This is choosing `4` from `5` (any `4` digit number chosen from `3, 4, 6, 8, 9` will be ` >1000`) plus `5` from `5` (any `5` digit number will be ` >1000`), where order is important.

So the number of ways we can arrange the given digits so that our resulting number is greater than `1000` such that no digit occurs more than once, is:

`P_4^5+P_5^5=(5!)/((5-4)!)+(5!)/((5-5)!)`

`=(5!)/(1!)+(5!)/(0!)`

`=240`