To find the number of integers greater than 6,000 that can be formed using the digits 3, 5, 6, 7, and 8 without repetition, we need to consider the possible arrangements of these digits. Here’s a detailed solution:
First, identify that a number greater than 6,000 formed with the given digits must have 5 digits or more.
Let's consider only 5-digit numbers since using all these digits gives the maximum possible formations:
The first digit, which determines whether the number is greater than 6,000, must be either 6, 7, or 8.
If the first digit is 6, the remaining four digits (3, 5, 7, 8) can be arranged in 4! ways.
If the first digit is 7, the remaining four digits (3, 5, 6, 8) can be arranged in 4! ways.
If the first digit is 8, the remaining four digits (3, 5, 6, 7) can be arranged in 4! ways.
Calculate the total number of 5-digit numbers:
$$ 3 \times 4! = 3 \times 24 = 72 $$
Since we are also interested in numbers with fewer digits (but these cannot be formed because the minimum 4-digit number with these digits is 6,735), re-evaluate any other possibility.
Hence, the total number of integers that can be formed, greater than 6,000 and utilizing the digits 3, 5, 6, 7, 8 exactly once, is 72.
