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Eight friends go to watch a movie but only 5 tickets were

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by Ian Stewart » Wed Mar 06, 2019 1:38 am
I don't like the wording of the question, because it's not altogether clear if order should matter. But that's the intention, so we have 8 choices for which person goes in the first seat, 7 for the next seat, and so on, and multiplying our choices gives the answer of (8)(7)(6)(5)(4).
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by Scott@TargetTestPrep » Fri Mar 08, 2019 6:50 am
BTGmoderatorLU wrote:Source: Veritas Prep

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*7*6*5*4
C. 5!
D. 8!/5!
E. 8*5

The OA is B
This is a permutation problem because the seating order is important. The number of ways for 8 friends to sit and watch a movie to which only 5 of them can obtain an admission ticket is 8P5 = 8 x 7 x 6 x 5 x 4.

Answer: B

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