If ${}^{n}C_{r-1} = 36$, ${}^{n}C_{r} = 84$ and ${}^{n}C_{r+1} = 126$, then find ${}^{r}C_{2}$.

If ⁿCr-1 = 36, ⁿCr = 84 and ⁿCr+1 = 126, then find rC₂.

NCERT Exemplar · Q15

{}^{n}C_{r-1} = 36

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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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Using ratios of consecutive binomial coefficients gives $n=9$ , $r=3$ , so ${}^{r}C_{2} = {}^{3}C_{2} = 3$ .

Use the ratio identity $\dfrac{{}^{n}C_{r}}{{}^{n}C_{r-1}} = \dfrac{n-r+1}{r}$ .

First ratio:

\frac{{}^{n}C_{r}}{{}^{n}C_{r-1}} = \frac{84}{36} = \frac{7}{3} \implies \frac{n-r+1}{r} = \frac{7}{3}.

So $3(n-r+1) = 7r$ , giving

3n + 3 = 10r. \qquad(1)

Second ratio:

\frac{{}^{n}C_{r+1}}{{}^{n}C_{r}} = \frac{126}{84} = \frac{3}{2} \implies \frac{n-r}{r+1} = \frac{3}{2}.

So $2(n-r) = 3(r+1)$ , giving

2n = 5r + 3. \qquad(2)

From (2), $n = \dfrac{5r+3}{2}$ . Substitute into (1): …

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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.If nCr−1=36{}^{n}C_{r-1} = 36nCr−1​=36, nCr=84{}^{n}C_{r} = 84nCr​=84 and nCr+1=126{}^{n}C_{r+1} = 126nCr+1​=126, then find rC2{}^{r}C_{2}rC2​.

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Q.If ${}^{n}C_{r-1} = 36$ , ${}^{n}C_{r} = 84$ and ${}^{n}C_{r+1} = 126$ , then find ${}^{r}C_{2}$ .

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