How many four-digit numbers can be formed with the digits 3, 5, 7, 8, 9 which are greater than 8000, if repetition of digits is not allowed?

Solution:

Numbers in excess of 8000 are required.

As a result, two numerals can be used in the thousand place: 8 or 9.

Assume four boxes; the first box can contain one of the two numbers 8 or 9, resulting in two possibilities: ²C₁

The numbers in the second box can be any of the four digits left, therefore ⁴C₁ is an option.

The numbers in the third box can be any of the three digits left, therefore ³C₁ is an option.

The numbers in the fourth box can be any of the two digits left, therefore ²C₁ is an option.

As a result, the total number of possible outcomes is ²C₁ × ⁴C₁ × ³C₁ × ²C₁ = 2 × 4 × 3 × 2 = 48.