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Using digits (0,1,2,3,4,5) without repetition, how many four-digit numbers divisible by (5) can be formed?

Answer and explanation

Correct answer: 108

For divisibility by 5, the units digit must be 0 or 5. If the units digit is 0, there are 5 choices (1 to 5) for the thousands digit, followed by \(4\times3\) arrangements for the two middle places: \(5\times4\times3=60\). If the units digit is 5, the thousands digit cannot be 0, so it has 4 choices (1 to 4), and the middle places can be filled in \(4\times3\) ways: \(4\times4\times3=48\). Hence, the total is \(60+48=108\). Option 96 is incorrect because the restriction on 0 in the thousands place is not the same in both ending-digit cases. Exam tip: When 0 is among the digits, handle the leading-digit restriction separately.

Related tags

Fundamental Counting PrinciplePermutationsFour Digit NumbersDivisibility By 5Leading ZeroFundamental Principle Of CountingPermutations And CombinationsMathematics

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