Two tour guides are leading six tourists. The guides decide to split up. Each tourist must choose one of the guides, but with the stipulation that each guide must take at least one tourist. How many different groupings of guides and tourists are possible?

$ \textbf{(A)}\ 56 \qquad \textbf{(B)}\ 58 \qquad \textbf{(C)}\ 60 \qquad \textbf{(D)}\ 62 \qquad \textbf{(E)}\ 64$
1. Let the tour guides be \(a\) and \(b\). Each of the six tourists must choose one of the two guides, but each guide must have at least one tourist.
2. We need to count the number of ways to distribute the six tourists between the two guides such that neither guide is left without any tourists.
3. First, calculate the total number of ways to distribute the six tourists between the two guides without any restrictions. Each tourist has 2 choices (either guide \(a\) or guide \(b\)), so there are \(2^6\) total ways to distribute the tourists.
   \[
   2^6 = 64
   \]
4. Next, we need to subtract the cases where one of the guides has no tourists. There are two such cases: all six tourists go with guide \(a\) or all six tourists go with guide \(b\). Each of these cases occurs exactly once.
   \[
   2 \text{ cases}
   \]
5. Therefore, the number of valid groupings where each guide has at least one tourist is:
   \[
   64 - 2 = 62
   \]

The final answer is \(\boxed{62}\)