How many 4-digit even numbers greater than 2000 can 1, 2, 3, 4, 5 form in a high school math problem (counting principle)? Please give a detailed answer.

If it is a four-digit even number greater than 2000, the ten thousand digits cannot be 1, and the one digit must be 2 or 4. Discuss in two situations:

(1) If there is no 1 in the four numbers, there are two choices: 2 and 4. If the other three bits are randomly arranged, there are 3X2=6 kinds; So there are 2X6= 12 kinds of * *.

(2) If there is 1 in the four numbers, and if 1 can only be thousands or hundreds, there are two choices; There are two choices: 2 and 4. 2; The remaining two bits can be randomly arranged among the remaining three bits, so there are 3X2=6 kinds; So there are 2X2X6=24 kinds.

So the general arrangement method is: 12+24=36.

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