If 20 lines are drawn in a plane such that no two of them are parallel and no three are concurrent, in how man

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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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Each unique pair of lines, under the given conditions, creates exactly one distinct intersection point. The problem reduces to finding the number of ways to choose 2 lines from 20, which is $\boxed{190}$ .

When lines are drawn in a plane, an intersection point is formed where two distinct lines cross each other. The problem asks for the total number of such points given specific conditions about the lines.

The core idea here is that to form an intersection point, you need exactly two lines. If you pick any two lines from the given set, they will either intersect or be parallel. The problem statement explicitly says "no two of them are parallel," which is crucial. This means that any pair of lines you choose will intersect.

Furthermore, the condition "no three are concurrent" is equally important. Concurrent means passing through the same point. If three or more lines were concurrent, say lines $L_1, L_2, L_3$ all passed through point $P$ , then the pair $(L_1, L_2)$ would form $P$ , the pair $(L_1, L_3)$ would form $P$ , and the pair $(L_2, L_3)$ would also form $P$ . In this scenario, three pairs of lines would yield only one intersection point. However, since no three lines are concurrent, every unique pair of lines will produce a distinct intersection point that is not shared by any other line.

Therefore, the problem simplifies to finding the number of ways to choose 2 lines from the total of 20 lines. The order in which we choose the lines does not matter (choosing $L_1$ then $L_2$ is the same as choosing $L_2$ then $L_1$ for forming an intersection point), so this is a combination problem.

Identify the total number of lines:

We are given $n = 20$ lines in the plane.
Determine the number of lines required for an intersection point:

An intersection point is formed by exactly two lines. So, we need to choose $k = 2$ lines at a time.
Apply the conditions:
- "No two of them are parallel": This ensures that every pair of chosen lines will intersect. …

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Q.If 202020 lines are drawn in a plane such that no two of them are parallel and no three are concurrent, in how many points will they intersect each other?

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Q.If $20$ lines are drawn in a plane such that no two of them are parallel and no three are concurrent, in how many points will they intersect each other?

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