Five digit number divisible by 3 is formed using 0, 1, 2, 3, 4 and 5 without repetition. Total numbe

Q: Wait, why do we only exclude 0 and 3? What about the other digits?

Great question! The sum of all six digits $\{0, 1, 2, 3, 4, 5\}$ is $15$. When we choose 5 digits, we are effectively removing one digit, $d$, from this sum. The sum of our 5-digit number becomes $15 - d$. For the number to be divisible by $3$, this sum must be a multiple of $3$. Since $15$ is already a multiple of $3$, $15 - d$ is divisible by $3$ if and only if $d$ is also a multiple of $3$. In our set, the only multiples of $3$ are $0$ and $3$. Thus, those are the only two digits we can exclude.

Q: Why can't we just use $5!$ for both cases? Why is the second case different?

That is a classic trap! In Case 1 (excluding $0$), all our available digits $\{1, 2, 3, 4, 5\}$ are non-zero, so any arrangement is a valid 5-digit number. However, in Case 2 (excluding $3$), our set is $\{0, 1, 2, 4, 5\}$. This set contains $0$. If we place $0$ in the first position (the ten-thousands place), the number becomes a 4-digit number, not a 5-digit number. Therefore, we must subtract the arrangements where $0$ is in the first position, or simply calculate the permutations by fixing the first digit to be non-zero.

Q: Is there a faster way to solve this without splitting into cases?

While you could potentially use more advanced combinatorial methods, splitting into cases based on the inclusion or exclusion of the digit $0$ is the most reliable and intuitive way to handle the 'zero constraint' in JEE problems. It prevents errors and ensures you account for every valid number systematically. Trust the process!

Question

Five digit number divisible by 3 is formed using 0, 1, 2, 3, 4 and 5 without repetition. Total number of such numbers are

Options:

(A) 312

(B) 3125

(C) 120

(D) 216

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The problem asks us to construct a 5-digit number using the set $\{0, 1, 2, 3, 4, 5\}$ such that the resulting number is divisible by $3$.

The fundamental rule for divisibility by $3$ states that a number is divisible by $3$ if and only if the sum of its digits is divisible by $3$.

The sum of all six available digits is:
$$S_{total} = 0 + 1 + 2 + 3 + 4 + 5 = 15$$

Since we are forming a 5-digit number, we must exclude exactly one digit $d$ from the set. The sum of the remaining five digits will be $S_5 = 15 - d$. For $S_5$ to be divisible by $3$, $d$ must be a multiple of $3$.

Within our set, the digits that are multiples of $3$ are $0$ and $3$. This leads us to two distinct, mutually exclusive cases.

### Case 1: Excluding the Digit $0$

If we exclude $0$, our set of digits becomes $\{1, 2, 3, 4, 5\}$. The sum of these digits is $15$, which is divisible by $3$.

Because the digit $0$ is not in this set, we do not need to worry about the leading digit constraint. We are arranging $5$ distinct digits in $5$ positions.

The number of ways to arrange these is given by $5!$:
$$5! = 5 	imes 4 	imes 3 	imes 2 	imes 1 = 120$$

Thus, there are **$120$** valid numbers in this case.

### Case 2: Excluding the Digit $3$

If we exclude $3$, our set of digits becomes $\{0, 1, 2, 4, 5\}$. The sum of these digits is $12$, which is divisible by $3$.

However, we must account for the leading digit constraint: a 5-digit number cannot begin with $0$.

For the first position (the ten-thousands place), we have $4$ choices: $\{1, 2, 4, 5\}$. Once the first digit is chosen, we have $4$ digits remaining (including $0$) to fill the remaining $4$ positions.

The number of valid arrangements is:
$$4 	imes 4! = 4 	imes 24 = 96$$

Thus, there are **$96$** valid numbers in this case.

### The Grand Synthesis

Since the two cases are mutually exclusive, we find the total count by summing the results from Case 1 and Case 2.

Total valid numbers = $120 + 96 = 216$.

By systematically applying the divisibility rule and carefully navigating the constraint of the leading zero, we conclude that the total number of such 5-digit numbers is **$216$**.

Animated Solution for Mathematics - Permutations and Combinations: Five digit number divisible by 3 is formed using 0, 1, 2, 3, 4 and 5 without repetition. Total number of such numbers are

Select Answer:

Visualized Solution

Analyzing the Problem

The Divisibility Rule of $3$

Calculating the Total Sum

Determining the Excluded Digit

Finding Multiples of $3$ in the Set

Case 1: Excluding the Digit $0$

Permutations for Case 1

Case 2: Excluding the Digit $3$

Handling the Zero Constraint

Permutations for Case 2

Combining Both Cases

Selecting the Correct Option

The Sigma Insight: Linear Permutations

Analyzing the Setup

Case 1

Case 2

The Grand Synthesis

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