Out of 18 points in a plane, no three are in the same line except five points which are collinear. Find the nu

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Concept understanding — Combinations Selection

Combinations: Choosing Without Ordering

Imagine you're picking a team of 3 players from a group of 5 friends: Alice, Bob, Charlie, Deepa, and Esha. The team {Alice, Bob, Charlie} is the same team as {Bob, Charlie, Alice} — the order you name them doesn't matter. What matters is which 3 people you pick.

That's the core idea of combinations: selection without regard to order.

The Intuition: Why Order Doesn't Matter

Let's contrast with permutations. If you were assigning positions — captain, vice-captain, treasurer — then {Alice as captain, Bob as vice-captain, Charlie as treasurer} is different from {Bob as captain, Alice as vice-captain, Charlie as treasurer}. Order matters there.

But for a plain team, a committee, a hand of cards, or a set of toppings on a pizza — order is irrelevant. You just care about which items are chosen.

Note

Key distinction: Permutations count arrangements (order matters). Combinations count selections (order doesn't matter).

From Permutations to Combinations

Suppose you want to choose 2 letters from {A, B, C}. If order mattered, you'd have these 6 permutations:

AB, BA, AC, CA, BC, CB

But if order doesn't matter, AB and BA are the same selection. So the distinct combinations are just:

{A, B}, {A, C}, {B, C} — only 3.

Notice the pattern: each combination of 2 items corresponds to $2! = 2$ permutations (because you can arrange those 2 items in 2 ways). So:

\text{Number of combinations} = \frac{\text{Number of permutations}}{r!}

Where $r$ is the number of items you're choosing.

The Precise Statement

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

This is read as "n choose r" and gives the number of ways to select $r$ distinct objects from a set of $n$ distinct objects, where order does not matter.

Conditions:

$n$ and $r$ are non-negative integers
$r \leq n$
The objects are distinct (no repetitions)

Why the Formula Works

Start with permutations of $r$ items from $n$ : $P(n, r) = \frac{n!}{(n-r)!}$ .

Each combination of $r$ items can be arranged in $r!$ different orders. So the number of combinations is the number of permutations divided by the number of ways to rearrange each selection:

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

Tip

A quick check: $\binom{n}{0} = 1$ (there's exactly one way to choose nothing), and $\binom{n}{n} = 1$ (one way to choose everything).

A Concrete Example

How many different 5-card hands can be dealt from a standard 52-card deck?

Here, $n = 52$ , $r = 5$ . The hand {A♠, K♥, Q♦, J♣, 10♠} is the same regardless of the order you receive the cards.

\binom{52}{5} = \frac{52!}{5! \cdot 47!} = \frac{52 \times 51 \times 50 \times 49 \times 48}{5 \times 4 \times 3 \times 2 \times 1} = 2,598,960

That's over 2.5 million possible hands — which is why poker is interesting. …

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The key idea is to count all possible lines from pairs of points, then subtract the overcount caused by the 5 collinear points (which would otherwise form many duplicate lines). The total number of distinct lines is 144.

Why This Approach Works

When you join any two points, you get a line. If no three points were collinear, every pair would give a unique line. But here, 5 points lie on the same straight line. Any pair chosen from those 5 points gives the same line — not 5 different lines. So we must count carefully: first count all possible pairs, then adjust for the collinear group.

The natural method is to use combinations: the number of lines = (total pairs of points) − (pairs within the collinear set that are overcounted) + 1 (for the one actual line they all lie on).

Step-by-Step Solution

1. Total points and the collinear group

We have 18 points in total. Five of them are collinear (all on one line). The remaining $18 - 5 = 13$ points have no three collinear among themselves, and also no three collinear with any point from the collinear set (except the 5 themselves).

2. Count all possible pairs of points

Any two distinct points determine a line. The total number of unordered pairs from 18 points is:

\binom{18}{2} = \frac{18 \times 17}{2} = 153

If no three points were collinear, this would be the answer. But we have a problem: the 5 collinear points produce many pairs that all give the same line.

3. Count pairs within the collinear set

From the 5 collinear points, the number of pairs is:

\binom{5}{2} = \frac{5 \times 4}{2} = 10

These 10 pairs would normally give 10 distinct lines. But in reality, they all lie on exactly one line.

4. Adjust the count

We started by counting all 153 pairs as if each gave a unique line. For the collinear group, we counted 10 lines where only 1 exists. So we must subtract the 9 extra lines we imagined: …

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Q.Out of 181818 points in a plane, no three are in the same line except five points which are collinear. Find the number of lines that can be formed joining the point.

Why This Approach Works

Step-by-Step Solution

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Q.Out of $18$ points in a plane, no three are in the same line except five points which are collinear. Find the number of lines that can be formed joining the point.

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