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Step-by-Step Solution
Step 1: Determine the total number of 6-digit integers (Sample Space)
We need to form 6-digit numbers using the digits {0, 1, 2, 3, 4, 5, 6} without repeating any digit. The first digit cannot be 0 (to ensure it is indeed a 6-digit number).
• Number of ways to choose the first digit (non-zero): 6
• Number of ways to arrange the remaining 6 digits in the next 5 positions: $6!$
Thus, the total number of 6-digit integers is:
$n(s) = 6 \times 6!$
Step 2: Understand the condition for divisibility by 3
A number is divisible by 3 if and only if the sum of its digits is divisible by 3. We will identify all possible ways to pick 6 digits from {0, 1, 2, 3, 4, 5, 6} so that their sum is a multiple of 3, and then count in how many ways they can form a valid 6-digit number.
Step 3: Enumerate all valid digit sets and count their permutations
We look for subsets of size 6 whose sum of digits is divisible by 3. Three such cases emerge:
Case I: Digits {1, 2, 3, 4, 5, 6}
Sum = 1 + 2 + 3 + 4 + 5 + 6 = 21, which is divisible by 3.
Number of ways to arrange these 6 digits into a 6-digit number = $6!$.
Case II: Digits {0, 1, 2, 4, 5, 6}
Sum = 0 + 1 + 2 + 4 + 5 + 6 = 18, which is divisible by 3.
However, the first digit cannot be 0, so we have:
• 5 choices for the first digit (from {1, 2, 4, 5, 6})
• $5!$ arrangements for the remaining 5 positions
Number of ways = $5 \times 5!$.
Case III: Digits {0, 1, 2, 3, 4, 5}
Sum = 0 + 1 + 2 + 3 + 4 + 5 = 15, which is divisible by 3.
Similarly, the first digit cannot be 0, so:
• 5 choices for the first digit (from {1, 2, 3, 4, 5})
• $5!$ arrangements for the remaining 5 positions
Number of ways = $5 \times 5!$.
Step 4: Calculate the total number of favorable cases
Summing up the favorable permutations:
$n(\text{favorable}) = 6! + \bigl(5 \times 5!\bigr) + \bigl(5 \times 5!\bigr)$
$= 6! + 2 \times \bigl(5 \times 5!\bigr)$
Substitute $6! = 720$ and $5! = 120$:
$n(\text{favorable}) = 720 + 2 \times (5 \times 120) = 720 + 2 \times 600 = 720 + 1200 = 1920.
Step 5: Compute the probability
The probability $P$ of the 6-digit number being divisible by 3 is:
$P = \dfrac{n(\text{favorable})}{n(s)}
= \dfrac{1920}{6 \times 6!}
= \dfrac{1920}{4320}
= \dfrac{4}{9}.$
Final Answer
$\displaystyle \frac{4}{9}$
