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(6) A store that sells movies contains 40 action movies, 20 comedy movies, 10 horror movies, and 30 romance movies. A. In how many ways can a person select, in order, 3 movies? [1] a B. In how many ways can a person select, in order, 3 movies such that the third movie is a romance movie? 12 a98 49210 303 C. In how many ways can a person select, without order. 2 comedy movies? 111 C(20,2) саф D. In how many ways can a person select, without order, 2 movies at least one of them is neither horror nor comedy? [3] XE -

User Sam Brinck
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Answer:

The solution to the student's questions involves understanding permutations and combinations in probability theory. Answers are derived through specific computations using these principles based on details provided about the store's movie collection.

Step-by-step explanation:

The questions are based on the principles of permutations and combinations in probability theory.

A. The number of ways to select 3 movies in order from the store's collection is equal to the permutation of 100 total movies taken 3 at a time, denoted as P(100,3). Computation gives us 970200.

B. For the scenario where the last movie has to be a romance movie, calculate the number of ways separately for the first two movies (any genre) and the last movie (romance). The final answer would be P(70,2) (for any genre excluding romance for the first two picks) multiplied by 30 (number of romance movies).

C. For selection without order of 2 comedy movies, it is a combination problem. It's represented as C(20,2) = 190.

D. For the scenario of selecting 2 movies with least one of them is neither horror nor comedy, first calculate total combinations of any 2 movies C(100,2), then subtract combinations of selecting from only comedy and horror movies. C(30,2).

User Mark Pearl
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