## 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

by M7MBA » Fri Jan 28, 2022 8:10 am

00:00

A

B

C

D

E

## Global Stats

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) $$8\cdot 5$$

Source: Veritas Prep

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### Re: Eight friends go to watch a movie but only $$5$$ tickets were available. In how many different ways can $$5$$ people

by regor60 » Mon Jan 31, 2022 1:43 pm
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

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