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In a two-day trip, a different city must be chosen each day. There are 5 cities and 3 hotel categories each day. If the capital is chosen on day two, only 2 hotel categories are available for day two. How many plans are possible?

Answer and explanation

Correct answer: 168

Use cases according to the city selected on day two. If the capital is selected on day two, day one can be any of the other 4 cities, day-one hotel choice has 3 possibilities, and day-two hotel choice has only 2 possibilities. This gives 4 × 3 × 2 = 24 plans. If a non-capital city is selected on day two, there are 4 choices for it, day one can be any of the remaining 4 cities, and both days have 3 hotel choices. This gives 4 × 4 × 3 × 3 = 144 plans. Adding disjoint cases gives 24 + 144 = 168, so option C is correct.

Related tags

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

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