A five-digit number divisible by 3 has to be | 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 to form

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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 to form 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 ways to arrange these $5$ digits 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 that can be formed is $5! - 4! = 120 - 24 = 96$ (subtracting cases where $0$ is in the first position).Total ways = $120 + 96 = 216$.

$A$ 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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