I have this question:
How many numbers greater than 40 000 can be formed using the digits 2, 3, 4, 5 and 6 if each digit is used only once in each number?
The first digit needs to either be 4, 5 or 6, so we can consider these bound together as 1 number for now, leaving us with three digits:
$$3! = 6$$
The first composite digit can be arranged in $3!$ ways also, giving us 36 overall. But the answer says 72. Why is this not correct?
There are three choices for the first digit. Once this choice has been made, there are four digits left. These can be arranged in $4!$ ways. Then $3\times 4!$ gives the correct answer.
Added Later: From what I can tell, you made two mistakes. First of all, after the choice of the first digit, you said there were three numbers left when in fact there are four. Secondly, you correctly identified that there are only three numbers which can be the first digit, but you calculated the number of ways you can arrange these three numbers. That's not relevant here as we are only concerned with which of the three numbers will be the first digit.