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fpsych
Junior Member
USA
3 Posts |
 Posted - 12/13/2011 : 15:21:07
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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. |
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