Understanding the Problem

We need to find the probability that a 4-digit number, formed using the digits 0, 1, 2, 3, 4 without repetition, is divisible by 6.

A number is divisible by 6 if it's divisible by both 2 and 3.

• Divisibility by 2: The last digit must be even (0, 2, or 4).
• Divisibility by 3: The sum of the digits must be divisible by 3.

Calculating Total Possible 4-Digit Numbers

The digits available are {0, 1, 2, 3, 4}. A 4-digit number cannot start with 0.

• The first digit has 4 choices (1, 2, 3, 4).
• The remaining 3 digits can be chosen and arranged from the remaining 4 digits in \(P(4, 3)\) ways.
• \(P(4, 3) = \frac{4!}{(4-3)!} = 4 \times 3 \times 2 = 24\).
• Total possible 4-digit numbers = \(4 \times 24 = 96\).

Calculating Favorable Outcomes (Divisible by 6)

First, find sets of 4 digits from {0, 1, 2, 3, 4} whose sum is divisible by 3.

• Set 1: {0, 1, 2, 3}. Sum = 6 (divisible by 3).
• Set 2: {0, 2, 3, 4}. Sum = 9 (divisible by 3).

Now, count the 4-digit numbers formed using these sets that are also divisible by 2.

Case 1: Digits {0, 1, 2, 3}

The number must end in 0 or 2.

• Ending in 0: The first three digits are permutations of {1, 2, 3}. Number of ways = 3! = 6.
• Ending in 2: The last digit is 2. The remaining digits are {0, 1, 3}. The first digit cannot be 0.
  • Total permutations of {0, 1, 3} for the first three places = 3! = 6.
  • Permutations starting with 0 (e.g., 0132) are not 4-digit numbers. There are 2! = 2 such numbers (0132, 0312).
  • Valid numbers ending in 2 = 6 - 2 = 4.
• Total numbers divisible by 6 from this set = 6 + 4 = 10.

Case 2: Digits {0, 2, 3, 4}

The number must end in 0, 2, or 4.

• Ending in 0: The first three digits are permutations of {2, 3, 4}. Number of ways = 3! = 6.
• Ending in 2: The last digit is 2. The remaining digits are {0, 3, 4}. The first digit cannot be 0.
  • Total permutations of {0, 3, 4} = 3! = 6.
  • Permutations starting with 0 (e.g., 0342) = 2! = 2.
  • Valid numbers ending in 2 = 6 - 2 = 4.
• Ending in 4: The last digit is 4. The remaining digits are {0, 2, 3}. The first digit cannot be 0.
  • Total permutations of {0, 2, 3} = 3! = 6.
  • Permutations starting with 0 (e.g., 0234) = 2! = 2.
  • Valid numbers ending in 4 = 6 - 2 = 4.
• Total numbers divisible by 6 from this set = 6 + 4 + 4 = 14.

Total Favorable Outcomes

Total numbers divisible by 6 = Numbers from Case 1 + Numbers from Case 2 = 10 + 14 = 24.

Calculating the Probability

Probability = \(\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}\)

Probability = \(\frac{24}{96} = \frac{1}{4}\)