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fpsych
Junior Member

USA
3 Posts

Posted - 12/13/2011 :  15:21:07  Show Profile  Reply with Quote
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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royhaas
Moderator

USA
3059 Posts

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