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*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.

1 L A T E S T R E P L I E S (Newest First)

royhaas

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