For how many three-digit whole numbers does the sum of the digits equal $25$?
$\text{(A)}\ 2 \qquad \text{(B)}\ 4 \qquad \text{(C)}\ 6 \qquad \text{(D)}\ 8 \qquad \text{(E)}\ 10$
Because $8+8+8=24$, it follows that one of the digits must be a $9$. The other two digits them have a sum of $25-9=16$. The groups of digits that produce a sum of $25$ are $799, 889$ and can be arranged as follows
\[799,979,997,889,898,988\]
The number of configurations is $\boxed{\text{(C)}\ 6}$.