The number of parallelograms that can be formed from a set of four parallel lines intersecting another set of

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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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A parallelogram is formed by choosing any 2 lines from one set and any 2 from the other set. With 4 parallel lines in one set and 3 in the other, the number is $\binom{4}{2} \times \binom{3}{2} = 6 \times 3 = 18$ .

Concept and intuition

A parallelogram is defined by two pairs of parallel sides. In this problem, we have two families of parallel lines: one set of 4 lines all parallel to each other, and another set of 3 lines all parallel to each other (but not parallel to the first set). Any parallelogram formed must have its opposite sides taken from the same family. So to make one parallelogram, you need to pick exactly 2 lines from the first set (these become one pair of opposite sides) and exactly 2 lines from the second set (these become the other pair of opposite sides). The intersection of these four chosen lines gives the four vertices of the parallelogram.

The key insight: the problem reduces to counting combinations, not permutations — the order in which you pick the lines doesn't matter, because the same set of four lines always produces the same parallelogram.

Step-by-step solution

Choose 2 lines from the first set of 4 parallel lines.

The number of ways to select any 2 distinct lines from 4 is given by the combination formula:

$\binom{4}{2} = \frac{4 \times 3}{2 \times 1} = 6$ .
Choose 2 lines from the second set of 3 parallel lines.

Similarly, the number of ways is:

$\binom{3}{2} = \frac{3 \times 2}{2 \times 1} = 3$ .
Multiply the two independent choices. …

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Q.The number of parallelograms that can be formed from a set of four parallel lines intersecting another set of three parallel lines is (A) 666 (B) 181818 (C) 121212 (D) 999

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Q.The number of parallelograms that can be formed from a set of four parallel lines intersecting another set of three parallel lines is
(A) $6$
(B) $18$
(C) $12$
(D) $9$

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