A $5$ 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

120

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96

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216

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220

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Solution

The correct option is B $216$
If a number is divisible by $3$ , then its sum of digits must be divisible by $3$ .
Thus selection of digit can be ${0, 1, 2, 4, 5}$ or ${1, 2, 3, 4, 5}$
Hence, required number of ways $= 4.4! + 5! = 96 + 120 = 216$

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A 5 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

The correct option is B 216If a number is divisible by 3, then its sum of digits must be divisible by 3.Thus selection of digit can be {0,1,2,4,5} or {1,2,3,4,5}Hence, required number of ways =4.4!+5!=96+120=216

A $5$ 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

The correct option is B $216$
If a number is divisible by $3$ , then its sum of digits must be divisible by $3$ .
Thus selection of digit can be ${0, 1, 2, 4, 5}$ or ${1, 2, 3, 4, 5}$
Hence, required number of ways $= 4.4! + 5! = 96 + 120 = 216$