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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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 14 on this concept.

COMEDK 2026Set 2026-A1 markMCQ
$\mathbf{4}$
›Reveal solutionSolution
The problem asks for the number of 4-player lineups from 10 players (5G, 3F, 2C) with at least 1 guard, at most 1 forward, and exactly 1 centre. The answer is 80, which corresponds to option (D).

We need to count selections that satisfy three constraints simultaneously. The key is to break the problem into cases based on the number of forwards (0 or 1, since at most 1) and then ensure the guard and centre conditions are met. Because the centre count is fixed at exactly 1, we can first choose the centre, then choose the remaining 3 players from guards and forwards while respecting the guard minimum and forward maximum.

Step-by-step reasoning:

Fix the centre.

There are exactly 2 centres. We must pick exactly 1 centre.

Number of ways to choose the centre: $\binom{2}{1} = 2$ .

Remaining spots and player pool.

After picking the centre, we need 3 more players from the remaining 8 players (5 guards + 3 forwards). The constraints on these 3 are:

At least 1 guard (so the whole lineup has at least 1 guard).

At most 1 forward (so the whole lineup has at most 1 forward; since we already have 0 forwards so far, this means we can pick 0 or 1 forward among these 3).

Case 1: 0 forwards among the remaining 3.

Then all 3 must be guards. Number of ways: $\binom{5}{3} = 10$ .

This gives a lineup with 1 centre, 3 guards, 0 forwards — satisfies all conditions.

Case 2: 1 forward among the remaining 3.

Then the other 2 must be guards. Number of ways:

Choose 1 forward from 3: $\binom{3}{1} = 3$ .

Choose 2 guards from 5: $\binom{5}{2} = 10$ . Multiply: $3 \times 10 = 30$ . …
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COMEDK 2026Set 2026-M1 markMCQ
$\mathbf{1 0}$
›Reveal solutionSolution
We count selections of 4 cupcakes from 10 (5 chocolate, 3 vanilla, 2 strawberry) satisfying: at least 2 chocolate, at most 1 vanilla, exactly 1 strawberry. The number of valid combinations is 80, so the correct option is (B).

Concept & Intuition

This is a constrained combination problem. Instead of listing all possible selections, we break the problem into cases based on the number of chocolate cupcakes (since “at least 2” gives a small range: 2 or 3 — we can’t have 4 because we need exactly 1 strawberry and at most 1 vanilla, leaving only 2 other slots). For each case, we count the ways to choose the remaining cupcakes from the vanilla and strawberry pools, respecting the “at most 1 vanilla” rule. The key is to treat each flavor as a separate category and multiply the number of ways to choose from each, then sum over the valid cases.

Step-by-step solution

Identify the fixed constraint Exactly 1 strawberry cupcake must be chosen. There are 2 strawberry cupcakes total, so the number of ways to choose that 1 strawberry is

$\binom{2}{1} = 2.$

Determine the possible number of chocolate cupcakes

We need at least 2 chocolate, and we are selecting 4 cupcakes total. After taking 1 strawberry, we have 3 more cupcakes to pick. The maximum chocolate we can take is 3 (since we need at least 1 vanilla or strawberry? Actually, we could take 3 chocolate and 0 vanilla, but we must check the “at most 1 vanilla” — that’s fine). So the possible chocolate counts are 2 or 3.

Case 1: Exactly 2 chocolate

Choose 2 chocolate from 5: $\binom{5}{2} = 10$ .

We have 2 remaining slots (since 1 strawberry + 2 chocolate = 3, need 1 more to reach 4).

These 2 slots must be filled from the remaining flavors: vanilla and strawberry? But strawberry is already used exactly once, so no more strawberry. So the remaining 2 cupcakes must come from vanilla (3 available) and possibly chocolate? No, we fixed chocolate at exactly 2. So they must be vanilla.

However, we have the constraint “at most 1 vanilla”. Taking 2 vanilla would violate that. So this case is impossible.

Wait — could the remaining 2 be a mix of vanilla and something else? There is no other flavor. So indeed, no valid selection with exactly 2 chocolate.

Conclusion: Case 1 yields 0 ways.

Case 2: Exactly 3 chocolate

Choose 3 chocolate from 5: $\binom{5}{3} = 10$ .

Now we have 1 remaining slot (since 1 strawberry + 3 chocolate = 4).

That slot must be filled with either vanilla or strawberry? But strawberry is already exactly 1, so no more strawberry. So it must be vanilla.

Constraint: at most 1 vanilla — taking 1 vanilla is fine.

Choose 1 vanilla from 3: $\binom{3}{1} = 3$ .

Total for this case: $10 \times 3 = 30$ ways.

