Key Concepts and Formulas

• Principle of Inclusion-Exclusion: For two sets A and B, $|A \cup B| = |A| + |B| - |A \cap B|$.
• Divisibility Rule for 3: A number is divisible by 3 if the sum of its digits is divisible by 3.
• Divisibility Rule for 5: A number is divisible by 5 if its last digit is 0 or 5.
• Permutations: The number of ways to arrange $n$ distinct objects in $r$ places is $P(n, r) = \frac{n!}{(n-r)!}$. For arranging $n$ distinct objects in $n$ places, it's $n!$.

Step-by-Step Solution

Step 1: Calculate the Number of Three-Digit Numbers Divisible by 3 (Set A)

We need to find the number of three-digit numbers formed using digits {1, 2, 3, 4, 5} without repetition, such that the number is divisible by 3. A number is divisible by 3 if the sum of its digits is divisible by 3.

1. Identify sets of three distinct digits whose sum is divisible by 3: We find all combinations of three distinct digits from {1, 2, 3, 4, 5} whose sum is divisible by 3.

   • {1, 2, 3}: Sum = $1 + 2 + 3 = 6$. (Divisible by 3)
   • {1, 2, 4}: Sum = $1 + 2 + 4 = 7$. (Not divisible by 3)
   • {1, 2, 5}: Sum = $1 + 2 + 5 = 8$. (Not divisible by 3)
   • {1, 3, 4}: Sum = $1 + 3 + 4 = 8$. (Not divisible by 3)
   • {1, 3, 5}: Sum = $1 + 3 + 5 = 9$. (Divisible by 3)
   • {1, 4, 5}: Sum = $1 + 4 + 5 = 10$. (Not divisible by 3)
   • {2, 3, 4}: Sum = $2 + 3 + 4 = 9$. (Divisible by 3)
   • {2, 3, 5}: Sum = $2 + 3 + 5 = 10$. (Not divisible by 3)
   • {2, 4, 5}: Sum = $2 + 4 + 5 = 11$. (Not divisible by 3)
   • {3, 4, 5}: Sum = $3 + 4 + 5 = 12$. (Divisible by 3)

   The sets of three distinct digits whose sum is divisible by 3 are: {1, 2, 3}, {1, 3, 5}, {2, 3, 4}, {3, 4, 5}. There are 4 such sets.

2. Form three-digit numbers from these sets: For each set of three distinct digits, we can arrange them in $3!$ ways to form different three-digit numbers. $3! = 3 \times 2 \times 1 = 6$ ways.

   Since there are 4 such sets, the total number of three-digit numbers divisible by 3 is: $|A| = \text{(Number of sets)} \times \text{(Permutations per set)} = 4 \times 6 = 24$

Step 2: Calculate the Number of Three-Digit Numbers Divisible by 5 (Set B)

To form a three-digit number divisible by 5 using digits {1, 2, 3, 4, 5} without repetition, the last digit must be 5.

1. Fix the last digit: The last digit (units place) must be 5. There is only 1 choice for this position. $\_ \_ \mathbf{5}$

2. Choose and arrange the first two digits: We have used one digit (5). The remaining digits are {1, 2, 3, 4}. We need to choose 2 digits from these 4 and arrange them in the first two positions (hundreds and tens place). For the hundreds place, we have 4 choices (any digit from {1, 2, 3, 4}). For the tens place, we have 3 choices (any remaining digit).

   The number of ways to arrange the first two digits is $4 \times 3 = 12$. This is also equivalent to $P(4, 2) = \frac{4!}{(4-2)!} = \frac{4 \times 3 \times 2 \times 1}{2 \times 1} = 12$.

   So, the total number of three-digit numbers divisible by 5 is: $|B| = 12$

Step 3: Calculate the Number of Three-Digit Numbers Divisible by Both 3 and 5 (Set $A \cap B$)

These numbers must satisfy both conditions: the sum of their digits is divisible by 3, AND their last digit is 5.

1. Identify sets of three digits that include 5 and whose sum is divisible by 3: From the sets identified in Step 1 ({1, 2, 3}, {1, 3, 5}, {2, 3, 4}, {3, 4, 5}), we select those that contain the digit 5:

   • {1, 3, 5} (Sum = 9, contains 5)
   • {3, 4, 5} (Sum = 12, contains 5) There are 2 such sets.

2. Form three-digit numbers with 5 as the last digit from these sets:

   • For the set {1, 3, 5}: The last digit is fixed as 5. The remaining digits {1, 3} can be arranged in the first two positions in $2! = 2 \times 1 = 2$ ways. (Numbers: 135, 315)
   • For the set {3, 4, 5}: The last digit is fixed as 5. The remaining digits {3, 4} can be arranged in the first two positions in $2! = 2 \times 1 = 2$ ways. (Numbers: 345, 435)

   The total number of three-digit numbers divisible by both 3 and 5 is: $|A \cap B| = 2 + 2 = 4$

Step 4: Apply the Principle of Inclusion-Exclusion

Now, we use the formula for the union of two sets: $|A \cup B| = |A| + |B| - |A \cap B|$ Substitute the values calculated in the previous steps: $|A \cup B| = 24 + 12 - 4$ $|A \cup B| = 36 - 4$ $|A \cup B| = 32$

Common Mistakes & Tips

• Repetition Constraint: Always ensure that the repetition of digits is not allowed while forming the numbers.
• Inclusion-Exclusion Principle: Remember to subtract the intersection of the two sets to avoid overcounting.
• Divisibility Rules: Apply divisibility rules correctly. For divisibility by 5, the last digit must be 5 (since 0 is not available).

Summary

We used the Principle of Inclusion-Exclusion to find the number of three-digit numbers divisible by either 3 or 5, formed using the digits {1, 2, 3, 4, 5} without repetition. We calculated the number of integers divisible by 3, divisible by 5, and divisible by both 3 and 5. Then, we applied the formula $|A \cup B| = |A| + |B| - |A \cap B|$ to get the final answer.

The final answer is \boxed{32}.