$$1000$$到$$9999$$之间有多少个整数四个数位上各不相同？  这个问题可以改为``\,''有多少四位正整数有四个不同的数，因为数字是$$1000$$到$$9999$$之间的四位整数，对于第一个数有$$9$$中选择，因为它不能是$$0$$，第二个数只有$$9$$种选择，因为它不同于第一个数，第三个数有$$8$$种选择，因为它不同于前两个数，第四个数有$$7$$种选择，因为它不同于前三个数，则$$9\times 9\times 8\times 7=4536$$．  故选$$\text{B}$$．  The question can be rephrased to \textquotesingle How many four-digit positive integers have four distinct digits?\textquotesingle, since numbers between $$1000$$ and $$9999$$ are four-digit integers. There are $$9$$ choices for the first number, since it cannot be $$0$$, there are only $$9$$ choices left for the second number since it must differ from the first, $$8$$ choices for the third number, since it must differ from the first two, and $$7$$ choices for the fourth number, since it must differ from all three. This means there are $$9\times9\times8\times7=\boxed{(\text{B})4536}$$ integers between $$1000$$ and $$9999$$ with four distinct digits. 
$$1000$$到$$9999$$之间有多少个整数四个数位上各不相同？．
Options:
a: $$3024$$ 
b: $$4536$$ 
c: $$5040$$ 
d: $$6480$$ 
e: $$6561$$ 