Q.Determine n if 2nC3:nC3=12:1.
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🔒 Start your 14-day free trial to unlock the full solution →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.
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:
Number of combinations=r!Number of permutations
Where r is the number of items you're choosing.
The Precise Statement
(rn)=r!(n−r)!n!
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≤n
- The objects are distinct (no repetitions)
Why the Formula Works
Start with permutations of r items from n: P(n,r)=(n−r)!n!.
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:
(rn)=r!P(n,r)=r!(n−r)!n!
A quick check: (0n)=1 (there's exactly one way to choose nothing), and (nn)=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.
(552)=5!⋅47!52!=5×4×3×2×152×51×50×49×48=2,598,960
That's over 2.5 million possible hands — which is why poker is interesting. …
Write out 2nC3 and nC3 using the combination formula, form the ratio, and solv …
Setting up nC32nC3=12 and simplifying gives n=5.
2nC3=62n(2n−1)(2n−2),nC3=6n(n−1)(n−2)
Given nC32nC3=112:
n(n−1)(n−2)2n(2n−1)(2n−2)=12
Since 2n−2=2(n−1):
n(n−1)(n−2)2n(2n−1)⋅2(n−1)=12⟹n(n−2)4n(2n−1)=12
Cancel n (nonzero) from numerator and denominator: …
Showing the 12 most recent of 21 on this concept.
- CBSE 2026Set ANNUAL1 markMCQQ.How many chords can be drawn through 21 points on a circle?(a) 420(b) 21P2(c) 21(d) 210
›Reveal solutionSolution
A chord is determined by an unordered pair of points, so count combinations (not permutations) of 21 points taken 2 at a time.
Every chord of a circle is uniquely determined by choosing 2 of the points on the circle (the order of choosing doesn't create a different chord — the chord from P to Q is the same as from Q to P). So this is a combinations problem:
21C2=2!19!21!=221×20=210 …
- CBSE 2025Set ANNUAL1 markMCQQ.2025C2025=(a) 2025(b) 0(c) 1(d) 2024
›Reveal solutionSolution
2025C2025=1.
The combination formula nCr=r!(n−r)!n!. For r=n: nCn=n!0!n!=n!×1n!=1 (using 0!=1), true for an …
- CBSE 2025Set ANNUAL1 markMCQQ.2025C1=(a) 0(b) 1(c) 2025(d) 2025!
›Reveal solutionSolution
2025C1=2025.
Using nCr=r!(n−r)!n! with r=1: nC1=1!(n−1)!n!=(n−1)!n(n−1)!=n.
…
- CBSE 2025Set ANNUAL1 markMCQQ.Relation between permutation and combination is(a) nCr=r!nPr(b) nCr×r!=1(c) r!nCr=nPr(d) None of these
›Reveal solutionSolution
A permutation counts ordered selections, a combination counts unordered ones; each combination corresponds to r! permutations (the orderings of its r chosen items), so nCr=nPr/r!.
nPr counts the number of ways to choose and arrange r objects out of n. Each unordered group of r objects (a combination) can be arranged in r! different orders (permutations).
So: …
- CBSE 2025Set ANNUAL1 markMCQQ.nCr+nCr−1=?(a) n+1Cr(b) n−1Cr(c) n+1Cr+1(d) None of these
›Reveal solutionSolution
nCr+nCr−1=n+1Cr is the standard Pascal's-rule identity for combinations.
This is a fundamental combinatorial identity (Pascal's Rule), which can be proved algebraically:
nCr+nCr−1=r!(n−r)!n!+(r−1)!(n−r+1)!n!
Taking r!(n−r+1)!n! as the common structure and combining the terms simplifies (via the standard derivation) to:
r!(n+1−r)!(n+1)!=n+1Cr
…
- CBSE 2025Set ANNUAL1 markMCQQ.How many different teams of 7 players can be chosen out of 10 players?(a) 720(b) 120(c) 70(d) None of these
›Reveal solutionSolution
A team of 7 from 10 players is 10C7, which by symmetry equals 10C3.
…
- CBSE 2025Set ANNUAL1 markMCQQ.Match the column: Column A entry 'Value of 5C2' — find the matching value from Column B.(a) 2(b) 8(c) 32(d) 1−tan2x2tanx(e) sin2x(f) 10(g) 20(h) 1+tan2x2tanx(i) 4
›Reveal solutionSolution
5C2=2!3!5!=10, matching Column B option (f).
…
- CBSE 2024Set ANNUAL1 markMCQQ.In how many ways a team of 3 boys and 4 girls can be selected from 5 boys and 4 girls?(a) 5C3×4C4(b) 5C3×4C3(c) 5C4×4C3(d) None of these
›Reveal solutionSolution
When two independent selections must both happen, multiply the number of ways for each (fundamental principle of counting).
We need to select a team of 3 boys and 4 girls from a pool of 5 boys and 4 girls.
Choosing the boys: we need 3 boys out of 5 available, which can be done in 5C3 ways.
…
- CBSE 2024Set ANNUAL1 markMCQQ.If nPr=720, nCr=120, then r=?(a) r=3(b) r=5(c) r=4(d) None of these
›Reveal solutionSolution
Permutations and combinations are related by nPr=nCr×r!; use this to solve for r! directly.
Recall the relationship between permutations and combinations:
nPr=nCr×r!
We are given nPr=720 and nCr=120. Substituting:
720=120×r! …
- CBSE 2024Set ANNUAL1 markMCQQ.If n=5 and r=3, then the value of nCr is:(a) 10(b) 30(c) 10!(d) 15!.
›Reveal solutionSolution
5C3=10.
The combination formula is nCr=r!(n−r)!n!. With n=5, r=3: …
- CBSE 2024Set hz1 markMCQQ.If n=10, r=3, then value of nCr is:(a) 100(b) 120(c) 110(d) 520
›Reveal solutionSolution
10C3=3!7!10!=120.
The formula for combinations is
nCr=r!(n−r)!n!
With n=10, r=3: …
- CBSE 2024Set ANNUAL1 markMCQQ.The value of nCn is:(a) 0(b) undefined(c) 1(d) 2
›Reveal solutionSolution
nCr=r!(n−r)!n!; putting r=n gives nCn=1.
Step 1. Use the formula nCr=r!(n−r)!n!.
…
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