$$2017~ AMC10B$$ P20  The number $$21!=51,090,942,171,709,440,000$$, has over $$60,000$$ positive integer divisors. One of them is chosen at random. What is the probability that it is odd?  数字21！=51,090,942,171,709,440,000，有超过$60,000$ 个的正整数因数。随机选择其中一个，它是奇数的概率是多少？
选项为:A:$$\frac{1}{21}$$  B:$$\frac{1}{19}$$  C:$$\frac{1}{18}$$  D:$$\frac{1}{2}$$  E:$$\frac{11}{21}$$  请回答其中一个选项.