2007 AMC 10A Problem 12

Count everything first, then take away the bad cases.

1. No rule yet. Tourist 1 picks a guide: $2$ choices. Tourist 2 picks: $2$ choices. And so on for all $6$ tourists. That gives
$$2 \times 2 \times 2 \times 2 \times 2 \times 2 = 64 \text{ ways}.$$
2. Bad cases. The rule fails only when one guide gets nobody. That happens in exactly $2$ ways: all six go with the first guide, or all six go with the second guide.
3. Take them away. $64 - 2 = 62$.

The answer is $\boxed{\textbf{(D)}\ 62}$.

"Each group gets at least one" is often easiest to handle by counting all the ways and removing the few that break the rule. Here the bad cases are just two, and they are easy to list. The guides are different people, so the two bad cases are different, which is why we subtract $2$ and not $1$.

Count by how many tourists go with the first guide. It can be $1, 2, 3, 4$ or $5$ (not $0$ and not $6$). The number of ways to choose that many tourists out of $6$ is
$$6 + 15 + 20 + 15 + 6 = 62.$$

Ignore the restriction first. Each of the six tourists has two choices of guide, and the choices are independent, so there are $2^6 = 64$ ways to assign tourists to guides.

Exactly two of those assignments violate the rule: all six choose the first guide, or all six choose the second. Remove them:
$$64 - 2 = 62.$$

The answer is $\boxed{\textbf{(D)}\ 62}$.