These are 10 points in a plane, out of these 6 are collinear, if N is the number of triangles formed

Q: Why can't we just pick 1 point from the line and 2 from outside?

That is actually a valid alternative method! You could pick 1 point from the 6 collinear points and 2 from the 4 non-collinear points, or 2 from the 6 collinear points and 1 from the 4 non-collinear points. However, the 'Total minus Invalid' method is generally faster and less prone to calculation errors in competitive exams.

Q: What if the points were in a circle? Would the answer change?

Yes, absolutely. If the points were on a circle, no three points would be collinear. In that case, every selection of 3 points would form a valid triangle, and the answer would simply be $\binom{10}{3} = 120$.

Q: Why do we subtract $\binom{6}{3}$? I thought we just ignore those points.

We subtract $\binom{6}{3}$ because $\binom{10}{3}$ counts *all* possible sets of 3 points, including those that happen to be collinear. Since a set of 3 collinear points does not form a triangle, we must remove those specific combinations from our total count to get the correct number of triangles.

Question

These are 10 points in a plane, out of these 6 are collinear, if $N$ is the number of triangles formed by joining these points. then:

Options:

(A) n \le 100

(B) 100 < n \le 140

(C) 140 < n \le 190

(D) n > 190

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a vast, empty plane with ten points scattered before you. Your goal is to connect these points to form triangles.

However, there is a critical geometric constraint: six of these points are trapped on a single straight line.

In geometry, three points can only form a triangle if they are **non-collinear**. If three points lie on the same line, they form a line segment, which is a degenerate triangle with an area of zero.

### The Naive Approach: The Total Count

Let us start by ignoring the constraint. If we have 10 points and we want to choose 3 to form a triangle, we calculate the number of combinations of 10 items taken 3 at a time, denoted as $\binom{10}{3}$.

The formula for combinations is given by:

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

Applying this to our set of 10 points:

$$
\binom{10}{3} = \frac{10 	imes 9 	imes 8}{3 	imes 2 	imes 1} = \frac{720}{6} = 120
$$

There are 120 ways to pick 3 points from 10. However, not every one of these 120 combinations forms a valid triangle.

### The Trap: The Collinear Constraint

This is where the JEE examiner tests your attention to detail. We have 6 points on a single line.

Any combination of 3 points chosen from these 6 will fail to form a triangle. We must calculate how many of our 120 combinations are "invalid" by calculating $\binom{6}{3}$:

$$
\binom{6}{3} = \frac{6 	imes 5 	imes 4}{3 	imes 2 	imes 1} = \frac{120}{6} = 20
$$

There are exactly 20 combinations that result in straight lines rather than triangles.

### The Resolution: Subtracting the Invalid

The path to the solution is now clear. We take our total combinations and subtract the invalid ones using the principle of **complementary counting**.

$$
N = \binom{10}{3} - \binom{6}{3}
$$

$$
N = 120 - 20 = 100
$$

We have found that there are exactly **100** valid triangles.

### Final Reflection

We have arrived at the final result of 100. This problem teaches us that in geometry and combinatorics, the most important step is identifying the constraints.

Never blindly apply a formula; always ask yourself if the calculation includes cases that are physically impossible. By doing so, you will navigate the traps of the JEE Advanced with confidence and precision.

Animated Solution for Mathematics - Permutations and Combinations: These are 10 points in a plane, out of these 6 are collinear, if $N$ is the number of triangles formed by joining these points. then:

Select Answer:

Visualized Solution

Visualizing the Points

Total Combinations of $3$ Points

Calculating $(3 10)$

Evaluating Total Combinations

Identifying Collinear Constraints

Setting Up Invalid Combinations

Calculating $(3 6)$

Final Calculation

Conclusion and Option Check

The Sigma Insight: Combinations and Selection

Analyzing the Setup

The Naive Approach

The Trap

The Resolution

Final Reflection

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