Step-by-step Solution:

Correct Answer: C (24)

To solve this problem, we need to find the number of 3-digit integers formed from the digits {1, 2, 3, 4, 5, 6} without repetition that are multiples of 6. This requires satisfying two conditions simultaneously.

Step 1: Understand the Divisibility Rules for 6

A number is a multiple of 6 if and only if it is divisible by both 2 and 3.

• Divisibility by 2: The number must be even, meaning its last digit must be an even number (2, 4, or 6).
• Divisibility by 3: The sum of its digits must be a multiple of 3.

We will first find all combinations of three distinct digits whose sum is a multiple of 3, and then from those combinations, we will count the number of even integers that can be formed.

Step 2: Find Combinations of Digits Divisible by 3

We need to find sets of three distinct digits from {1, 2, 3, 4, 5, 6} whose sum is a multiple of 3. The possible sums are 6, 9, 12, and 15.

• Sum = 6: {1, 2, 3}
• Sum = 9: {1, 2, 6}, {1, 3, 5}, {2, 3, 4}
• Sum = 12: {1, 5, 6}, {2, 4, 6}, {3, 4, 5}
• Sum = 15: {4, 5, 6}

This gives us 8 possible sets of digits to work with.

Step 3: Count the Even Numbers for Each Combination

Now, for each set, we count how many 3-digit even numbers can be formed. An even number must end in an even digit.

1. {1, 2, 3}: Even digits available: {2}. The last digit must be 2. The other two digits (1, 3) can be arranged in $2! = 2$ ways. (Numbers: 132, 312) → 2 integers
2. {1, 2, 6}: Even digits: {2, 6}. (2 choices for the last digit) $\times$ (2! ways to arrange the other two) = $2 \times 2 = 4$ integers. → 4 integers
3. {1, 3, 5}: Even digits: None. → 0 integers
4. {2, 3, 4}: Even digits: {2, 4}. $2 \times 2! = 4$ integers. → 4 integers
5. {2, 4, 6}: Even digits: {2, 4, 6}. $3 \times 2! = 6$ integers. → 6 integers
6. {1, 5, 6}: Even digits: {6}. $1 \times 2! = 2$ integers. → 2 integers
7. {3, 4, 5}: Even digits: {4}. $1 \times 2! = 2$ integers. → 2 integers
8. {4, 5, 6}: Even digits: {4, 6}. $2 \times 2! = 4$ integers. → 4 integers

Step 4: Calculate the Total Number of Integers

Finally, we sum the counts from all the valid combinations:

Total = $2 + 4 + 0 + 4 + 6 + 2 + 2 + 4 = 24$

There are 24 such 3-digit integers.