Finally, I understand. The topic is: If there are 2 and 3 in three numbers, then 2 comes before 3. How many of these three numbers are there?

Yes, if, then, it's not 2 and 3.

Then consider whether there are 2 and 3.

1 has two digits and three digits.

therefore

There are 2 and 3 in the three digits, so there is still one digit missing. Choose one from the remaining numbers, that is, C3 1=3. Choose 1 from three digits.

There are two situations before 3, 2 is a hundred digits or 2 is a ten digits, and 3 is a digit.

The first situation: there are 100 places, and the remaining two are arranged. A22=2。

The second case: 2 is ten digits and 3 is one digit, so there is only one case: 1.

So C3 1(A22+ 1)=9.

Such a three-digit number has no 2 and 3.

(1) has two or three. Choose two C32 from the remaining three digits, and all three digits are arranged as A33. There are two or three situations, so multiply by two.

Is c (3 3,2 2) × a (3 3,3) × 2 = 36.

(2) No.2, No.3 and the remaining 3 digits are all arranged.

Is a (3 3,3) = 6.

9+36+6=5 1