Let A denote the event that a 6-digit integer formed by 0,1,2,3,4,5,6 without repetitions,be divisib

Q: Wait, why do we have to split this into cases? Can't we just calculate the total permutations and divide?

That is a common trap! The constraint 'divisible by $3$' depends on the *set* of digits chosen. Because the set of digits changes based on which digit you exclude, the number of valid permutations for each set is different. Specifically, the presence of '$0$' changes the number of valid arrangements for the first digit. You cannot treat all sets as identical; you must partition the problem based on the excluded digit.

Q: Why does excluding '$0$' make the calculation easier?

When you exclude '$0$', the set of digits is $\{1, 2, 3, 4, 5, 6\}$. Since there is no '$0$' in this set, every position can be filled by any of the remaining digits without violating the 'no leading zero' rule. It is a simple permutation of $6$ distinct items, which is $6!$.

Q: How do I handle the 'first digit cannot be 0' constraint when 0 is in the set?

Think of it as a two-step process. For the first slot, you have $5$ choices (excluding '$0$'). Once that is filled, you have $5$ digits remaining (including '$0$', which is now allowed in the other positions). So, you have $5$ choices for the first slot and $5!$ ways to arrange the rest. That gives $5 \times 5!$.

Question

Let A denote the event that a 6-digit integer formed by 0,1,2,3,4,5,6 without repetitions,be divisible by 3. Then probability of event A is equal to :

Options:

(A) $9/50$

(B) $4/9$

(C) $3/7$

(D) $11/27$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

To determine the total number of ways to form a $6$-digit number using the set $\{0, 1, 2, 3, 4, 5, 6\}$, we must respect the constraint that the first digit cannot be $0$.

For the first slot, we have $6$ choices (digits $1$ through $6$). Once the first digit is placed, we have $6$ remaining digits (including $0$) to fill the remaining $5$ slots.

The number of ways to arrange these is given by $^6P_5$. Thus, the total sample space $n(S)$ is:

$$n(S) = 6 	imes 6! = 4320$$

### The Divisibility Insight

A number is divisible by $3$ if and only if the sum of its digits is a multiple of $3$. The sum of all available digits is $0 + 1 + 2 + 3 + 4 + 5 + 6 = 21$.

Since we are choosing $6$ digits out of $7$, we are effectively excluding exactly one digit, which we denote as $x$. The sum of the chosen digits is $21 - x$.

For the sum to be divisible by $3$, $x$ must be a multiple of $3$. Within our set, the possible values for $x$ are $0, 3,$ or $6$.

### Evaluating the Cases

We must analyze these three scenarios separately because the presence of $0$ affects the restriction on the first digit.

**Case 1: Exclude $0$**
The set of digits is $\{1, 2, 3, 4, 5, 6\}$. Since $0$ is not present, there are no restrictions on the first digit. The number of arrangements is:

$$6! = 720$$

**Case 2: Exclude $3$**
The set of digits is $\{0, 1, 2, 4, 5, 6\}$. Here, $0$ is present, so the first slot cannot be $0$. We have $5$ choices for the first slot and $5!$ ways to arrange the remaining digits:

$$5 	imes 5! = 600$$

**Case 3: Exclude $6$**
The set of digits is $\{0, 1, 2, 3, 4, 5\}$. The logic is identical to Case 2, as $0$ is present and the first slot cannot be $0$. This yields:

$$5 	imes 5! = 600$$

### Final Calculation

The total number of favorable outcomes $n(A)$ is the sum of the valid arrangements from all three cases:

$$n(A) = 720 + 600 + 600 = 1920$$

The probability $P(A)$ is the ratio of favorable outcomes to the total sample space:

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

By simplifying this fraction, we arrive at the final result:

**$$P(A) = \frac{4}{9}$$**

Animated Solution for Mathematics - Probability: Let A denote the event that a 6-digit integer formed by 0,1,2,3,4,5,6 without repetitions,be divisible by 3. Then probability of event A is equal to :

Select Answer:

Visualized Solution

Defining the Sample Space $n (S)$

Filling the First Place

Calculating Total Cases $n (S)$

Divisibility by $3$ Rule

Sum of All Available Digits

Identifying Digits to Exclude

Case 1: Excluding Digit $0$

Case 2: Excluding Digit $3$

Case 3: Excluding Digit $6$

Total Favorable Cases $n (A)$

Final Probability Calculation

The Sigma Insight: Classical Definition of Probability

Analyzing the Setup

The Divisibility Insight

Evaluating the Cases

Final Calculation

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