Question

所有小于 $10{,}000$ 的正奇整数的乘积是多少？

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

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

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

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

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

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

从 $1$ 到 $10000$ 的所有整数的乘积是 $10000!$ ，所以所有奇数的乘积等于 $10000!$ 除以所有偶数的乘积。

偶数的乘积可分解为 $\begin{gathered} 2 \cdot 4 \cdots 10000 \\ = 2^{5000}(1 \cdot 2 \cdots 5000) \\ = 2^{5000} \cdot 5000!. \end{gathered}$

因此奇整数的乘积是 $\dfrac{10000!}{2^{5000} \cdot 5000!}.$

因此，正确答案是 D。

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.