Question

What is the product of all positive odd integers less than $10{,}000?$

$\dfrac{10000!}{(5000!)^2}$

$\dfrac{10000!}{2^{5000}}$

$\dfrac{9999!}{2^{5000}}$

$\dfrac{10000!}{2^{5000} \cdot 5000!}$

$\dfrac{5000!}{2^{5000}}$

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.