Q.If nC9=nC8, find nC17.
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.
-
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.
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k=n/2 (when n is even): Here k=n−k, so the symmetry says (n/2n)=(n/2n). It is trivially true, but it tells you that the middle binomial coefficient is the largest one — the symmetry is about a central peak.
A Common Mistake to Avoid
Do not confuse this with the symmetry of permutations. For permutations, P(n,k)=(n−k)!n! and P(n,n−k) are not equal. The symmetry property is unique to combinations because order does not matter.
Quick Check
If (512)=792, what is (712)?
Answer: (712)=792, because 7=12−5.
No calculation needed — just the symmetry property.
The Combinations Symmetry Property is a standard result taught alongside the NCERT Class 11 Permutations and Combinations chapter, and it frequently appears in "combinations formula and properties" or "nCr = nC(n-r) proof" searches by CBSE and JEE aspirants. Recognising this identity quickly is a common time-saving trick tested in Class 11/12 mathematics important questions and competitive exam MCQs.
The key idea is the symmetry property of combinations:
nCr=nCn−r.
Given nC9=nC8, we can apply the property:
- By symmetry, nC9=nCn−9 and nC8=nCn−8.
- Since the two are equal, either 9=8 (impossible) or 9=n−8 (the complementary pair match).
- Solving 9=n−8 gives n=17.
Now find nC17=17C17. By definition, 17C17=1.
The value is 1.
By the symmetry nCa=nCb with a=b⇒a+b=n, the equality gives n=9+8=17, so nC17=17C17=1.
1. Use the combination identity. If nCa=nCb then either a=b or a+b=n.
2. Apply it here. Since 9=8, the second case must hold:
9+8=n⇒n=17
3. Evaluate the required combination.
nC17=17C17=1
because there is exactly one way to choose all 17 objects from 17.
nC17=1.
- KEAM 2026Set eng-2026-04204 marksMCQQ.If nC4=1365 then the value of n is equal to (A) 13 (B) 14 (C) 15 (D) 12 (E) 16
›Reveal solutionSolution
(4n)=1365 gives n(n−1)(n−2)(n−3)=1365⋅24=32760. Four consecutive integers with this product are 15,14,13,12, so n=15.
We have
(4n)=24n(n−1)(n−2)(n−3)=1365.
So
n(n−1)(n−2)(n−3)=1365×24=32760.
Seek four consecutive integers with product 32760. Testing n=15:
15⋅14⋅13⋅12=32760.✓
Hence n=15.
✓Final answerThe correct option is (C).
- KEAM 2025Set eng-2025-04264 marksMCQQ.21C1+21C2+…+21C10= (A) 220 (B) 221 (C) 221−1 (D) 221−2 (E) 220−1
›Reveal solutionSolution
The 22 coefficients split into two equal halves; each half sums to 220; subtract (021)=1 to get 220−1.
Symmetry. (k21)=(21−k21), so ∑k=010(k21)=∑k=1121(k21). Each equals half of 221, i.e. 220.
Trim the k=0 term. (121)+⋯+(1021)=220−(021)=220−1.
✓Final answerThe correct option is (E).
- KEAM 2025Set eng-2025-04294 marksMCQQ.The value of 6Cr5Cr when the numerator and denominator take their greatest value, is (A) 2 (B) 21 (C) 1 (D) 65 (E) 56
›Reveal solutionSolution
6Cr is greatest at r=3 (=20), and 5Cr is greatest at r=3 (=10); their ratio is 2010=21.
The greatest binomial coefficient occurs at the middle term. 6Cr is maximum at r=3 with 6C3=20. At r=3, 5C3=10 is also the maximum value of 5Cr (equal to 5C2).
So taking the common r=3 where numerator and denominator are greatest:
6C35C3=2010=21.
✓Final answerThe correct option is (B).
- KEAM 2024Set eng-2024-06064 marksMCQQ.The value of the sum 15C6+14C6+13C6+12C6+11C6+10C6 is equal to (A) 15C7−10C6 (B) 15C7−10C7 (C) 16C7−10C7 (D) 16C7−10C6 (E) 16C7−11C6
›Reveal solutionSolution
Use ∑n=6N(6n)=(7N+1).
∑n=615(6n)=(716) and ∑n=69(6n)=(710). Subtracting, ∑n=1015(6n)=(716)−(710).
✓Final answerThe correct option is (C).
- KEAM 2022Set eng-2022-P2-B14 marksMCQQ.If nC5+nC6=51C6, then the value of n is equal to (A) 49 (B) 50 (C) 45 (D) 46 (E) 51
›Reveal solutionSolution
n=50.
Concept and Intuition
Pascal's identity: nCr−1+nCr=n+1Cr.
Step-by-Step Solution
- nC5+nC6=n+1C6.
- Set equal to 51C6: n+1C6=51C6.
- Therefore n+1=51, so n=50.
Common Mistakes
- Misremembering Pascal's rule (indices must match as r−1 and r).
- Solving n=51 instead of n+1=51.
✓Final answerThe correct option is (B) — 50.
ANSWER: B
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