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Subjects
Class 11MathematicsPermutations and CombinationsFundamental principle of countingHard
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An exam code has (3) letters followed by (2) digits. Letters may repeat but digits cannot repeat. If the code must contain at least one vowel among the letters, how many codes are possible?

Answer and explanation

Correct answer: (421200)

Total letter arrangements are (26^3) and no-vowel arrangements are (21^3). Multiply the difference by (10 \times 9) digit choices.

Related tags

PermutationsCombinationsCountingHardFundamental Principle Of CountingPermutations And CombinationsMathematicsClass 11 Mcq
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