A five-digit number divisible by 3 is to be formedusing the digits 0, 1, 2, 3, 4, and 5, without repetition. The total numberof ways this can done isA. 216B. 600C. 240D. 3125

A five-digit number divisible by 3 is to be formedusing the digits 0,

1.	A five-digit number divisible by 3 is to be formedusing the digits 0, 1, 2, 3, 4, and 5, without repetition. The total numberof ways this can done isA. 216B. 600C. 240D. 3125
Answer» Correct Answer - A we know that, a number is divisible by 3, when sum of digits in the number must be divisible by 3. So, if we consider the digits 0, 1, 2, 4, 5, then ( 0 + 1 + 2 + 4 + 5) = 12) We see that, sum is divisible by 3. Therefore, five-digit numbers using the digit `0, 1, 2, 4, 5 = 4 xx 4 xx 3 xx 2 xx 1 = 96` `{:(4, 4, 3, 2, 1):}` and if we consider the digit 1, 2, 3, 4, 5, then `(1 + 2 + 3 + 4 + 5 = 15)` This sum is also divisible by 3. So, five - digit number can be formed using the digit 1, 2, 3, 4, 5 is `5"!"` ways. Total number of ways `= 96 + 5"!" = 96 + 120 = 216`

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A five-digit number divisible by 3 is to be formedusing the digits 0, 1, 2, 3, 4, and 5, without repetition. The total numberof ways this can done isA. 216B. 600C. 240D. 3125

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