(1) When all five greeting cards are sent wrong, and there are no pairs of mistakes, then the student 1 has four ways to make mistakes. For example, the student of 1 gets the No.2 card, but the No.2 student will neither get the card of 1 nor get the No.2 card. There are only three ways for the No.3 student to make mistakes. (2) When all five greeting cards are sent by mistake, and there are mistakes in pairs, then the other three cards are also wrong; When determining a group of students who are wrong with each other, there are only two situations in which the remaining three students are wrong with each other; Choose 2 out of 5 students, that is, 2 out of 5, * * * has the choice of 10; 2× 10=20 species; So a * * * has: 24+20=44 kinds.

Solution: According to the analysis, there are two situations:

4×3×2× 1+5×4÷2×2,

=24+20,

=44 (species);

A: There are 44 possible situations.

So the answer is: 44.

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