A board game spinner is divided into three regions labeled $A$, $B$ and $C$. The probability of the arrow stopping on region $A$ is $\frac{1}{3}$ and on region $B$ is $\frac{1}{2}$. The probability of the arrow stopping on region $C$ is:

$\text{(A)}\ \frac{1}{12} \qquad \text{(B)}\ \frac{1}{6} \qquad \text{(C)}\ \frac{1}{5} \qquad \text{(D)}\ \frac{1}{3} \qquad \text{(E)}\ \frac{2}{5}$
1. The total probability of the arrow stopping on any of the three regions \(A\), \(B\), or \(C\) must sum to 1. This is because the spinner must stop on one of the regions.
   \[
   P(A) + P(B) + P(C) = 1
   \]

2. We are given the probabilities for regions \(A\) and \(B\):
   \[
   P(A) = \frac{1}{3}
   \]
   \[
   P(B) = \frac{1}{2}
   \]

3. Substitute the given probabilities into the total probability equation:
   \[
   \frac{1}{3} + \frac{1}{2} + P(C) = 1
   \]

4. To combine the fractions, find a common denominator. The least common multiple of 3 and 2 is 6:
   \[
   \frac{1}{3} = \frac{2}{6}
   \]
   \[
   \frac{1}{2} = \frac{3}{6}
   \]

5. Substitute these equivalent fractions back into the equation:
   \[
   \frac{2}{6} + \frac{3}{6} + P(C) = 1
   \]

6. Combine the fractions:
   \[
   \frac{2}{6} + \frac{3}{6} = \frac{5}{6}
   \]
   \[
   \frac{5}{6} + P(C) = 1
   \]

7. Solve for \(P(C)\) by subtracting \(\frac{5}{6}\) from 1:
   \[
   P(C) = 1 - \frac{5}{6}
   \]
   \[
   P(C) = \frac{6}{6} - \frac{5}{6}
   \]
   \[
   P(C) = \frac{1}{6}
   \]

8. Therefore, the probability of the arrow stopping on region \(C\) is:
   \[
   P(C) = \frac{1}{6}
   \]

The final answer is \(\boxed{\frac{1}{6}}\)