pick $3$ points from side $CA$, we have failed.

We must identify these 'invalid' selections and remove them from our total of $364$.

### Filtering the Noise

Let us calculate the number of invalid selections for each side.

For side $AB$, we have $3$ points. The number of ways to choose $3$ points from these $3$ is ${}^{3}C_{3} = 1$. This is one 'flat' triangle that we must discard.

For side $BC$, we have $5$ points. The number of ways to choose $3$ points from these $5$ is ${}^{5}C_{3}$. Using our formula:

$$ {}^{5}C_{3} = \frac{5 	imes 4 	imes 3}{3 	imes 2 	imes 1} = \frac{5 	imes 4}{2 	imes 1} = 10 $$

So, there are $10$ invalid combinations on side $BC$.

For side $CA$, we have $6$ points. The number of ways to choose $3$ points from these $6$ is ${}^{6}C_{3}$:

$$ {}^{6}C_{3} = \frac{6 	imes 5 	imes 4}{3 	imes 2 	imes 1} = 5 	imes 4 = 20 $$

There are $20$ invalid combinations on side $CA$.

### The Final Synthesis

Now, we bring it all together. We started with a grand total of $364$ possible selections. We identified the 'noise'—the invalid, collinear selections—which sum up to $1 + 10 + 20 = 31$.

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

$$ 	ext{Valid Triangles} = {}^{14}C_{3} - ({}^{3}C_{3} + {}^{5}C_{3} + {}^{6}C_{3}) $$

$$ 	ext{Valid Triangles} = 364 - (1 + 10 + 20) $$

$$ 	ext{Valid Triangles} = 364 - 31 = 333 $$

And there it is. **$333$ valid triangles**. It is a beautiful, symmetric result.

Remember, in JEE Advanced, the math is rarely just about calculation; it is about understanding the constraints of the system. You have successfully navigated the trap of collinearity and arrived at the truth. Keep this mindset—always look for the constraints, and you will never be lost.

Animated Solution for Mathematics - Permutations and Combinations: If the sides AB, BC and CA of a triangle ABC have 3,5 and 6 interior points respectively, then the total number of triangles that can be constructed using these points as vertices, is equal to :

Select Answer:

Visualized Solution

Visualizing the Problem

Total Available Points

Forming a Triangle

Calculating $^{14} C_{3}$

The Collinearity Constraint

Invalid Triangles on $A B$

Invalid Triangles on $B C$

Invalid Triangles on $C A$

Total Valid Triangles

Final Result

The Sigma Insight: Combinations and Selection

Analyzing the Setup

The Universe of Possibilities

The Collinear Trap

Filtering the Noise

The Final Synthesis

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