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.

Showing the 12 most recent of 17 on this concept.

CBSE 2026Set ANNUAL1 markMCQ
$^nC_8 = {}^nC_2$
(a) 54
(b) 10
(c) 45
(d) None of these
›Reveal solutionSolution

Use the identity $^nC_r = {}^nC_s \implies r+s=n$ (for $r\ne s$ ) to find $n$ , then compute $^nC_2$ .

The combination identity states: if $^nC_r = {}^nC_s$ with $r\ne s$ , then $r + s = n$ . Here $r=8, s=2$ , so:

$n = 8 + 2 = 10$ …

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CBSE 2026Set ANNUAL1 markMCQ
$Q. If ⁿC₈ = ⁿC₂, then the value of ⁿC₃ is:$
(a) 720
(b) 360
(c) 120
(d) None of these
›Reveal solutionSolution

nC8 = nC2 forces n = 10 (since nCr = nC(n-r) implies r + s = n); then compute 10C3 = 120.

Using the identity $^nC_r = {}^nC_s \iff r+s=n$ (or $r=s$ ), and since $8\ne 2$ , we must have:

$8 + 2 = n \implies n = 10$

Now: …

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CBSE 2026Set 1A1 markMCQ
$^{12}C_{s+1} = {}^{12}C_{2s-5}$
(1) $4$
(2) $8$
(3) $12$
(4) $6$
›Reveal solutionSolution

$^{12}C_{s+1}={}^{12}C_{2s-5}$ gives $s=6$ .

If $^{n}C_a={}^{n}C_b$ then either $a=b$ or $a+b=n$ .

Case 1: $s+1=2s-5 \Rightarrow s=6$ .

…

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CBSE 2025Set ANNUAL1 markMCQ
${}^nC_8 = {}^nC_6$
(a) 350
(b) 346
(c) 364
(d) 580
›Reveal solutionSolution

Using ${}^nC_r = {}^nC_s \implies r+s=n$ (for $r\neq s$ ), $n=8+6=14$ . Then ${}^{14}C_3 = 364$ .

${}^nC_8 = {}^nC_6$

Since ${}^nC_r = {}^nC_{n-r}$ , having ${}^nC_8={}^nC_6$ with $8\ne6$ means $8=n-6$ , i.e.

$n = 8+6 = 14$

Now compute: …

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CBSE 2025Set ANNUAL1 mark
$Q. If ⁿC₆ = ⁿC₄, then the value of n is .............$
›Reveal solutionSolution

When two different combination values from the same $n$ are equal, their lower indices add up to $n$ .

Given $^nC_6 = {}^nC_4$ .

…

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CBSE 2024Set ANNUAL1 markMCQ
${}^{n}C_{12} = {}^{n}C_{8}$
(a) 66
(b) 190
(c) 120
(d) None of these
›Reveal solutionSolution

Use the symmetry property $^nC_r = {}^nC_{n-r}$ to first find $n$ , then compute the required combination.

We are given $^nC_{12} = {}^nC_8$ . Using the identity $^nC_r = {}^nC_{n-r}$ , this equality holds when either $12 = 8$ (false) or $12 = n - 8$ , giving:

$n = 12 + 8 = 20$

…

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CBSE 2023Set ANNUAL1 markMCQ
$^{n}C_{5} = {}^{n}C_{7}$
(a) 5
(b) 7
(c) 12
(d) 0
›Reveal solutionSolution

The identity $^nC_r = {}^nC_{n-r}$ means two equal combination values (with different lower indices) must have indices summing to $n$ .

A key combinatorics identity is:

$^{n}C_{r} = {}^{n}C_{n-r}$

…

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CBSE 2023Set ANNUAL1 markMCQ
${}^{n}C_7 = {}^{n}C_5$
(a) 2
(b) 10
(c) 12
(d) 35
›Reveal solutionSolution

Use the identity ${}^nC_r={}^nC_{n-r}$ .

By the identity ${}^nC_r={}^nC_{n-r}$ , ${}^nC_7={}^nC_{n-7}$ .

…

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CBSE 2022Set ANNUAL1 mark
$Q. If ⁿC₆ = ⁿC₈, then the value of n is ............ .$
›Reveal solutionSolution

Since $6\ne 8$ , the equal-combination property gives $n = 6 + 8 = 14$ .

For combinations, $^nC_r = {}^nC_s$ implies either $r = s$ , or $r + s = n$ .

…

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CBSE 2022Set ANNUAL1 mark
$^{n}C_8 = {}^{n}C_2$
›Reveal solutionSolution

$^nC_2=45$ .

Using $^nC_r={}^nC_s\Rightarrow r+s=n$ (for $r\ne s$ ): here $^nC_8={}^nC_2$ gives

$n=8+2=10.$

…

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CBSE 2022Set ANNUAL1 markMCQ
${}^{n}C_{12} = {}^{n}C_{8}$
(a) 11
(b) 14
(c) 20
›Reveal solutionSolution

${}^nC_{12}={}^nC_8$ implies $n=12+8=20$ .

For combinations, ${}^nC_r = {}^nC_s$ holds if either $r=s$ , or $r+s=n$ (since ${}^nC_r = {}^nC_{n-r}$ ).

…

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CBSE 2022Set ANNUAL1 markMCQ
$^nC_{12} = {}^nC_8$
(a) 20
(b) 12
(c) 6
(d) 8
›Reveal solutionSolution

$n=20$ .

If $^nC_r={}^nC_s$ with $r e s$ , then $r+s=n$ . Here $12+8=20$ , so $n=20$ .

…

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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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