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! _____ (124)!4!
12! _____ 8!4!
12*11*10*9 ___________ 8*7*6*5
3*11*5*9 _________ 2*7*3*5
1485 ______ 210
The 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. 

