## 7. Three-digit numbers

How many three-digit numbers are there where the digit in the hundreds place is greater than the digit in the tens place, and the digit in the tens place is greater than the digit in the units place?

Result: 120
## First solution.

Three different digits can be chosen in $10 \cdot 9 \cdot 8=720$ ways because the first digit can be chosen in 10 ways, the second, which must be different from the first, in 9 ways, and the third, which must be different from the first two, in 8 ways.

Notice that we have six different ways to arrange these three digits $a, b, c: a b c, a c b, b a c, b c a, c a b$, $c b a$. However, in exactly one of these six arrangements, the digits are in descending order.

Therefore, the total number of the desired three-digit numbers is $720: 6=120$.