nC_{r-1} = 36, nC_r = 84 and nC_{r+1} = 126, then r is: (a) 1 (b) 2 (c) 3 (d) None of these.

Q: Wait, why can't we just expand the factorials like $n! / (r!(n-r)!)$?

You absolutely could, but it would be a mathematical marathon! Expanding those factorials leads to complex algebraic expressions that are prone to calculation errors. The ratio property $\frac{^nC_r}{^nC_{r-1}} = \frac{n-r+1}{r}$ is a shortcut derived specifically to bypass that complexity. In JEE, efficiency is just as important as accuracy.

Q: How do I know which ratio to take? Could I have taken $\frac{^nC_{r+1}}{^nC_{r-1}}$?

You could, but it would be much harder. Taking the ratio of *consecutive* terms is the standard strategy because it results in the cleanest algebraic cancellation. If you skip a term, you lose the simplicity of the identity. Always look for the 'consecutive' pattern first!

Q: What if I get non-integer values for $n$ or $r$?

That is a great 'trap-catcher' question! In the context of binomial coefficients $^nC_r$, $n$ and $r$ must be non-negative integers, and $n \ge r$. If you ever calculate a non-integer, it is a clear signal to re-check your algebraic steps, specifically your cross-multiplication or simplification of the ratios.

Question

$^nC_{r-1} = 36, ^nC_r = 84$ and $^nC_{r+1} = 126$, then $r$ is:

Options:

(A) 1

(B) 2

(C) 3

(D) None of these.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineers! Today, we are diving into a problem that might look like a tedious calculation at first glance, but it is actually a beautiful exercise in pattern recognition.

We are given three consecutive binomial coefficients: $^nC_{r-1} = 36$, $^nC_r = 84$, and $^nC_{r+1} = 126$. Our mission is to find the value of $r$.

When you see a sequence like this, your first instinct might be to reach for the factorial definition of combinations. I want you to pause. If you start writing out the factorials and comparing them, you are walking into a trap of endless algebra. Instead, let us embrace the elegance of the ratio property.

### The Power of the Ratio

The secret weapon in our toolkit is the identity:

$$\frac{^nC_r}{^nC_{r-1}} = \frac{n-r+1}{r}$$

Think of this as a bridge that allows us to leap over the factorial wall. By taking the ratio of consecutive terms, the factorials cancel out, leaving us with a clean, linear relationship.

Let us apply this to our first pair: $^nC_r$ and $^nC_{r-1}$. We have:

$$\frac{^nC_r}{^nC_{r-1}} = \frac{84}{36} = \frac{7}{3}$$

Now, we equate this to our formula:

$$\frac{n-r+1}{r} = \frac{7}{3}$$

Cross-multiplying, we get $3(n-r+1) = 7r$, which simplifies to $3n - 3r + 3 = 7r$, or:

$$3n - 10r = -3 \quad 	ext{(Equation 1)}$$

### The Second Leap

Now, we repeat the process for the next pair: $^nC_{r+1}$ and $^nC_r$. The ratio is:

$$\frac{^nC_{r+1}}{^nC_r} = \frac{126}{84} = \frac{3}{2}$$

Applying our ratio formula—but being careful to replace $r$ with $r+1$—we get:

$$\frac{n-(r+1)+1}{r+1} = \frac{n-r}{r+1}$$

Equating this to $\frac{3}{2}$, we cross-multiply: $2(n-r) = 3(r+1)$, which simplifies to $2n - 2r = 3r + 3$, or:

$$2n - 5r = 3 \quad 	ext{(Equation 2)}$$

### The Final Victory

We have successfully transformed a daunting combinatorics problem into a simple system of two linear equations:
1) $3n - 10r = -3$
2) $2n - 5r = 3$

To solve this, let us use the elimination method. If we multiply Equation 2 by 2, we get $4n - 10r = 6$.

Now, subtract Equation 1 from this new equation:

$$(4n - 10r) - (3n - 10r) = 6 - (-3)$$

The $r$ terms vanish, leaving us with $n = 9$.

With $n$ in hand, finding $r$ is a breeze. Substitute $n=9$ into Equation 2:

$$2(9) - 5r = 3$$

$$18 - 5r = 3$$

$$5r = 15$$

**$r = 3$**

We have arrived at the solution! This problem teaches us that in JEE Advanced, the most complex-looking expressions often yield to the simplest properties if you know where to look. Keep this ratio property in your arsenal, and you will tackle any binomial problem with confidence.

Animated Solution for Mathematics - Permutations and Combinations: $^{n} C_{r - 1} = 36,^{n} C_{r} = 84$ and $^{n} C_{r + 1} = 126$ , then $r$ is:

Select Answer:

Visualized Solution

Given Combinations

The Ratio Property of $^{n} C_{r}$

Setting up the First Ratio

Forming Equation 1

Setting up the Second Ratio

Forming Equation 2

The System of Linear Equations

Solving for $n$

Final Calculation for $r$

Conclusion

The Sigma Insight: Combinations and Selection

Analyzing the Setup

The Power of the Ratio

The Second Leap

The Final Victory

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Animated Solution for Mathematics - Permutations and Combinations: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } nCr−1​=36,nCr​=84 and nCr+1​=126, then r is: