Several sets of prime numbers, such as $\{7,83,421,659\}$ use each of the nine nonzero digits exactly once. What is the smallest possible sum such a set of primes could have?
$\text{(A) }193 \qquad \text{(B) }207 \qquad \text{(C) }225 \qquad \text{(D) }252 \qquad \text{(E) }447$
Neither of the digits $4$, $6$, and $8$ can be a units digit of a prime. Therefore the sum of the set is at least $40 + 60 + 80 + 1 + 2 + 3 + 5 + 7 + 9 = 207$.
We can indeed create a set of primes with this sum, for example the following sets work: $\{ 41, 67, 89, 2, 3, 5 \}$ or $\{ 43, 61, 89, 2, 5, 7 \}$.
Thus the answer is $207\implies \boxed{\mathrm{(B)}}$.