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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Q. 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

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. $\newline$ Sum of digits = $0 + 1 + 2 + 3 + 4 + 5 = 15$
Check 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 $(1, 2, 3, 4, \text{ or } 5)$ .
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).
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.
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: $\newline$ $5$ (choices for the first digit) $\times$ $4$ (choices for the second digit) $\times$ $3$ (choices for the third digit) $\times$ $2$ (choices for the fourth digit) $\times$ $1$ (choice for the fifth digit) $3$ $3$