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.
How many numbers between 450 and 700 can be formed using only digits 3,4,5,6,7, and 8?
0
$\begingroup$
combinatorics
1 Answers
1
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.