AMC12 2007 A

We can find the number of increasing arithmetic sequences of length 3 possible from 0 to 9, and then find all the possible permutations of these sequences. This gives us a total of $2 + 4 + 6 + 8 = 20$ sequences. There are $3! = 6$ to permute these, for a total of $120$. However, we note that the conditions of the problem require three-digit numbers, and hence our numbers cannot start with zero. There are $2! \cdot 4 = 8$ numbers which start with zero, so our answer is $120 - 8 = 112 \Longrightarrow \mathrm{(C)}$.