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

We are required to simplify the expression $$\displaystyle \sum_{r=0}^{25}\;{^{50}C_r}\;{^{50-r}C_{25-r}}$$ and compare it with $$K\;{^{50}C_{25}}$$ in order to determine the constant $$K$$.

First, recall the following standard combinatorial identity, which we will state explicitly before using it:

For any non-negative integers $$n,\;k,\;r$$ with $$0\le r\le k\le n$$, we have $$^{n}C_{r}\;^{\,n-r}C_{k-r}=^{n}C_{k}\;^{k}C_{r}.$$

To see why this holds, write every factorial in the binomial coefficients:

$$^{n}C_{r}\;^{\,n-r}C_{k-r}=\frac{n!}{r!(n-r)!}\;\cdot\;\frac{(n-r)!}{(k-r)!\,(n-k)!} =\frac{n!}{r!\,(k-r)!\,(n-k)!}.$$

On the other hand, $$^{n}C_{k}\;^{k}C_{r}= \frac{n!}{k!(n-k)!}\;\cdot\;\frac{k!}{r!(k-r)!} =\frac{n!}{r!\,(k-r)!\,(n-k)!},$$ which is the same expression. Hence the identity is proved and can be used safely.

Now we apply this identity with the specific values $$n=50$$ and $$k=25$$. For each term of the given sum we therefore have

$$^{50}C_{r}\;^{\,50-r}C_{25-r}=^{50}C_{25}\;^{25}C_{r}.$$

Substituting this result into the entire summation, we obtain

$$\sum_{r=0}^{25}{^{50}C_{r}}\;{^{50-r}C_{25-r}} =\sum_{r=0}^{25}\Bigl(^{50}C_{25}\;^{25}C_{r}\Bigr).$$

The constant $$^{50}C_{25}$$ is independent of the index $$r$$ and can be pulled outside the summation sign:

$$\sum_{r=0}^{25}{^{50}C_{r}}\;{^{50-r}C_{25-r}} =^{50}C_{25}\;\sum_{r=0}^{25}{^{25}C_{r}}.$$

Next, recall another well-known result: the sum of all binomial coefficients of order $$n$$ is a power of two. Explicitly,

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

Here $$n=25$$, so

$$\sum_{r=0}^{25}{^{25}C_{r}} = 2^{25}.$$

Substituting this back into our expression gives

$$\sum_{r=0}^{25}{^{50}C_{r}}\;{^{50-r}C_{25-r}} =^{50}C_{25}\;\bigl(2^{25}\bigr).$$

By direct comparison with the form $$K\;^{50}C_{25}$$, we see that the multiplicative constant must be

$$K = 2^{25}.$$

Hence, the correct answer is Option A.