2001 AMC 12 Problem 5

The product of every integer from $1$ to $10000$ is $10000!,$ so the product of the odd ones is $10000!$ divided by the product of the even ones. The even numbers factor as $$\begin{gathered} 2 \cdot 4 \cdots 10000 \\ = 2^{5000}(1 \cdot 2 \cdots 5000) \\ = 2^{5000} \cdot 5000!. \end{gathered}$$ Therefore the product of the odd integers is $$\dfrac{10000!}{2^{5000} \cdot 5000!}.$$ Thus, the correct answer is D.