Eight friends go to watch a movie but only \(5\) tickets were available. In how many different ways can \(5\) people sit

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Eight friends go to watch a movie but only \(5\) tickets were available. In how many different ways can \(5\) people sit and watch the movie?

A) \(8C5\)

B) \(8\cdot 7 \cdot 6\cdot 5\cdot 4\)

C) \(5!\)

D) \(\dfrac{8!}{5!}\)

E) \(8\cdot 5\)

Answer: B

Source: Veritas Prep

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If the question were asked "how many ways can the friends attend the movie ?", then this could be considered a combination problem, in which order doesn't matter, namely
8!/3!5! = 56

But because the question indicates that how they sit is important, this becomes a permutation problem in which the 5 people selected above are then permuted

8!/3!5! * 5! = 8!/3!=
B