# How many numbers between 450 and 700 can be formed using only digits 3,4,5,6,7, and 8?

Can you show me a way to solve this? How exactly would I be able to solve this, and what is the most elegant solution? I would like to know if it is possible to be solved without broken up into many cases.

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You probably do need to do a little breaking into cases. The most straightforward approach, I think, is to note first that the first digit must be $4,5$, or $6$: it cannot be $7$, because you can’t use $0$.
Suppose that the first digit is $5$ or $6$: then each of the other two digits can be any of $3,4,5,6,7$, and $8$, so there are $2\cdot 6^2=72$ such numbers beginning with $5$ or $6$.
If the first digit is $4$, the second digit must be $5,6,7$, or $8$, but the last digit may be any of the six legal digits, so in this case we get $4\cdot 6=24$ numbers.
The total of the two cases is then $72+24=96$ numbers meeting the requirements.