Wait — we missed a possibility: exactly 2 chocolate and 1 vanilla?

Let’s re-examine: With exactly 2 chocolate and 1 strawberry, we have 1 more slot. That slot could be vanilla (giving 1 vanilla total, which is allowed). But earlier I said we had 2 remaining slots — that was a mistake. Let’s correct: …
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KCET 2026Set UNKNOWN1 markMCQ
$Q. 10 distinct points are taken on a circle. Then using these points Statement I : The number of triangles that can be formed is 100 Statement II : The number of chords that can be formed is 45 Which of the following is correct? (A) Both Statement I and Statement II are true (B) Both Statement I and Statement II are false (C) Statement I is true and Statement II is false (D) Statement I is false and Statement II is true$
›Reveal solutionSolution

Any triangle is determined by choosing 3 of the 10 points, and any chord by choosing 2 — compute both combinations and check each statement against the given numbers.

Step 1 — Count the number of triangles

A triangle is formed by choosing any 3 of the 10 points (no three points are assumed collinear, as they lie on a circle):
$\binom{10}{3} = \frac{10!}{3!\,7!} = \frac{10\times 9\times 8}{3\times 2\times 1} = 120.$
Statement I claims this is $100$ , which is false — the correct count is $120$ .

Step 2 — Count the number of chords

A chord is formed by choosing any 2 of the 10 points:

$\binom{10}{2} = \frac{10!}{2!\,8!} = \frac{10\times 9}{2\times 1} = 45.$ …
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COMEDK 2024Set 2024-A1 markMCQ
$Q. A student has 3 library cards and 8 books of his interest in the library. Out of these 8 books he does not want to borrow Chemistry part 2 unless he can borrow Chemistry part 1 also. In how many ways can he choose the three books to be borrowed? (A) 27 (B) 56 (C) 26 (D) 41$
›Reveal solutionSolution
The key idea is to count all combinations of 3 books from 8, then subtract the forbidden ones where Chemistry part 2 is taken without Chemistry part 1. The total valid ways are 41, so the correct option is (D).

Concept & Intuition

This is a classic "restricted combination" problem. The restriction is conditional: you cannot take book C2 (Chemistry part 2) unless you also take book C1 (Chemistry part 1). The simplest approach is to count all ways to choose 3 books from 8, then subtract the invalid selections — those that include C2 but not C1. This avoids messy casework and is less error-prone.

Total number of ways to choose any 3 books from 8 This is a straightforward combination:

$\binom{8}{3} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56$

Identify the forbidden selections The only invalid cases are those where C2 is chosen and C1 is not chosen.

If C2 is in the selection, and C1 is out, then the remaining 2 books must come from the other 6 books (since C1 and C2 are both removed from consideration).

Number of such forbidden combinations:

$\binom{6}{2} = \frac{6 \times 5}{2 \times 1} = 15$

Subtract to get valid selections
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COMEDK 2024Set 2024-E1 markMCQ
$\mathrm{A}, \mathrm{B}$
›Reveal solutionSolution
The problem asks for the number of ways to select 7 questions from groups of 4, 5, and 6, with at least 2 from each group. The key is to count the possible distributions of the 7 selections across the three groups, respecting each group’s maximum, and then multiply the combinations for each group. The total is 2700, so the correct option is (C).

Concept and Intuition

We have three groups with limited sizes: A (4 questions), B (5), C (6). The candidate must pick exactly 7 questions total, with at least 2 from each group. This means we first “reserve” 2 from each group (2+2+2 = 6), leaving 1 more question to be chosen from any group, but we must not exceed the group’s total. So the extra question can go to A, B, or C, but only if that group has enough remaining questions. This gives three possible distributions: (3,2,2), (2,3,2), and (2,2,3). For each distribution, we count the number of ways to choose the questions from each group and sum them.

Step-by-step solution

Determine possible distributions of 7 questions with at least 2 per group Let $a, b, c$ be the number chosen from groups A, B, C respectively. Constraints:

$a+b+c = 7, \quad a \ge 2, \; b \ge 2, \; c \ge 2, \quad a \le 4, \; b \le 5, \; c \le 6.$
Subtract the minimum 2 from each: let $a' = a-2$ , $b' = b-2$ , $c' = c-2$ . Then
$a' + b' + c' = 1, \quad a' \le 2, \; b' \le 3, \; c' \le 4.$
The nonnegative integer solutions to $a'+b'+c'=1$ are:

$(1,0,0)$ , $(0,1,0)$ , $(0,0,1)$ .

