Analyzing the 4-digit natural numbers:

a) How many of them have all different digits?

b) How many have the digit 1 exactly once and all different digits?

c) How many have the digit 1?
Solution

a) To choose the thousand unit, we have 9 possibilities, since we cannot use zero, as it would not be a four-digit number; to choose the hundred unit, we have 9 possibilities, as we have already used one of the digits; for the ten unit, there are 8 possibilities; and for the unit, there are 7 possibilities. Therefore, there are $9 \cdot 9 \cdot 8 \cdot 7=4,536$ possibilities in total.

b) Let's divide it into cases:

I) Fixing the 1 in the thousand unit, and not using it again, there are 9$\cdot$8$\cdot$7 = 504 possibilities.

II) Fixing the 1 now in the hundred, ten, or unit place, there are $8 \cdot 8 \cdot 7=448$ possibilities in each of these cases (remembering that in the thousand unit of this case, we cannot use 0 or 1), that is, there are $3 \cdot 448=$ 1,344 possibilities.

Therefore, the total number of possibilities is $504+1,344=1,848$.

c) The number of 4-digit numbers that do NOT contain the digit 1 is $8 \cdot 9 \cdot 9 \cdot 9=5,832$. Since there are 9,000 four-digit numbers, of these, $9,000-5,832=3,168$ contain the digit 1.