If Σ[r=0 to 25] {⁵⁰Cᵣ · ⁵⁰⁻ʳC₂₅₋ᵣ} = K(⁵⁰C₂₅), then K is equal to : (A) 2²⁵ - 1 (B) (25)² (C) 2²⁵ (D

Q: Wait, why do we cancel $(50-r)!$ in the first place? Is that always allowed?

Yes, absolutely! In algebraic expressions involving factorials, if you have the same factorial term in both the numerator and the denominator, they represent the same product of integers. As long as the factorial is defined (which it is for non-negative integers), you can cancel them just like you would cancel $x$ in $\frac{x}{x}$. It is the most powerful way to simplify complex combinatorial products.

Q: Why can't we just use energy conservation... wait, wrong subject. Why can't we just use the identity $^{n}C_{r} \cdot ^{n-r}C_{k} = ^{n}C_{k} \cdot ^{n-k}C_{r}$?

You absolutely can! That is a brilliant observation. That identity is a fantastic shortcut for this exact type of problem. However, for many students, remembering that specific identity under exam pressure is difficult. The factorial expansion method we used is the 'first principles' approach—it is robust, reliable, and works even if you forget the shortcut.

Q: Why is the sum of $^{25}C_{r}$ from $r=0$ to $25$ equal to $2^{25}$?

That comes from the Binomial Theorem! Recall that $(1+x)^{n} = \sum_{r=0}^{n} ^{n}C_{r} x^{r}$. If you set $x=1$, the expression becomes $(1+1)^{n} = \sum_{r=0}^{n} ^{n}C_{r} (1)^{r}$, which simplifies to $2^{n} = \sum_{r=0}^{n} ^{n}C_{r}$. It is one of the most fundamental and useful identities in combinatorics.

Question

If $\sum_{r=0}^{25} \{^{50}C_r \cdot ^{50-r}C_{25-r}\} = K(^{50}C_{25})$, then K is equal to :

Options:

(A) $2^{25} - 1$

(B) $(25)^2$

(C) $2^{25}$

(D) $2^{24}$

Sigma Percentile · Accepted Answer

### The Unmasking of the Summation

Combinatorics problems often look like a dense forest of symbols, but beneath the surface, they are governed by elegant, simple symmetries. Today, we are tackling a problem that might seem intimidating at first glance:

$$\sum_{r=0}^{25} \{^{50}C_{r} \cdot ^{50-r}C_{25-r}\} = K(^{50}C_{25})$$

It looks like a mess of indices and combinations, but by the end of this, you will see the beauty in how these terms dance together to simplify into something remarkably clean.

### Phase 1: The Factorial Expansion

Our first step is to strip away the notation. We know that the combination formula is our best friend:

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

Let's apply this to the two terms inside our summation.

First, we have $^{50}C_{r}$, which expands to:

$$\frac{50!}{r!(50-r)!}$$

Next, we have $^{50-r}C_{25-r}$. Applying the same logic, the numerator is $(50-r)!$. The denominator is the product of the lower index factorial, $(25-r)!$, and the difference factorial, $(50-r - (25-r))!$.

Look closely at that difference: $50 - r - 25 + r = 25$. So, the second term simplifies to:

$$\frac{(50-r)!}{(25-r)!25!}$$

### Phase 2: The Great Cancellation

Now, let's bring these two pieces together inside the summation:

$$\sum_{r=0}^{25} \left( \frac{50!}{r!(50-r)!} \cdot \frac{(50-r)!}{(25-r)!25!} \right)$$

Do you see it? The $(50-r)!$ in the denominator of the first fraction and the $(50-r)!$ in the numerator of the second fraction are begging to be canceled. When we remove them, the expression becomes much lighter:

$$\sum_{r=0}^{25} \frac{50!}{r!(25-r)!25!}$$

This is the turning point. We have successfully removed the variable $r$ from the factorial terms that were causing the most trouble.

### Phase 3: The Art of Targeting

We are told this sum equals $K(^{50}C_{25})$. Let's look at our target, $^{50}C_{25}$, which is:

$$\frac{50!}{25!25!}$$

Our current expression has $\frac{50!}{25!}$ in it. We are missing a $25!$ in the denominator.

In mathematics, when you need something, you create it—as long as you balance the equation. We multiply and divide the inside of the summation by $25!$:

$$\sum_{r=0}^{25} \frac{50!}{25!25!} \cdot \frac{25!}{r!(25-r)!}$$

Since $\frac{50!}{25!25!}$ is just $^{50}C_{25}$ and it does not depend on $r$, we can pull it right out of the summation sign.

### Phase 4: The Final Victory

Now we are left with:

$$^{50}C_{25} \cdot \sum_{r=0}^{25} \frac{25!}{r!(25-r)!}$$

That internal term, $\frac{25!}{r!(25-r)!}$, is the definition of $^{25}C_{r}$. So the expression is now $^{50}C_{25} \cdot \sum_{r=0}^{25} ^{25}C_{r}$.

We know the classic identity for the sum of binomial coefficients:

$$\sum_{r=0}^{n} ^{n}C_{r} = 2^{n}$$

With $n=25$, our sum is simply $2^{25}$. Comparing this to $K(^{50}C_{25})$, we see that **$K = 2^{25}$**.

Animated Solution for Mathematics - Binomial Theorem: If $\sum_{r = 0}^{25} {^{50} C_{r} \cdot^{50 - r} C_{25 - r}} = K (^{50} C_{25})$ , then K is equal to :

Select Answer:

Visualized Solution

Analyze the Given Expression

Expand $^{50} C_{r}$ using Factorials

Expand $^{50 - r} C_{25 - r}$ using Factorials

Combine the Expanded Terms

Identify the Target Constant

Rearrange to Form $^{50} C_{25}$

Recognize the Internal Combination

Apply the Binomial Sum Property

Final Comparison and Result

The Sigma Insight: Properties of Binomial Coefficients

The Unmasking of the Summation

Phase 1

Phase 2

Phase 3

Phase 4

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Animated Solution for Mathematics - Binomial Theorem: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If ∑r=025​{50Cr​⋅50−rC25−r​}=K(50C25​), then K is equal to :

Select Answer:

Visualized Solution

Analyze the Given Expression

Expand .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 50Cr​ using Factorials

Expand .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 50−rC25−r​ using Factorials