The side AB, BC and CA of a triangle ABC have 3, 4 and 5 interior points respectively on them. The n

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

You absolutely could, but that would be a combinatorial nightmare! You would have to account for picking one point from one side, one from another, and one from the third, or two from one side and one from another... it gets messy very quickly. Complementary counting—calculating the total and removing the 'bad' cases—is the elegant, professional way to handle this.

Q: Why are the points on the same side considered 'invalid'? Don't they form a triangle with the other vertices?

Remember, we are selecting three points from the set of interior points provided. If all three points you select happen to lie on the same line segment (a side of the triangle), they are collinear. By definition, three collinear points cannot form a triangle; they just form a line segment. We are looking for the number of triangles formed using these specific points as vertices.

Q: How do I know when to use permutations versus combinations?

This is the classic JEE dilemma! Ask yourself: Does the order of selection matter? If I pick point X, then Y, then Z, is that a different triangle than picking Z, then Y, then X? No, it is the same triangle. Since the order of vertices does not change the triangle, we use combinations ($^nC_r$).

Question

The side $AB, BC$ and $CA$ of a triangle $ABC$ have 3, 4 and 5 interior points respectively on them. The number of triangles that can be constructed using these interior points as vertices is .........

Sigma Percentile · Accepted Answer

### The Geometry of Choice: A Combinatorial Journey

Welcome, future engineer. Today, we are not just solving a math problem; we are exploring the delicate balance between geometry and combinatorics.

We have a triangle, $ABC$, and scattered along its sides are interior points—3 on $AB$, 4 on $BC$, and 5 on $CA$. Our mission is to find out how many unique triangles we can construct using these points as vertices.

This problem is a beautiful lesson in **Complementary Counting**, a technique that separates the masters from the novices.

### Phase 1: The Universe of Possibilities

First, let us look at the big picture. We have a collection of points. How many are there in total?

We have 3 points on side $AB$, 4 on side $BC$, and 5 on side $CA$. Summing these up, we get $n = 3 + 4 + 5 = 12$ points.

Imagine these 12 points floating in space. If we were to pick any 3 of them, how many ways could we do it? This is a classic selection problem where order does not matter, so we use the combination formula:

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

For our case, we need to choose 3 points out of 12, which is $\binom{12}{3}$. Calculating this, we get:

$$\binom{12}{3} = \frac{12 	imes 11 	imes 10}{3 	imes 2 	imes 1} = 220$$

This number, 220, represents our 'Total Universe' of selections. It includes every possible combination of three points. But here is the catch—not every combination creates a triangle.

### Phase 2: The Collinear Trap

In 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.' We have inadvertently included these 'flat' triangles in our count of 220. We must identify and remove them.

Let us look at each side individually:

*   On side $AB$, we have 3 points. If we select all 3 of these points, they are collinear. The number of ways to choose 3 points from these 3 is $\binom{3}{3} = 1$. This is one invalid case.

*   On side $BC$, we have 4 points. If we select any 3 of these 4 points, they will all lie on the line $BC$, forming a line segment. The number of ways to choose 3 points from these 4 is $\binom{4}{3} = 4$. These are 4 more invalid cases.

*   On side $CA$, we have 5 points. If we select any 3 of these 5 points, they lie on the line $CA$. The number of ways to choose 3 points from these 5 is $\binom{5}{3} = 10$. These are 10 more invalid cases.

### Phase 3: The Elegant Subtraction

Now, we bring it all together. We have our total universe of 220 combinations.

We have identified the 'forbidden' combinations that fail to form a triangle: 1 from side $AB$, 4 from side $BC$, and 10 from side $CA$. The total number of invalid cases is:

$$1 + 4 + 10 = 15$$

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

$$	ext{Number of Triangles} = 220 - 15 = 205$$

This gives us **205**. It is a clean, satisfying result.

We started with a large, chaotic set of possibilities and used logical constraints to prune away the invalid ones, leaving us with the exact number of triangles that can be formed. Remember this process—it is the hallmark of a great problem solver.

You do not always need to count the 'good' things; sometimes, it is much more powerful to count the 'bad' things and take them away. Keep practicing, keep visualizing, and keep falling in love with the logic behind the numbers.

Animated Solution for Mathematics - Permutations and Combinations: The side $A B, B C$ and $C A$ of a triangle $A B C$ have 3, 4 and 5 interior points respectively on them. The number of triangles that can be constructed using these interior points as vertices is .........

Enter Numerical Value:

Visualized Solution

Visualizing the Setup

Points on Side $A B$

Points on Side $B C$

Points on Side $C A$

Total Available Points

Selecting 3 Points to Form a Triangle

Calculating Total Combinations

The Collinear Trap

Invalid Cases on Side $A B$

Invalid Cases on Side $B C$

Invalid Cases on Side $C A$

Total Invalid Cases

Calculating the Final Answer

The Sigma Insight: Combinations and Selection

The Geometry of Choice

Phase 1

Phase 2

Phase 3

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