1. We need to evaluate

$$S=\sum_{r=0}^{25} {^{50}C_r\,^{50-r}C_{25-r}}$$

and compare it with

$$S=K\,^{50}C_{25}.$$

So our goal is to find $K$.

2. Use the standard combinatorial identity:

$${}^nC_r\,{}^{n-r}C_{m-r} = {}^nC_m\,{}^mC_r.$$

Let us verify it quickly:

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

Also,

$${}^nC_m\,{}^mC_r =\frac{n!}{m!(n-m)!}\cdot \frac{m!}{r!(m-r)!} =\frac{n!}{r!(m-r)!(n-m)!}.$$

Hence they are equal.

3. Apply this identity with $n=50$ and $m=25$:

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

Therefore,

$$S=\sum_{r=0}^{25} \left({}^{50}C_{25}\,{}^{25}C_r\right) = {}^{50}C_{25}\sum_{r=0}^{25} {}^{25}C_r.$$

4. Now use the binomial identity:

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

So,

$$S = {}^{50}C_{25}\cdot 2^{25}.$$

Comparing with

$$S=K\,{}^{50}C_{25},$$

we get

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

5. Match with the options:

• A: $2^{24}$
• B: $2^{25}-1$
• C: $2^{25}$
• D: $(2^5)^2=2^{10}$ or possibly ambiguous, but not $2^{25}$

Hence the correct option is C.

So,

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