Example 1-17 5 girls and 7 boys are to form a group of 5 people, with the requirement that boy A and girl B cannot be in the group at the same time. How many schemes are there?
The number of combinations of choosing 5 out of 12 people is
$$
\binom{12}{5}=11 \times 9 \times 8=792,
$$

The number of schemes where both A and B participate is $\binom{10}{3}=\frac{10 \times 9 \times 8}{3 \times 2}=120$, so the required number of schemes is $792-120=672$.