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In how many different ways can robert order 3 objects from a set of 8 objects?

User Tomako
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Explanation:

If the order of the objects doesn't matter (ABC is considered the same as CBA), then the number of combinations is:

₈C₃ = 8! / (3! (8−3)!) = 56

If the order does matter (ABC is considered different than CBA), then the number of permutations is:

₈P₃ = 8! / (8−3)! = 336

User Ofnowhere
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