A five-digit number divisible by 3 is to be f | vedclass

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(A) number is divisible by $3$ if the sum of its digits is divisible by $3$. The sum of all given digits is $0+1+2+3+4+5 = 15$. Since we need a $5$-di

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$216$

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(A) number is divisible by $3$ if the sum of its digits is divisible by $3$. The sum of all given digits is $0+1+2+3+4+5 = 15$. Since we need a $5$-digit number,we must exclude one digit such that the sum of the remaining $5$ digits is divisible by $3$. Case $1$: Exclude $0$. The remaining digits are ${1, 2, 3, 4, 5}$. The sum is $15$,which is divisible by $3$. The number of $5$-digit numbers formed is $5! = 120$. Case $2$: Exclude $3$. The remaining digits are ${0, 1, 2, 4, 5}$. The sum is $12$,which is divisible by $3$. The number of $5$-digit numbers formed is $5! - 4! = 120 - 24 = 96$. Note: If we exclude any other digit (like $1, 2, 4, 5$),the sum of the remaining digits will not be divisible by $3$. Total ways $= 120 + 96 = 216$.

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

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