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.
- COMEDK 2026Set 2026-A1 markMCQQ.
[!FORMULA] If nC13,nC14 and nC15 are in arithmetic progression, then the positive integer value of ’ n ’ can be
(A) 34 (B) 14 (C) 24 (D) 41›Reveal solutionSolution
The problem uses the condition that three consecutive binomial coefficients are in arithmetic progression, which leads to a quadratic equation in n. Solving it yields two possible positive integer values, and only one matches the given options: n=14.
We are told that nC13, nC14, and nC15 are in arithmetic progression (AP). That means the middle term is the average of the other two:
2⋅nC14=nC13+nC15.
The key idea: Binomial coefficients have a well-known recurrence relation nCr=nCr−1⋅rn−r+1. Using this, we can rewrite the AP condition as an equation in n without having to compute large factorials.
Let’s work through it step by step.
- Write the AP condition
2⋅nC14=nC13+nC15.
- Express everything in terms of nC13 Using the recurrence:
nC14=nC13⋅14n−13,
nC15=nC14⋅15n−14=nC13⋅14n−13⋅15n−14.
- Substitute into the AP equation Factor out nC13 (which is nonzero for n≥15):
2⋅14n−13=1+14⋅15(n−13)(n−14).
- Clear denominators Multiply both sides by 14⋅15=210:
2⋅15⋅(n−13)=210+(n−13)(n−14).
Simplify left side:
30(n−13)=210+(n−13)(n−14).
- Expand and rearrange Left: 30n−390. Right: 210+(n2−27n+182)=n2−27n+392. So:
30n−390=n2−27n+392.
Bring all terms to one side:
0=n2−27n+392−30n+390=n2−57n+782.
- Solve the quadratic
n2−57n+782=0.
Discriminant: Δ=572−4⋅782=3249−3128=121.
So:
n=257±121=257±11.
This gives:
n=268=34orn=246=23.
- Check which values are valid
Both are positive integers. But we must also ensure that the binomial coefficients are defined: n≥15 for C15 to exist. Both 34 and 23 satisfy that. However, the problem asks for the positive integer value of n that can be from the given options. The options are 34, 14, 24, 41.
- n=34 is option (A).
- n=23 is not listed. So the only matching option is 34.
Watch outA common mistake is to forget that n=23 also satisfies the AP condition, but since it’s not among the choices, we must pick the one that is. Always check the options!
TipThe quadratic n2−57n+782=0 can also be factored as (n−23)(n−34)=0 — a quick check saves time.
✓Final answerThe correct option is (A).
ANSWER: A
- KCET 2023Set A-21 markMCQQ.A bag contains 2n+1 coins. It is known that n of these coins have head on both sides whereas the other n+1 coins are fair. One coin is selected at random and tossed. If the probability that toss results in heads is 4231 then the value of n is (A) 6 (B) 8 (C) 10 (D) 5
›Reveal solutionSolution
Apply the law of total probability over the two kinds of coin, then solve the resulting linear equation for n.
Step 1 — Set up the two cases.
The bag has 2n+1 coins, each equally likely to be picked:
- n double-headed coins: P(pick)=2n+1n, and P(H∣double-headed)=1.
- n+1 fair coins: P(pick)=2n+1n+1, and P(H∣fair)=21.
Step 2 — Law of total probability.
Because the two coin types partition the sample space,
P(H)=2n+1n⋅1+2n+1n+1⋅21
Step 3 — Simplify.
P(H)=2n+11(n+2n+1)=2n+11⋅22n+n+1=2(2n+1)3n+1
Step 4 — Equate to the given value and solve.
2(2n+1)3n+1=4231
42(3n+1)=62(2n+1)
126n+42=124n+62 ⇒ 2n=20 ⇒ n=10
Check (n=10): 21 coins, P(H)=2110+2111⋅21=4220+11=4231 ✓
✓Final answerThe correct option is (C) — 10.
ANSWER: C
- COMEDK 2023Set 2023-M1 markMCQQ.
[!FORMULA] Find nC21, if nC10=nC12
(A) 1 (B) 21 (C) 22 (D) 2›Reveal solutionSolution
nCr=nCs with r=s forces n=r+s=22, so 22C21=22.
If nC10=nC12 and 10=12, then by the property nCr=nCs⇒r=s or r+s=n, we get
n=10+12=22.
Hence
nC21=22C21=22C1=22.
✓Final answerThe correct option is (C) — 22
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