If nCr denotes the number of combination of n things taken r at a time, then the expression nCr+1 +

Q: Why do we split $2 \times ^nC_r$ into two separate terms? Isn't that just making the expression longer?

That is a great observation! While it does make the expression look longer, it is a strategic move. We are essentially 'un-simplifying' the expression to reveal the structure required for Pascal's Identity. By splitting it, we create two distinct pairs that each perfectly match the form $^nC_k + ^nC_{k-1}$, allowing us to collapse the expression systematically.

Q: Can't we just use the factorial formula $^nC_r = \frac{n!}{r!(n-r)!}$ to solve this?

You absolutely can! Using the factorial definition is a valid, robust method. However, it involves significant algebraic manipulation, finding common denominators, and simplifying complex fractions, which increases the risk of calculation errors. Using Pascal's Identity is the 'JEE way'—it is faster, more elegant, and less prone to those pesky arithmetic mistakes.

Q: How do I know which terms to group together?

The secret lies in the lower indices. Pascal's Identity requires the lower indices to be consecutive (like $r$ and $r-1$). When you look at the expression $^nC_{r+1} + ^nC_{r-1} + ^nC_r + ^nC_r$, you see that $r$ is consecutive with both $r-1$ and $r+1$. Therefore, grouping $(^nC_{r-1} + ^nC_r)$ and $(^nC_r + ^nC_{r+1})$ is the most logical path to satisfy the identity.

Question

If $^nC_r$ denotes the number of combination of $n$ things taken $r$ at a time, then the expression $^nC_{r+1} + ^nC_{r-1} + 2 	imes ^nC_r$ equals

Options:

(A) ^{n+1}C_{r+1}

(B) ^{n+2}C_r

(C) ^{n+2}C_{r+1}

(D) ^{n+1}C_r

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler on the JEE journey. Today, we are going to demystify a problem that, at first glance, looks like a messy collection of binomial coefficients.

We are tasked with simplifying the expression:

$${}^{n}C_{r+1} + {}^{n}C_{r-1} + 2 	imes {}^{n}C_{r}$$

When you see an expression like this, it is easy to feel overwhelmed by the indices. But remember, in combinatorics, complexity is often just a mask for a beautiful, underlying symmetry.

### The Strategic Split

The first step in our journey is to look at the middle term, $2 	imes {}^{n}C_{r}$. In many JEE problems, the key to unlocking the solution is to manipulate the terms so they fit a known identity.

Here, we can split $2 	imes {}^{n}C_{r}$ into two identical parts: ${}^{n}C_{r} + {}^{n}C_{r}$. By rewriting the expression as:

$${}^{n}C_{r+1} + {}^{n}C_{r-1} + {}^{n}C_{r} + {}^{n}C_{r}$$

We have essentially created the building blocks we need to proceed.

### The Power of Pascal's Identity

Now, let us introduce our most powerful tool: **Pascal's Identity**, which states:

$${}^{n}C_{k} + {}^{n}C_{k-1} = {}^{n+1}C_{k}$$

This identity is the heartbeat of binomial coefficients. It tells us that if we add two consecutive terms from the same row of Pascal's triangle, we get a term in the next row.

With this in mind, let us group our terms:
*   Group 1: $({}^{n}C_{r-1} + {}^{n}C_{r})$
*   Group 2: $({}^{n}C_{r} + {}^{n}C_{r+1})$

### The Final Collapse

Applying Pascal's Identity to the first group, ${}^{n}C_{r-1} + {}^{n}C_{r}$, we get ${}^{n+1}C_{r}$. Applying it to the second group, ${}^{n}C_{r} + {}^{n}C_{r+1}$, we get ${}^{n+1}C_{r+1}$.

Our original, four-term expression has now collapsed into just two terms:

$${}^{n+1}C_{r} + {}^{n+1}C_{r+1}$$

Look at these two new terms; they also fit Pascal's Identity perfectly. The upper indices are both $n+1$, and the lower indices $r$ and $r+1$ are consecutive.

Applying the identity one last time, we arrive at the final result:

$${}^{n+2}C_{r+1}$$

Just like that, the entire expression simplifies to a single, elegant term: **${}^{n+2}C_{r+1}$**. This is the beauty of mathematics—taking a complex, sprawling problem and, through a few strategic steps, revealing its simple, core truth.

Animated Solution for Mathematics - Permutations and Combinations: If $^{n} C_{r}$ denotes the number of combination of $n$ things taken $r$ at a time, then the expression $^{n} C_{r + 1} +^{n} C_{r - 1} + 2 \times^{n} C_{r}$ equals

Select Answer:

Visualized Solution

The Given Expression

Splitting the Middle Term

Grouping the Terms

Pascal's Identity

Applying Identity to First Group

Applying Identity to Second Group

The Simplified Expression

Final Application of Identity

The Final Result

Conclusion

The Sigma Insight: Combinations and Selection

Analyzing the Setup

The Strategic Split

The Power of Pascal's Identity

The Final Collapse

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