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fpsych
Junior Member

USA
3 Posts

 Posted - 12/13/2011 :  15:21:07 Can't figure out if these are permutations or combinations.How many ways can an adviser choose 4 students from a class of 12 if they are all assigned the same task? How many ways can the students be chosen if they are each given a different task?Here's what I've done so far. I tried solving the first part as a combination, since the order didn't seem to matter. 12C4 (subscript for the numbers)12!_____(12-4)!4!12!_____8!4!12*11*10*9___________8*7*6*53*11*5*9_________2*7*3*51485______210The textbook answer for this part is 495.It's 11,880 for the second part.Which procedure is correct for this? If I could figure that out, I'd get the right answer, I think.Thanks.

royhaas
Moderator

USA
3057 Posts

 Posted - 12/14/2011 :  08:20:00 Your arithmetic is wrong, since 4!=24.
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