A five-digit number divisible by 3 has to be formed using the numerals 0, 1, 2, 3, 4 and 5 without repetition. The total number of ways in which this can be done is

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Calculate Digit Sum: To form a five-digit number divisible by 3, the sum of its digits must be divisible by 3. The available digits are 0, 1, 2, 3, 4, and 5. Let's find the sum of these digits. Sum of digits = 0 + 1 + 2 + 3 + 4 + 5 = 15Check Divisibility: Since the sum of all available digits is 15, which is divisible by 3, any combination of these digits without repetition will result in a number divisible by 3. Now, we need to count the number of five-digit numbers that can be formed.Count Choices: The first digit cannot be 0, as we need a five-digit number. So, there are 5 choices for the first digit (1, 2, 3, 4, or 5).Calculate Total Ways: After choosing the first digit, we have 4 remaining choices for the second digit (since we cannot repeat digits and we cannot use the digit already used for the first digit).For the third digit, we have 3 remaining choices, for the fourth digit, we have 2 remaining choices, and for the fifth digit, we have 1 remaining choice.To find the total number of ways to arrange these digits, we multiply the number of choices for each position. So, we have: 5 (choices for the first digit) * 4 (choices for the second digit) * 3 (choices for the third digit) * 2 (choices for the fourth digit) * 1 (choice for the fifth digit) = 5 * 4 * 3 * 2 * 1 = 120

A five-digit number divisible by $3$ has to be formed using the numerals $0$ , $1$ , $2$ , $3$ , $4$ and $5$ without repetition. The total number of ways in which this can be done is

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A five-digit number divisible by 333 has to be formed using the numerals 000, 111, 222, 333, 444 and 555 without repetition. The total number of ways in which this can be done is

A five-digit number divisible by $3$ has to be formed using the numerals $0$ , $1$ , $2$ , $3$ , $4$ and $5$ without repetition. The total number of ways in which this can be done is