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A meal package includes one drink, one snack, and one dessert. There are 5 drinks, 6 snacks, and 4 desserts. If coffee cannot be paired with chocolate dessert, how many packages are possible?

Answer and explanation

Correct answer: 114

Without the restriction, the multiplication principle gives 5×6×4 = 120 packages, because one drink, one snack, and one dessert are selected independently. The forbidden condition fixes both the drink as coffee and the dessert as chocolate. The snack can still be any of the 6 snacks, so exactly 6 packages are invalid. Subtracting them gives 120 − 6 = 114 valid packages. Therefore option A is correct. The value 120 ignores the incompatibility, while 116 and 118 subtract the wrong number of forbidden combinations. The restriction removes only the coffee–chocolate pair, not every package containing either item.

Related tags

Fundamental Principle Of CountingComplementary CountingRestricted PackagesFundamental Principle Of CountingPermutations And CombinationsMathematicsClass 11 Mcq

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