kevinmurillo09
kevinmurillo09 kevinmurillo09
  • 23-07-2017
  • Mathematics
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A group of 8 friends plans to watch a movie, but they only have 5 tickets. How many different combinations of 5 friends could possibly recieve the tickets

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mreijepatmat
mreijepatmat mreijepatmat
  • 23-07-2017
Since the order doesn't matter, then this is a combination of 5 friends to be chosen among 8 movies:

⁸C₅ = (8!)/(8-5)!(5!) = 56 ways
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