【Example 3】Divide 10 people into two groups, Group A and Group B, with each group having at least 1 person. How many different ways are there to do this?
Since every person either belongs to Group A or Group B, one of the two must be true, which means each person has to make a choice between A and B. Therefore, this is a problem of repeated permutations, with a total of $2^{10}=1024$ different ways of making choices. However, considering that it includes the two extreme cases where all 10 people choose Group A or all 10 people choose Group B, in these cases, the other group becomes an empty group, which does not meet the requirements. Therefore, there are a total of $1024-2=$ 1022 different ways of grouping.