Key Concepts and Formulas

• Binomial Coefficient Definition: ${^nC_r} = \frac{n!}{r!(n-r)!}$
• Binomial Theorem Identity: $\sum_{r=0}^{n} {^nC_r} = 2^n$
• Factorial Manipulation: Understanding how to simplify and cancel terms in factorial expressions.

Step-by-Step Solution

Step 1: Expressing Binomial Coefficients in terms of Factorials

We begin by expressing the binomial coefficients in the given summation using their factorial definitions. This allows us to manipulate and simplify the expression algebraically. Recall that ${^nC_r} = \frac{n!}{r!(n-r)!}$.

The given sum is: $S = \sum\limits_{r = 0}^{25} {\left\{ {{}^{50}{C_r}.{}^{50 - r}{C_{25 - r}}} \right\}}$

Expressing each binomial coefficient in terms of factorials: ${^{50}{C_r}} = \frac{50!}{r!(50-r)!}$ ${^{50 - r}{C_{25 - r}}} = \frac{(50-r)!}{(25-r)!((50-r) - (25-r))!} = \frac{(50-r)!}{(25-r)!(25)!}$

Substituting these back into the summation, we get: $S = \sum\limits_{r = 0}^{25} {\left\{ {\frac{50!}{r!(50-r)!} \cdot \frac{(50-r)!}{(25-r)!(25)!}} \right\}}$

Step 2: Simplifying the Product of Factorials

Now, we simplify the expression inside the summation by canceling the common term $(50-r)!$ in the numerator and denominator. This crucial simplification makes the expression more manageable.

$S = \sum\limits_{r = 0}^{25} {\frac{50!}{r!\cancel{(50-r)!}} \cdot \frac{\cancel{(50-r)!}}{(25-r)!(25)!}}$ After cancellation, we have: $S = \sum\limits_{r = 0}^{25} {\frac{50!}{r!(25-r)!(25)!}}$

Step 3: Rearranging to Isolate ${^{50}C_{25}}$

Our objective is to express the sum in the form $K \left( {^{50}{C_{25}}} \right)$. To do this, we need to isolate and factor out the term ${^{50}{C_{25}}}$ from the summation. Recall that ${^{50}{C_{25}}} = \frac{50!}{25!25!}$.

We rewrite the general term inside the summation to introduce the ${^{50}{C_{25}}}$ term: $\frac{50!}{r!(25-r)!(25)!} = \frac{50!}{25!25!} \cdot \frac{25!}{r!(25-r)!}$ This allows us to express the sum as: $S = \sum\limits_{r = 0}^{25} {\left\{ {{}^{50}{C_{25}} \cdot \frac{25!}{r!(25-r)!}} \right\}}$ Since ${^{50}{C_{25}}}$ is independent of the summation index $r$, we can factor it out: $S = {^{50}{C_{25}}} \sum\limits_{r = 0}^{25} {\frac{25!}{r!(25-r)!}}$

Step 4: Recognizing Another Binomial Coefficient

We now focus on the expression inside the summation: $\frac{25!}{r!(25-r)!}$. This exactly matches the definition of the binomial coefficient ${^{25}{C_r}}$.

Therefore, we rewrite the sum as: $S = {^{50}{C_{25}}} \sum\limits_{r = 0}^{25} {{^{25}C_r}}$

Step 5: Applying the Binomial Theorem Identity

We have now arrived at a standard summation that can be evaluated using the binomial theorem identity. The identity states that the sum of all binomial coefficients for a given $n$ is $2^n$, i.e., $\sum\limits_{r = 0}^{n} {{^nC_r}} = 2^n$.

In our case, $n = 25$. Therefore, the summation term becomes: $\sum\limits_{r = 0}^{25} {{^{25}C_r}} = 2^{25}$

Substituting this back into our expression for $S$: $S = {^{50}{C_{25}}} \cdot 2^{25}$

Step 6: Determining K

The problem states that $S = K\left( {^{50}{C_{25}}} \right)$. Comparing this with our derived expression for $S$: $K\left( {^{50}{C_{25}}} \right) = {^{50}{C_{25}}} \cdot 2^{25}$ We can see that the value of $K$ is $2^{25}$.

Common Mistakes & Tips

• Careless Factorial Manipulation: Ensure accurate expansion and cancellation of factorial terms. Misinterpreting $(n-r)!$ can lead to errors.
• Forgetting the Binomial Theorem: Memorize and practice using the binomial theorem and related identities, especially $\sum_{r=0}^{n} {^nC_r} = 2^n$.
• Double-Check Index Limits: Verify that the summation index limits are consistent with the binomial coefficient definitions.

Summary

This problem demonstrates a systematic approach to simplifying summations involving binomial coefficients. The key steps involved expressing binomial coefficients in terms of factorials, simplifying the expression through cancellation, factoring out the desired term, recognizing standard binomial coefficient sums, and applying the binomial theorem identity. The value of $K$ is found to be $2^{25}$.

Final Answer

The final answer is $\boxed{2^{25}}$, which corresponds to option (C).