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A three-digit lock code is formed and then one letter is added. Digits are distinct, the number is divisible by (5), and if the last digit is (0), the letter cannot be a vowel. How many codes can be formed?

Answer and explanation

Correct answer: (3176)

When the last digit is (0), choices are (9\times8\times21), and when it is (5), choices are (8\times8\times26). Start divisibility conditions from the last digit.

Related tags

Permutations-And-CombinationsFundamental-PrincipleCountingHardFundamental Principle Of CountingPermutations And CombinationsMathematicsClass 11 Mcq

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