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18 January, 00:49

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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  1. 18 January, 04:20
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    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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