<s>[INST] If $K = \frac{2}{1}\times \frac{4}{3}\times \cdots \times \frac{100}{99}.$ Then value of $\lfloor K \rfloor$ 
Let $K = \frac{2}{1}\times \frac{4}{3}\times \frac{6}{5}\times \frac{8}{7}\times \cdots \times \frac{100}{99}.$ Then what is the value of $\lfloor K \rfloor$, where $\lfloor x \rfloor$ is the floor function?

My Attempt:
By factoring out powers of $2$, we can write
$$
\begin{align}
K &= 2^{50}\times \left(\frac{1}
{1}\times \frac{2}{3}\times \frac{3}{5}\times \frac{4}{7}\times \frac{5}{9}\times\cdots\times \frac{49}{97}\times \frac{50}{99}\right)\\
&= 2^{50}\cdot 2^{25}\times \left(\frac{1\cdot 3 \cdot 5\cdots49}{1\cdot 3 \cdot 5\cdots 49}\right)\times \left(\frac{1}{51\cdot 53\cdot 55\cdots99}\right)\\
&= \frac{2^{75}}{51\cdot 53\cdot 55\cdots99}
\end{align}
$$
How can I solve for $K$ from here?
 [/INST] I'm sure this answer is not rigorous, but perhaps someone can make it so.
By manipulating factorials we get
$$\frac{1}{K}=\frac{1}{2^{100}}\binom{100}{50}=P(X\,{=}\,50)\ ,$$
where $X$ is a binomial random variable with $n=100$ and $p=\frac{1}{2}$.  Approximating $X$ by a normal random variable $Y$ in the usual way, we have $Y\sim N(50,5^2)$ and so
$$\frac{1}{K}\approx P(49.5\,{<}\,Y\,{<}\,50.5)=P(-0.1\,{<}\,Z\,{<}\,0.1)
  \approx0.0796\ .$$
Hence
$$K\approx12.56$$
and so $\lfloor K\rfloor=12$.
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