Three Digit Numbers Above $560$ Formed From $3,4,5,6,7$ Is there a straight forward way of calculating the number of three digit numbers greater than 560 that can be formed from the numbers $3,4,5,6$, and $7$. I found it to be $30$ but I did it in a round about way. Is there a better method? Repeated digits are not allowed.
 A: I think going through the cases is straightforward. That is, 
To have a three digit number greater than $560$ we need the first two digits to be at least $56$. 
If the number starts with $56$ then there are 3 choices for the final digit.
If the number starts with $57$ then there are 3 choices for the final digit.
If the number starts with $6$ then there are $4\times 3$ choices for the last two digits.
If the number starts with $7$ then there are $4\times 3$ choices for the last two digits.
This list of cases are all the possible cases, so the total number is $30$ as you say. 
A: By symmetry, the median of all three-digit numbers with non-repeated digits $3$ to $7$ is $555$. Since there are no such numbers between $555$ and $560$, you want exactly half of them. There's a total of $5\cdot4\cdot3$ of them, and half of that is $30$.
A: pick a number for the first must be a $(5, 6 ,7)$ then pick a second number, if $5$ is first our choices are $(6,7)$ this has probability $1/3$. if $6$ or $7$ is chosen then any number can come next so $3×(2/3)×4 ×3+3×(1÷3)×2×3= total$ 
