To write a three-digit number with the given digits, I need to choose one digit for the hundreds place, one for the tens place and one for the ones place. The order of the digits matters, so I can use the permutation formula to count the number of ways to do this. The permutation formula says that the number of ways to arrange $n$ objects in $r$ positions is $nPr = n! / (n-r)!$, where $n!$ is the factorial of $n$. In this case, I have $n = 4$ digits to choose from and $r = 3$ positions to fill, so the number of ways is $4P3 = 4! / (4-3)! = 4! / 1! = 4! = 4 \times 3 \times 2 \times 1 = 24$. Therefore, there are 24 distinct three-digit numbers that can be written with the digits $1$, $2$, $3$ and $4$ without repeating any digit. 
How many distinct three-digit numbers can be written with the digits $1$, $2$, $3$ and $4$ if no digit may be used more than once in a three-digit number?