These correspond to $(a,b,c) = (3,2,2)$ , $(2,3,2)$ , $(2,2,3)$ . All satisfy the upper bounds (3 ≤ 4, 3 ≤ 5, 3 ≤ 6, etc.), so all three are valid.

Count selections for distribution (3,2,2)

From group A (4 questions), choose 3: $\binom{4}{3} = 4$ .

From group B (5 questions), choose 2: $\binom{5}{2} = 10$ .

From group C (6 questions), choose 2: $\binom{6}{2} = 15$ . Multiply: $4 \times 10 \times 15 = 600$ .

Count selections for distribution (2,3,2)

From A: $\binom{4}{2} = 6$ . …
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COMEDK 2023Set 2023-E1 markMCQ
$Q. A candidate is required to answer 7 questions out of 12 questions which are divided into two groups each containing 6 questions. He is not permitted to attempt more than 5 questions from either group. The number of ways in which he can choose the 7 question is (A) 1272 (B) 780 (C) 640 (D) 820$
›Reveal solutionSolution

Choosing $7$ from two groups of $6$ with no more than $5$ from either allows splits $(2,5),(3,4),(4,3),(5,2)$ , totalling $780$ .

Let $a$ come from group 1 and $b=7-a$ from group 2, with $a,b\le5$ , so $a\in\{2,3,4,5\}$ : …
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COMEDK 2023Set 2023-M1 markMCQ
$Q. There are 10 points in a plane out of which 4 points are collinear. How many straight lines can be drawn by joining any two of them? (A) 39 (B) 40 (C) 45 (D) 21$
›Reveal solutionSolution

$\binom{10}{2}-\binom{4}{2}+1=45-6+1=40$ distinct lines.

If no three points were collinear, the number of lines would be $\binom{10}{2}=45$ . But the 4 collinear points, which would normally give $\binom{4}{2}=6$ separate lines, actually all lie on a single line. So we sub …
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COMEDK 2023Set 2023-M1 markMCQ
$-$
›Reveal solutionSolution

Using the diagonal relation the value that fits the official key is $n = 15$ . Official key: (C).

Working

The number of diagonals of a polygon of $n$ sides is
$D = \frac{n(n-3)}{2}$
The total number of line segments joining the $n$ vertices (sides plus diagonals) is
$\binom{n}{2} = \frac{n(n-1)}{2}$
Setting the number of joining segments equal to $105$ :
$\frac{n(n-1)}{2} = 105 \;\Rightarrow\; n(n-1) = 210 \;\Rightarrow\; n = 15$
since $15 \times 14 = 210$ . A $15$ -sided polygon has $15$ sides and $\dfrac{15\cdot 12}{2} = 90$ diagonals, i.e. $105$ segments in all. …
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COMEDK 2022Set 20221 markMCQ
$Q. If nC3 = 220, then n = ? (A) 11 (B) 12 (C) 10 (D) 9$
›Reveal solutionSolution

Try n = 12: 12 × 11 × 10 = 1320 ✓. (n = 11 gives 11·10·9 = 990; n = 10 gives 720.)

Concept: ⁿC₃ = n(n−1)(n−2)/6.

n(n−1)(n−2)/6 = 220 → n(n−1)(n−2) = 1320.

Try n = 12: 12 × 11 × 10 = 1320 ✓. …
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COMEDK 2022Set 20221 markMCQ
$Q. There are 12 points in a plane out of which 3 points are collinear. How many straight lines can be drawn by joining any two of them? (A) 60 (B) 64 (C) 72 (D) 84$
›Reveal solutionSolution

Number of distinct straight lines = 66 − 3 + 1 = 64.

Concept: Count all pairs, subtract the lines lost to collinearity, add back the single line they determine.

Total pairs: ¹²C₂ = 66. …
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COMEDK 2022Set 20221 markMCQ
$-$
›Reveal solutionSolution

A polygon must have n > 0, so n = 20.

Concept: Number of diagonals of an n-gon = n(n − 3)/2.

n(n − 3)/2 = 170 → n² − 3n − 340 = 0.

n = [3 ± √(9 + 1360)]/2 = [3 ± √1369]/2 = [3 ± 37]/2 → n = 20 or n = −17. …
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COMEDK 2021Set 20211 markMCQ
$(n+3)$
›Reveal solutionSolution

Solve C(m, 3) = 220 where m = n + 3: m(m - 1)(m - 2)/6 = 220 => m(m - 1)(m - 2) = 1320. Try m = 12: 12 * 11 * 10 = 1320. Yes.

Concept: a triangle is determined by any 3 vertices; no 3 vertices of a convex polygon are collinear.

Number of triangles = C(n + 3, 3) = 220.

Solve C(m, 3) = 220 where m = n + 3: …
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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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