How many three-digit numbers have at least one $2$ and at least one $3$?
$\text{(A) }52 \qquad \text{(B) }54  \qquad \text{(C) }56 \qquad \text{(D) }58 \qquad \text{(E) }60$
We can do this problem with some simple case work.
Case 1: The hundreds place is not $2$ or $3.$
This means that the tens place and ones place must be $2$ and $3$ respectively or $3$ and $2$ respectively. This case covers $1, 4, 5, 6, 7, 8,$ and $9,$ so it gives us $2 \cdot 7 = 14$ cases.
Case 2: The hundreds place is $2.$
This means that $3$ must be in the tens place or ones place. Starting with cases in which the tens place is not $3$, we get $203, 213, 223, 243, 253, 263, 273, 283,$ and $293.$  With cases in which the tens place is $3$, we have $230-239$, or $10$ more cases. This gives us $9 + 10=19$ cases.
Case 3: The hundreds place is $3.$
This case is almost identical to the second case, just swap the $2$s with $3$s and $3$s with $2$s in the reasoning and its the same, giving us an additional $19$ cases.
Addition up all of these cases gives $14+19+19=52$ cases, or $\boxed{\textbf{(A) }52}.$