Key Concepts and Formulas

• Fundamental Principle of Counting (Multiplication Rule): If there are $m$ ways to do one thing, and $n$ ways to do another, then there are $m \times n$ ways to do both. This extends to multiple events.
• Permutations: The number of ways to arrange $r$ objects from a set of $n$ distinct objects is given by $P(n, r) = \frac{n!}{(n-r)!}$. When arranging all $n$ objects, the number of permutations is $n!$.

Step-by-Step Solution

Step 1: Analyze the problem and identify possible cases.

We are given the digits {3, 5, 6, 7, 8} and need to find the number of integers greater than 6000 that can be formed using these digits without repetition. Since we have 5 distinct digits, we can form 4-digit and 5-digit numbers. A 4-digit number will be greater than 6000 only if its first digit is 6, 7, or 8. All 5-digit numbers formed using these digits will be greater than 6000. Thus we have two cases:

• Case 1: 4-digit numbers greater than 6000
• Case 2: 5-digit numbers

Step 2: Calculate the number of 4-digit numbers greater than 6000 (Case 1).

A 4-digit number can be represented as $\_ \_ \_ \_$.

• Substep 2.1: Determine the number of choices for the first digit (thousands place). For the number to be greater than 6000, the first digit must be 6, 7, or 8. Therefore, there are 3 choices for the first digit.

• Substep 2.2: Determine the number of choices for the remaining digits (hundreds, tens, and units places). After choosing the first digit, we have 4 digits remaining. We need to arrange these 4 digits in the remaining 3 places. The number of ways to arrange 3 digits from a set of 4 is given by the permutation formula: $P(4, 3) = \frac{4!}{(4-3)!} = \frac{4!}{1!} = 4 \times 3 \times 2 = 24$.

• Substep 2.3: Apply the Fundamental Principle of Counting. The total number of 4-digit numbers greater than 6000 is the product of the number of choices for the first digit and the number of ways to arrange the remaining 3 digits: $3 \times 24 = 72$.

Step 3: Calculate the number of 5-digit numbers (Case 2).

A 5-digit number can be represented as $\_ \_ \_ \_ \_$.

• Substep 3.1: Determine the number of choices for each digit. Since any 5-digit number formed using the given digits will be greater than 6000, we need to find the number of ways to arrange all 5 distinct digits in 5 places. This is simply the number of permutations of 5 objects, which is $5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$.

Step 4: Calculate the total number of integers greater than 6000.

The total number of integers greater than 6000 is the sum of the number of 4-digit numbers greater than 6000 and the number of 5-digit numbers: $72 + 120 = 192$

Common Mistakes & Tips

• Carefully consider the constraints: The condition "greater than 6000" dictates which cases to consider (4-digit and 5-digit numbers).
• Avoid double-counting: Ensure that each possible integer is counted only once by correctly applying permutations and the Fundamental Principle of Counting.
• Remember the "no repetition" rule: Once a digit is used, it cannot be reused.

Summary

We divided the problem into two cases: forming 4-digit numbers greater than 6000 and forming 5-digit numbers. We calculated the number of possibilities for each case using permutations and the fundamental principle of counting. Summing the results from both cases gave us the total number of integers greater than 6000 that can be formed using the given digits without repetition.

The final answer is \boxed{192}, which corresponds to option (D).