Let 0 ≤ r ≤ n. If n+1Cr+1 : nCr : n-1Cr-1 = 55 : 35 : 21, then 2n + 5r is equal to:
(1) 50
(2) 62
(3

Q: Wait, why can't I just expand the factorials? It seems more straightforward.

While expanding factorials is technically correct, it is a trap in JEE Advanced. You would end up with a massive, messy algebraic expression that is prone to calculation errors. The property $\frac{{}^{n}C_{r}}{{}^{n-1}C_{r-1}} = \frac{n}{r}$ is designed to bypass that complexity entirely. Always look for the elegant path first!

Q: How do I know which ratio to pick first?

You can pick any two consecutive terms from the given ratio. The key is to create two independent equations. By picking the first two terms and then the last two terms, you get two distinct equations that allow you to solve for the two unknowns, $n$ and $r$.

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

That is a huge red flag! Since $n$ and $r$ are parameters of combinations, they must be non-negative integers. If you find a fraction, re-check your cross-multiplication or your simplification of the ratios. It is a great way to self-verify your work.

Question

Let $0 \le r \le n$. If ${}^{n+1}C_{r+1} : {}^{n}C_r : {}^{n-1}C_{r-1} = 55 : 35 : 21$, then $2n + 5r$ is equal to:

Options:

(A) 50

(B) 62

(C) 55

(D) 60

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are staring at a problem that looks like a mountain of factorials. You are given the ratio:

$${}^{n+1}C_{r+1} : {}^{n}C_r : {}^{n-1}C_{r-1} = 55 : 35 : 21$$

Your first instinct might be to write out the factorial definitions, such as $\frac{(n+1)!}{(r+1)!(n-r)!}$. Stop! Take a breath.

In the world of JEE Advanced, the most complex-looking problems often have the most elegant, hidden shortcuts. We are not here to do brute-force arithmetic; we are here to uncover the geometric and algebraic beauty of the binomial coefficients.

### The Divide and Conquer Strategy

The secret to this problem lies in breaking the three-part ratio into two manageable, bite-sized pieces. We have three terms, so let us look at the first two:

$$\frac{{}^{n+1}C_{r+1}}{{}^{n}C_r} = \frac{55}{35}$$

Simplifying this fraction by dividing both numerator and denominator by $5$ gives us $\frac{11}{7}$. Now, we have a clean, simple ratio.

### The Magic Property

There is a powerful identity in combinatorics that acts as a skeleton key for these types of problems:

$$\frac{{}^{n}C_{r}}{{}^{n-1}C_{r-1}} = \frac{n}{r}$$

Notice the beauty of this relationship? The upper index $n$ and the lower index $r$ are both reduced by exactly $1$. When you see this pattern, you should immediately think of this property.

Applying this to our first ratio, $\frac{{}^{n+1}C_{r+1}}{{}^{n}C_r}$, we see that it perfectly matches the form $\frac{n+1}{r+1}$. So, our equation becomes:

$$\frac{n+1}{r+1} = \frac{11}{7}$$

Cross-multiplying gives us $7(n+1) = 11(r+1)$, which simplifies to $7n + 7 = 11r + 11$, or:

$$7n - 11r = 4$$

This is our first solid pillar of truth.

### The Second Pillar

We are not done yet. We still have the second part of the ratio:

$$\frac{{}^{n}C_{r}}{{}^{n-1}C_{r-1}} = \frac{35}{21}$$

Simplifying this fraction by dividing by $7$, we get $\frac{5}{3}$. Again, applying our magic property, this simplifies to:

$$\frac{n}{r} = \frac{5}{3}$$

This gives us a beautiful, direct relationship: $n = \frac{5r}{3}$.

### The Final Victory

Now, we have a system of two linear equations: $7n - 11r = 4$ and $n = \frac{5r}{3}$. Substituting the second into the first, we get:

$$7\left(\frac{5r}{3}\right) - 11r = 4$$

Multiplying through by $3$ to clear the denominator, we get $35r - 33r = 12$, which leads us to $2r = 12$, or **$r = 6$**.

With $r = 6$ in hand, finding $n$ is trivial:

$$n = \frac{5(6)}{3} = 10$$

Finally, the question asks for $2n + 5r$. Substituting our values, we get:

$$2(10) + 5(6) = 20 + 30 = 50$$

We have conquered the mountain, not by brute force, but by understanding the underlying structure of the binomial coefficients. The final answer is **$50$**.

Animated Solution for Mathematics - Permutations and Combinations: Let $0 \leq r \leq n$ . If $^{n + 1} C_{r + 1} :^{n} C_{r} :^{n - 1} C_{r - 1} = 55 : 35 : 21$ , then $2 n + 5 r$ is equal to:

Select Answer:

Visualized Solution

Given Ratio

Key Property of Combinations

Applying the Property

Deriving Equation 1

Analyzing the Second Ratio

Deriving Equation 2

Solving for $r$

Solving for $n$

Final Calculation

The Sigma Insight: Combinations and Selection

Analyzing the Setup

The Divide and Conquer Strategy

The Magic Property

The Second Pillar

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; } Let 0≤r≤n. If n+1Cr+1​:nCr​:n−1Cr−1​=55:35:21, then 2n+5r is equal to: