Every body in a room shakes hands with everybody else. The total number of hand shakes is 66. The total number

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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

\binom{n}{k} = \binom{n}{n-k}

Where $\binom{n}{k}$ (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 $0 \leq k \leq n$ .

Why It Works (The Algebraic Proof)

The formula for combinations is:

\binom{n}{k} = \frac{n!}{k!(n-k)!}

Now compute $\binom{n}{n-k}$ :

\binom{n}{n-k} = \frac{n!}{(n-k)!(n-(n-k))!} = \frac{n!}{(n-k)!k!}

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

$k = 0$ : $\binom{n}{0} = 1$ (there is exactly one way to choose nothing). By symmetry, $\binom{n}{n} = 1$ (one way to choose everything). Both make sense — you either take nothing or take all.
$k = 1$ : $\binom{n}{1} = n$ . Symmetry gives $\binom{n}{n-1} = n$ . 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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The problem asks for the number of people in a room given the total number of handshakes. Since each handshake involves two distinct people and the order doesn't matter, this is a combination problem. We set up the equation $\binom{n}{2} = 66$ and solve for $n$ , finding that there are 12 people.

When people shake hands, each handshake involves exactly two individuals. The crucial insight here is that the order in which these two individuals are chosen does not matter. If person A shakes hands with person B, it's the same handshake as person B shaking hands with person A. This characteristic — selecting a group of items where the order of selection is irrelevant — is the definition of a combination.

If there are $n$ people in a room, and every person shakes hands with every other person, we are essentially choosing a group of 2 people from the $n$ available people for each handshake. The total number of handshakes is therefore given by the combination formula $\binom{n}{2}$ .

k

Let's apply this understanding to solve the problem.

Define the variable and set up the equation:

Let $n$ be the total number of persons in the room.

The total number of handshakes is given as $66$ .

Using the combination formula for choosing 2 people out of $n$ :

$\binom{n}{2} = 66$
Expand the combination formula:

We know that $\binom{n}{2} = \frac{n(n-1)}{2}$ .

So, the equation becomes:

$\frac{n(n-1)}{2} = 66$
Solve the equation for $n$ :

Multiply both sides by $2$ :

$n(n-1) = 132$

Expand the left side:

$n^2 - n = 132$

Rearrange into a standard quadratic equation form:

$n^2 - n - 132 = 0$

Now, we need to solve this quadratic equation. We can do this by factoring, using the quadratic formula, or by inspection. We are looking for two consecutive integers whose product is $132$ .

We can list factors of $132$ :

$1 \times 132$

$2 \times 66$

$3 \times 44$

$4 \times 33$

$6 \times 22$

$11 \times 12$ …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

COMEDK 2026Set 2026-A1 markMCQ
$\text { If }{ }^{\mathrm{n}} C_{13},{ }^{\mathrm{n}} C_{14} \text { and }{ }^{\mathrm{n}} C_{15} \text { are in arithmetic progression, then the positive integer value of ' } \mathbf{n} \text { ' can be }$
›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 ${}^n C_{13}$ , ${}^n C_{14}$ , and ${}^n C_{15}$ are in arithmetic progression (AP). That means the middle term is the average of the other two:
$2 \cdot {}^n C_{14} = {}^n C_{13} + {}^n C_{15}.$
The key idea: Binomial coefficients have a well-known recurrence relation ${}^n C_r = {}^n C_{r-1} \cdot \frac{n - r + 1}{r}$ . 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

$2 \cdot {}^n C_{14} = {}^n C_{13} + {}^n C_{15}.$

Express everything in terms of ${}^n C_{13}$ Using the recurrence:

${}^n C_{14} = {}^n C_{13} \cdot \frac{n - 13}{14},$ ${}^n C_{15} = {}^n C_{14} \cdot \frac{n - 14}{15} = {}^n C_{13} \cdot \frac{n - 13}{14} \cdot \frac{n - 14}{15}.$

Substitute into the AP equation Factor out ${}^n C_{13}$ (which is nonzero for $n \ge 15$ ):

$2 \cdot \frac{n - 13}{14} = 1 + \frac{(n - 13)(n - 14)}{14 \cdot 15}.$

Clear denominators Multiply both sides by $14 \cdot 15 = 210$ :

$2 \cdot 15 \cdot (n - 13) = 210 + (n - 13)(n - 14).$
Simplify left side:
$30(n - 13) = 210 + (n - 13)(n - 14).$

Expand and rearrange Left: $30n - 390$ . Right: $210 + (n^2 - 27n + 182) = n^2 - 27n + 392$ . So:

$30n - 390 = n^2 - 27n + 392.$
Bring all terms to one side:
$0 = n^2 - 27n + 392 - 30n + 390 = n^2 - 57n + 782.$

Solve the quadratic

$n^2 - 57n + 782 = 0.$
Discriminant: $\Delta = 57^2 - 4 \cdot 782 = 3249 - 3128 = 121$ .

So:
$n = \frac{57 \pm \sqrt{121}}{2} = \frac{57 \pm 11}{2}.$
This gives:
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KCET 2023Set A-21 markMCQ
$2n + 1$
›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: $P(\text{pick})=\dfrac{n}{2n+1}$ , and $P(H\mid\text{double-headed})=1$ .

$n+1$ fair coins: $P(\text{pick})=\dfrac{n+1}{2n+1}$ , and $P(H\mid\text{fair})=\dfrac12$ .

Step 2 — Law of total probability.

Because the two coin types partition the sample space,
$P(H)=\frac{n}{2n+1}\cdot 1+\frac{n+1}{2n+1}\cdot\frac12$
Step 3 — Simplify.

$P(H)=\frac{1}{2n+1}\left(n+\frac{n+1}{2}\right)=\frac{1}{2n+1}\cdot\frac{2n+n+1}{2}=\frac{3n+1}{2(2n+1)}$ …
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COMEDK 2023Set 2023-M1 markMCQ
$\text { Find }{ }^n C_{21} \text {, if }{ }^n C_{10}={ }^n C_{12}$
›Reveal solutionSolution

$^nC_r={}^nC_s$ with $r\neq s$ forces $n=r+s=22$ , so $^{22}C_{21}=22$ .

If $^nC_{10}={}^nC_{12}$ and $10\neq12$ , then by the property $^nC_r={}^nC_s\Rightarrow r=s$ or $r+s=n$ , we get

$n=10+12=22.$ …
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Q.Every body in a room shakes hands with everybody else. The total number of hand shakes is 666666. The total number of persons in the room is (A) 111111 (B) 121212 (C) 131313 (D) 141414

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Q.Every body in a room shakes hands with everybody else. The total number of hand shakes is $66$ . The total number of persons in the room is
(A) $11$
(B) $12$
(C) $13$
(D) $14$

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