Question 228329
If you were to randomly select 3 movies from a selection of 29 movies, how many different lists of 3 could you have?
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You were counting arrangements of three objects.
You want to count GROUPS of three objects.
29C3 = 29!/[(29-3)!*3!]
= (29*28*27)/(1*2*3) = 29*14*9 = 3654
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Cheers,
Stan H.