Q.If 12Cr+1=12C3r−5, then find r.
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🔒 Start your 14-day free trial to unlock the full solution →Concept understanding — Combinations Symmetry Property
The Intuition: Two Ways to Choose
Imagine you have a group of 10 friends, and you need to pick 3 of them to form a committee. One way to think about this is: you are choosing the 3 people who will be on the committee. But there is another, equally valid way to think about it: you are rejecting the 7 people who will not be on the committee.
Choosing 3 to include is the same decision as choosing 7 to exclude. Every time you pick a set of 3, you automatically determine the set of 7 who are left out. There is a perfect one-to-one match between the two choices.
This is the heart of the symmetry property: the number of ways to choose k items from n is exactly the same as the number of ways to choose n−k items from n.
The Precise Statement
(kn)=(n−kn)
Where (kn) (read "n choose k") is the number of combinations — the number of distinct subsets of size k you can pick from a set of n distinct objects.
This holds for any non-negative integers n and k where 0≤k≤n.
Why It Works (The Algebraic Proof)
The formula for combinations is:
(kn)=k!(n−k)!n!
Now compute (n−kn):
(n−kn)=(n−k)!(n−(n−k))!n!=(n−k)!k!n!
The denominator is just k!(n−k)! written in a different order. Since multiplication is commutative, the two expressions are identical.
The symmetry is purely algebraic, but the intuition is what makes it memorable: choosing k to keep is the same as choosing n−k to discard.
Special Cases That Make Sense
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k=0: (0n)=1 (there is exactly one way to choose nothing). By symmetry, (nn)=1 (one way to choose everything). Both make sense — you either take nothing or take all.
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k=1: (1n)=n. Symmetry gives (n−1n)=n. Choosing 1 person to include is the same as choosing n−1 people to exclude — there are n choices in either case. …
Two combinations nCx and nCy are equal only when x=y or x+y=n; applying this to the given equation produces two p …
Use the property nCx=nCy⟺x=y or x+y=n; apply this to x=r+1, y=3r−5, n=12.
Given 12Cr+1=12C3r−5.
By the standard property of combinations, this holds if either:
Case 1: r+1=3r−5⇒2r=6⇒r=3.
Check: r+1=4, 3r−5=4 — both equal 4, valid (and 0≤4≤12).
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Showing the 12 most recent of 17 on this concept.
- CBSE 2026Set ANNUAL1 markMCQQ.If nC8=nC2 then the value of nC2 is(a) 54(b) 10(c) 45(d) None of these
›Reveal solutionSolution
Use the identity nCr=nCs⟹r+s=n (for r=s) to find n, then compute nC2.
The combination identity states: if nCr=nCs with r=s, then r+s=n. Here r=8,s=2, so:
n=8+2=10 …
- CBSE 2026Set ANNUAL1 markMCQQ.If ⁿC₈ = ⁿC₂, then the value of ⁿC₃ is:(a) 720(b) 360(c) 120(d) None of these
›Reveal solutionSolution
nC8 = nC2 forces n = 10 (since nCr = nC(n-r) implies r + s = n); then compute 10C3 = 120.
Using the identity nCr=nCs⟺r+s=n (or r=s), and since 8=2, we must have:
8+2=n⟹n=10
Now: …
- CBSE 2026Set 1A1 markMCQQ.If 12Cs+1=12C2s−5, then the value of s is -(1) 4(2) 8(3) 12(4) 6
›Reveal solutionSolution
12Cs+1=12C2s−5 gives s=6.
If nCa=nCb then either a=b or a+b=n.
Case 1: s+1=2s−5⇒s=6.
…
- CBSE 2025Set ANNUAL1 markMCQQ.If nC8=nC6 then the value of nC3 is(a) 350(b) 346(c) 364(d) 580
›Reveal solutionSolution
Using nCr=nCs⟹r+s=n (for r=s), n=8+6=14. Then 14C3=364.
nC8=nC6
Since nCr=nCn−r, having nC8=nC6 with 8=6 means 8=n−6, i.e.
n=8+6=14
Now compute: …
- CBSE 2025Set ANNUAL1 markQ.If ⁿC₆ = ⁿC₄, then the value of n is .............
›Reveal solutionSolution
When two different combination values from the same n are equal, their lower indices add up to n.
Given nC6=nC4.
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- CBSE 2024Set ANNUAL1 markMCQQ.If nC12=nC8, then nC18=(a) 66(b) 190(c) 120(d) None of these
›Reveal solutionSolution
Use the symmetry property nCr=nCn−r to first find n, then compute the required combination.
We are given nC12=nC8. Using the identity nCr=nCn−r, this equality holds when either 12=8 (false) or 12=n−8, giving:
n=12+8=20
…
- CBSE 2023Set ANNUAL1 markMCQQ.If nC5=nC7, then n=(a) 5(b) 7(c) 12(d) 0
›Reveal solutionSolution
The identity nCr=nCn−r means two equal combination values (with different lower indices) must have indices summing to n.
A key combinatorics identity is:
nCr=nCn−r
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- CBSE 2023Set ANNUAL1 markMCQQ.If nC7=nC5, then the value of n is(a) 2(b) 10(c) 12(d) 35
›Reveal solutionSolution
Use the identity nCr=nCn−r.
By the identity nCr=nCn−r, nC7=nCn−7.
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- CBSE 2022Set ANNUAL1 markQ.If ⁿC₆ = ⁿC₈, then the value of n is ............ .
›Reveal solutionSolution
Since 6=8, the equal-combination property gives n=6+8=14.
For combinations, nCr=nCs implies either r=s, or r+s=n.
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- CBSE 2022Set ANNUAL1 markQ.If nC8=nC2, find nC2.
›Reveal solutionSolution
nC2=45.
Using nCr=nCs⇒r+s=n (for r=s): here nC8=nC2 gives
n=8+2=10.
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- CBSE 2022Set ANNUAL1 markMCQQ.Fill in the blank with the correct option: if nC12=nC8, then n=____(a) 11(b) 14(c) 20
›Reveal solutionSolution
nC12=nC8 implies n=12+8=20.
For combinations, nCr=nCs holds if either r=s, or r+s=n (since nCr=nCn−r).
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- CBSE 2022Set ANNUAL1 markMCQQ.If nC12=nC8, then n is(a) 20(b) 12(c) 6(d) 8
›Reveal solutionSolution
n=20.
If nCr=nCs with res, then r+s=n. Here 12+8=20, so n=20.
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