If a five-digit number divisible by 3 is to b | vedclass

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(D) number is divisible by $3$ if the sum of its digits is divisible by $3$. We need to choose $5$ digits out of ${0, 1, 2, 3, 4, 5}$ such that their

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$216$

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(D) number is divisible by $3$ if the sum of its digits is divisible by $3$. We need to choose $5$ digits out of ${0, 1, 2, 3, 4, 5}$ such that their sum is divisible by $3$. The sum of all digits is $0+1+2+3+4+5 = 15$. To get a sum of $5$ digits divisible by $3$,we must exclude a subset of digits whose sum is divisible by $3$. The possible subsets of size $1$ to exclude are ${0}$ (sum $0$) or ${3}$ (sum $3$).$(i)$ Excluding ${0}$: The digits are ${1, 2, 3, 4, 5}$. The number of $5$-digit numbers is $5! = 120$.(ii) Excluding ${3}$: The digits are ${0, 1, 2, 4, 5}$. The first digit cannot be $0$. The number of ways to arrange these is $4 	imes 4! = 4 	imes 24 = 96$.Total ways $= 120 + 96 = 216$.

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

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