If nPr = nPr+1 and nCr = nCr-1, then the value of r is equal to: (1) 1 (2) 4 (3) 2 (4) 3

Q: Wait, why can we just cancel $n!$ from both sides? Is it possible that $n! = 0$?

That is a fantastic question! In the world of factorials, $n!$ is defined for non-negative integers. Since $n!$ is always a positive integer (specifically $n! \ge 1$ for $n \ge 0$), it can never be zero. Therefore, dividing both sides by $n!$ is perfectly safe and mathematically sound. You are never dividing by zero, so your logic remains rock solid!

Q: Why does $^nC_r = ^nC_{r-1}$ lead to $n+1 = 2r$? I thought the formula was more complex.

It looks complex because of the factorials, but the algebra simplifies beautifully. When you write out $\frac{n!}{r!(n-r)!} = \frac{n!}{(r-1)!(n-r+1)!}$, you are essentially comparing the 'growth' of the denominators. By canceling the common terms like $(r-1)!$ and $(n-r)!$, you are left with $\frac{1}{r} = \frac{1}{n-r+1}$. Cross-multiplying this gives $n-r+1 = r$, which rearranges to $n+1 = 2r$. It is just a matter of keeping your terms organized!

Q: Is there a geometric way to think about $^nC_r = ^nC_{r-1}$?

Yes! Think about Pascal's Triangle. The identity $^nC_r = ^nC_{n-r}$ is the symmetry property. While our problem uses $^nC_r = ^nC_{r-1}$, this is essentially asking for the point where two adjacent entries in a row of Pascal's Triangle are equal. It is a specific algebraic constraint that forces $n$ and $r$ into a very rigid relationship. You are essentially finding the 'peak' or the balance point of the row.

Question

If $^nP_r = ^nP_{r+1}$ and $^nC_r = ^nC_{r-1}$, then the value of $r$ is equal to:

Options:

(A) 1

(B) 4

(C) 2

(D) 3

Sigma Percentile · Accepted Answer

### The Symphony of Symmetry: Unlocking Permutations and Combinations

Welcome, my dear student. Today, we are not just solving a problem; we are embarking on a journey through the elegant architecture of counting.

Permutations and Combinations are the bedrock of probability, and they possess a hidden, rhythmic beauty. When you see equations like $^nP_r = ^nP_{r+1}$ and $^nC_r = ^nC_{r-1}$, do not let the notation intimidate you. These are not just symbols; they are stories of selection and arrangement.

Let us peel back the layers together.

### Phase 1: The Permutation Puzzle

Let us start with the first condition: $^nP_r = ^nP_{r+1}$. The permutation formula, $^nP_r = \frac{n!}{(n-r)!}$, tells us how many ways we can arrange $r$ items from a set of $n$.

When we equate $^nP_r$ and $^nP_{r+1}$, we are essentially saying that the number of ways to arrange $r$ items is the same as arranging $r+1$ items. Let us look at the math:

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

Immediately, we see the $n!$ on both sides. Since $n!$ is never zero, we can strike it out with confidence. We are left with:

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

Here is where the magic happens. We know that $(n-r)! = (n-r) 	imes (n-r-1)!$. By substituting this into our equation, we get:

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

By canceling the $(n-r-1)!$ term, we are left with the simple, elegant relation: $n-r = 1$, or **$n = r+1$**. This is our first anchor point.

### Phase 2: The Combination Conundrum

Now, we turn our attention to the second condition: $^nC_r = ^nC_{r-1}$. The formula is $^nC_r = \frac{n!}{r!(n-r)!}$.

When we set $^nC_r = ^nC_{r-1}$, we are looking for the point where the number of ways to choose $r$ items is equal to the number of ways to choose $r-1$ items. Let us write it out:

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

Again, the $n!$ vanishes. We use our 'ladder' logic again, knowing $r! = r 	imes (r-1)!$ and $(n-r+1)! = (n-r+1) 	imes (n-r)!$. Substituting these in:

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

Look at the symmetry! We can cancel $(r-1)!$ and $(n-r)!$ from both sides. What remains is a beautiful, clean fraction:

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

Cross-multiplying gives us $n-r+1 = r$, which simplifies to **$n+1 = 2r$**. We have our second anchor point!

### Phase 3: The Synthesis

We have arrived at the final act. We have two equations:
1. $n = r+1$
2. $n+1 = 2r$

This is a system of linear equations. We simply substitute the first into the second. Replacing $n$ with $r+1$ in the second equation gives us:

$$(r+1) + 1 = 2r$$

$$r + 2 = 2r$$

Subtracting $r$ from both sides, we find the treasure we have been seeking: **$r = 2$**.

### Conclusion: The Elegance of the Result

Isn't it satisfying? We started with intimidating factorial expressions, and through careful, step-by-step simplification, we arrived at a single integer.

The lesson here is not just about the value of $r$; it is about the process. When you face a daunting problem, break it down, use the properties of factorials to simplify the landscape, and always trust your algebra.

Keep practicing, keep questioning, and keep falling in love with the logic of the universe. You are doing great!

Animated Solution for Mathematics - Permutations and Combinations: If $^{n} P_{r} =^{n} P_{r + 1}$ and $^{n} C_{r} =^{n} C_{r - 1}$ , then the value of $r$ is equal to:

Select Answer:

Visualized Solution

Introduction to the Problem

The Permutation Formula

Applying the Formula to $^{n} P_{r} =^{n} P_{r + 1}$

Simplifying the Permutation Equation

Expanding the Larger Factorial

Deriving the First Relation

The Combination Formula

Applying the Formula to $^{n} C_{r} =^{n} C_{r - 1}$

Simplifying the Combination Equation

Deriving the Second Relation

Substitution and Final Calculation

Solving for $r$

Conclusion and Key Takeaways

The Sigma Insight: Combinations and Selection

The Symphony of Symmetry

Phase 1

Phase 2

Phase 3

Conclusion

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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; } If nPr​=nPr+1​ and nCr​=nCr−1​, then the value of r is equal to: