AMC12 2001 A

AMC12 2001 A · Q5

AMC12 2001 A · Q5. It mainly tests Exponents & radicals, Combinations.

What is the product of all positive odd integers less than $10000$?

所有小于 $10000$ 的正奇整数的乘积是多少？

(A) $\frac{10000!}{(5000)!^2}$ $\frac{10000!}{(5000)!^2}$

(B) $\frac{10000!}{25000}$ $\frac{10000!}{25000}$

(D) $\frac{10000!}{25000 \cdot 5000!}$ $\frac{10000!}{25000 \cdot 5000!}$

(E) $\frac{25000}{25000}$ $\frac{25000}{25000}$

Answer

Correct choice: (D)

正确答案：（D）

Solution

$1 \cdot 3 \cdot 5 \cdots 9999 = \dfrac{1 \cdot 2 \cdot 3 \cdot 4 \cdots 10000}{2 \cdot 4 \cdot 6 \cdots 10000}= \dfrac{10000!}{2^{5000} \cdot 1 \cdot 2 \cdot 3 \cdots 5000}= \dfrac{10000!}{2^{5000}\cdot5000!}$ Therefore the answer is $\boxed{\textbf{(D)} \dfrac{10000!}{2^{5000} \cdot 5000!}}$.

$1 \cdot 3 \cdot 5 \cdots 9999 = \dfrac{1 \cdot 2 \cdot 3 \cdot 4 \cdots 10000}{2 \cdot 4 \cdot 6 \cdots 10000}= \dfrac{10000!}{2^{5000} \cdot 1 \cdot 2 \cdot 3 \cdots 5000}= \dfrac{10000!}{2^{5000}\cdot5000!}$ 因此答案是 $\boxed{\textbf{(D)} \dfrac{10000!}{2^{5000} \cdot 5000!}}$。

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