7. (10 points) The school organizes 482 people to go on an outing, renting 42-seat buses and 20-seat minibuses. If it is required that each person has one seat and each seat is occupied by one person, then there are $\qquad$ rental options.
【Analysis】Let there be $x$ 42-seat buses and $y$ 20-seat minibuses. According to the problem, we have:
$42 x + 20 y = 482$, we need to find the integer solutions of this equation, which will give us the answer.
【Solution】Let there be $x$ 42-seat buses and $y$ 20-seat minibuses. According to the problem, we have:
$42 x + 20 y = 482$, dividing both sides by 2 gives: $21 x + 10 y = 241$
The unit digit of $10 y$ is 0, so the unit digit of $21 x$ must be 1. Therefore, $x$ can only be 1 or 11. When $x = 1$, $y = 22$; when $x = 11$, $y = 1$.

Therefore, there are 2 rental plans.
Answer: There are 2 rental plans.