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.
Note
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. …
Use the symmetry property nCr=nCn−r to rewrite terms, then group and apply Pascal's identity twice; the expression telescopes to 0.
The symmetry property of combinations tells us that choosing r objects from n is the same as choosing which n−r objects to leave behind: nCr=nCn−r. This seemingly simple fact is the key to unlocking this problem. Once we rewrite the higher-index terms using symmetry, the expression will reveal a hidden structure.
Let me apply the symmetry property 15Cr=15C15−r to the terms with larger indices:
15C8=15C15−8=15C7
15C9=15C15−9=15C6
Now I can rewrite the original expression by substituting these equivalent forms:
Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.
COMEDK 2026Set 2026-A1 markMCQ
Q.
[!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:
Q.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,