Key Concepts and Formulas

1. Probability of an Event A: The probability of an event $A$ is given by the ratio of the number of outcomes favorable to $A$ to the total number of possible outcomes in the sample space $S$. $P(A) = \frac{\text{Number of favorable outcomes for A}}{\text{Total number of possible outcomes}} = \frac{n(A)}{n(S)}$

2. Divisibility Rule for 3: A number is divisible by 3 if and only if the sum of its digits is divisible by 3. This rule is fundamental for identifying the favorable outcomes.

3. Permutations with Restrictions: The number of ways to arrange $n$ distinct objects is $n!$. When forming multi-digit numbers using a set of digits that includes '0', a critical restriction is that '0' cannot be the leading digit. This requires careful consideration using the multiplication principle, often by filling the first digit place separately.

Step-by-Step Solution

Let the given set of digits be $D = \{0, 1, 2, 3, 4, 5, 6\}$. We need to form 6-digit integers using these digits without repetition.

Step 1: Determine the total number of possible 6-digit integers ($n(S)$). We need to form a 6-digit number using 6 distinct digits chosen from the set $D$. The primary constraint for forming a 6-digit number is that the first digit (hundreds of thousands place) cannot be '0'.

• For the 1st digit (leftmost position): Since '0' cannot be the first digit, we have 6 choices available from the non-zero digits $\{1, 2, 3, 4, 5, 6\}$.
• For the 2nd digit: One digit has been used for the first place. Now, '0' is available again, along with the remaining 5 non-zero digits. So, we have 6 remaining choices.
• For the 3rd digit: Two digits have been used. We are left with 5 choices from the remaining digits.
• For the 4th digit: Three digits have been used. We are left with 4 choices.
• For the 5th digit: Four digits have been used. We are left with 3 choices.
• For the 6th digit: Five digits have been used. We are left with 2 choices.

By the multiplication principle, the total number of distinct 6-digit integers that can be formed is: $n(S) = 6 \times 6 \times 5 \times 4 \times 3 \times 2$ This can be written more compactly as: $n(S) = 6 \times (6 \times 5 \times 4 \times 3 \times 2 \times 1) = 6 \times 6!$ $n(S) = 6 \times 720$ $n(S) = 4320$

Step 2: Determine the number of favorable outcomes ($n(A)$) for Event A (divisible by 3). For a 6-digit integer to be divisible by 3, the sum of its digits must be divisible by 3. First, let's find the sum of all digits in the given set $D = \{0, 1, 2, 3, 4, 5, 6\}$: $S_{\text{total}} = 0 + 1 + 2 + 3 + 4 + 5 + 6 = 21$ We need to form a 6-digit number using 6 distinct digits chosen from this set of 7 digits. This means we must exclude exactly one digit from the original set $D$. Let the excluded digit be $x$. The sum of the 6 digits chosen will be $S_{\text{chosen}} = S_{\text{total}} - x = 21 - x$. For $S_{\text{chosen}}$ to be divisible by 3, and knowing that 21 is divisible by 3, it follows that $x$ must also be divisible by 3. The digits in set $D$ that are divisible by 3 are $\{0, 3, 6\}$. Therefore, to ensure the 6-digit number is divisible by 3, we must exclude either 0, 3, or 6 from the original set of 7 digits. This leads to three distinct cases for the set of digits used to form the number.

Case 1: The digit '0' is excluded.

• The set of 6 digits used is $D_1 = \{1, 2, 3, 4, 5, 6\}$.
• The sum of these digits is $1+2+3+4+5+6 = 21$, which is divisible by 3.
• Since '0' is not in this set, all 6 digits are non-zero. Any permutation of these 6 distinct digits will form a valid 6-digit number.
• Number of ways for Case 1 = $6! = 720$.

Case 2: The digit '3' is excluded.

• The set of 6 digits used is $D_2 = \{0, 1, 2, 4, 5, 6\}$.
• The sum of these digits is $0+1+2+4+5+6 = 18$, which is divisible by 3.
• We need to form a 6-digit number using these digits. Here, '0' is present in the set, so the first digit cannot be '0'.
  • For the 1st digit: We have 5 choices (cannot use '0', so pick from $\{1, 2, 4, 5, 6\}$).
  • For the remaining 5 digits: After selecting the first digit, we have 5 unused digits left (including '0'). These 5 digits can be arranged in the remaining 5 positions in $5!$ ways.
• Number of ways for Case 2 = $5 \times 5! = 5 \times 120 = 600$.

Case 3: The digit '6' is excluded.

• The set of 6 digits used is $D_3 = \{0, 1, 2, 3, 4, 5\}$.
• The sum of these digits is $0+1+2+3+4+5 = 15$, which is divisible by 3.
• Similar to Case 2, '0' is present in this set, so the first digit cannot be '0'.
  • For the 1st digit: We have 5 choices (cannot use '0', so pick from $\{1, 2, 3, 4, 5\}$).
  • For the remaining 5 digits: We have 5 unused digits left (including '0'). These 5 digits can be arranged in the remaining 5 positions in $5!$ ways.
• Number of ways for Case 3 = $5 \times 5! = 5 \times 120 = 600$.

The total number of favorable outcomes, $n(A)$, is the sum of outcomes from these three mutually exclusive cases: $n(A) = 720 + 600 + 600$ $n(A) = 1920$

Step 3: Calculate the probability of Event A ($P(A)$). Now we have both the total number of possible outcomes and the number of favorable outcomes. $P(A) = \frac{n(A)}{n(S)}$ Substitute the values we calculated: $P(A) = \frac{1920}{4320}$ To simplify the fraction, we divide both the numerator and the denominator by their common factors: $P(A) = \frac{192}{432}$ (Dividing by 10) $P(A) = \frac{96}{216}$ (Dividing by 2) $P(A) = \frac{48}{108}$ (Dividing by 2) $P(A) = \frac{24}{54}$ (Dividing by 2) $P(A) = \frac{12}{27}$ (Dividing by 2) $P(A) = \frac{4}{9}$ (Dividing by 3)

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Common Mistakes & Tips

• Leading Zero Restriction: Always remember that '0' cannot be the first digit of a multi-digit number. This is a common pitfall that affects both the total number of arrangements and specific favorable cases.
• Systematic Case Analysis: When conditions lead to multiple scenarios (e.g., excluding different digits), break them down into distinct cases. Ensure these cases are mutually exclusive and collectively exhaustive.
• Divisibility Rule Application: Correctly apply the divisibility rule for 3. A quick check of the sum of digits is crucial before proceeding with permutations.

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Summary To determine the probability, we first calculated the total number of distinct 6-digit integers that could be formed from the given digits, carefully accounting for the restriction of '0' not being the leading digit, yielding $n(S) = 4320$. Next, we identified the sets of 6 digits whose sum is divisible by 3, which involved excluding one digit from the original set such that the excluded digit itself was a multiple of 3 (0, 3, or 6). For each such valid set of 6 digits, we calculated the number of distinct 6-digit integers that could be formed, again applying the '0' restriction where applicable. Summing these favorable outcomes gave us $n(A) = 1920$. Finally, the probability was computed as the ratio $n(A)/n(S)$, which simplified to $\frac{4}{9}$.

The final answer is \boxed{\frac{4}{9}} which corresponds to option (A).