1. Forming a 6-digit number

We form a 6-digit integer using digits from $\{0,1,2,3,4,5,6\}$ without repetition.

Since there are 7 digits available and the number must have 6 digits, exactly one digit is omitted.

Also, the first digit cannot be $0$.

2. Total number of 6-digit integers

We count all 6-digit numbers formed from $\{0,1,2,3,4,5,6\}$ without repetition.

• Total permutations of 6 digits chosen from 7 digits: $^7P_6 = 7! = 5040$
• But this includes arrangements starting with $0$.

Count those starting with $0$:

• Fix first digit as $0$.
• Arrange remaining 5 places using any 5 of the remaining 6 digits: $^6P_5 = 6! = 720$

Hence total valid 6-digit integers: $5040 - 720 = 4320$

3. Condition for divisibility by 3

A number is divisible by $3$ if the sum of its digits is divisible by $3$.

Sum of all digits $0,1,2,3,4,5,6$ is $0+1+2+3+4+5+6 = 21$

If one digit $d$ is omitted, then the sum of digits of the 6-digit number is $21-d$

For divisibility by $3$: $21-d \equiv 0 \pmod 3$ Since $21 \equiv 0 \pmod 3$, this requires $d \equiv 0 \pmod 3$

Among $\{0,1,2,3,4,5,6\}$, the digits divisible by $3$ are: $0,3,6$

So favorable cases occur when the omitted digit is one of $0,3,6$.

4. Count favorable numbers casewise

Case 1: Omit $0$

Then digits used are $1,2,3,4,5,6$. All are nonzero, so every arrangement gives a valid 6-digit number.

Number of such numbers: $6! = 720$

Case 2: Omit $3$

Then digits used are $0,1,2,4,5,6$. Total arrangements: $6! = 720$ Invalid ones starting with $0$: $5! = 120$ So valid numbers: $720-120=600$

Case 3: Omit $6$

Then digits used are $0,1,2,3,4,5$. Similarly, valid numbers: $6! - 5! = 720-120=600$

Thus total favorable numbers: $720+600+600 = 1920$

5. Required probability

$P(A)=\frac{\text{favorable}}{\text{total}} = \frac{1920}{4320}$

Simplify: $\frac{1920}{4320} = \frac{4}{9}$

6. Check options

• A: $\frac{4}{9}$ ✔
• B: $\frac{9}{56}$ ✘
• C: $\frac{11}{27}$ ✘
• D: $\frac{3}{7}$ ✘

So the correct option is A.