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
-
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.
-
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.
- AHSEC Higher Secondary (HS) 1st Year Examination 2026Set ANNUAL1 markMCQQ.If nCr+nCr+1=n+1Cx. Then x=? (A) r−1 (B) r (C) r+1 (D) n
›Reveal solutionSolution
Pascal's rule: nCr+nCr+1=n+1Cr+1, hence x=r+1.
Pascal's identity states
nCr+nCr+1=n+1Cr+1.
Comparing with n+1Cx, we get x=r+1.
✓Final answer(C) r+1.
- AHSEC Higher Secondary (HS) 1st Year Examination 2025Set ANNUAL1 markMCQQ.If nC9=nC8, find nC17.(a) 1(b) 0(c) 17(d) None
›Reveal solutionSolution
n=17, hence nC17=17C17=1.
The property nCa=nCb holds either when a=b or when a+b=n. Since 9e8, we must have
n=9+8=17.
Therefore nC17=17C17=1 (choosing all 17 objects can be done in exactly one way).
✓Final answer(a) 1.
- AHSEC Higher Secondary (HS) 1st Year Examination 2020Set ANNUAL1 markQ.If nC4=nC5, then find the value of nC2.
›Reveal solutionSolution
nC4=nC5 forces n=9; then 9C2=36.
We use the standard combination identity: if nCr=nCs with r=s, then r+s=n.
Here nC4=nC5 with 4=5, so:
n=4+5=9
Now evaluate nC2=9C2:
9C2=2!7!9!=2×19×8=272=36
✓Final answernC2=36
- AHSEC Higher Secondary (HS) 1st Year Examination 2018Set ANNUAL1 markQ.If nCx=nCy and x=y, then x+y=?
›Reveal solutionSolution
If nCx=nCy with x=y, then x+y=n.
The key property of combinations is nCr=nCn−r, and for a fixed n, nCr takes each value at most twice — once at r and once at n−r (these coincide only when r=n−r). Since we're told x=y, the equality nCx=nCy must come from this symmetry:
y=n−x⟹x+y=n.
✓Final answerx+y=n.
🎓Unlock everything free for 14 days
- ✓Full step-by-step solutions
- ✓Concept-first explanations
- ✓Methods, shortcuts & mistakes
- ✓PYQ mapping + timed mock tests
Full access for 14 days. No credit card required.