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02-10-2011, 08:43 PM
You can answer any 8 questions from the 12 questions on an exam. In how many different ways can you choose the 8 questions, assuming that the order in which you choose the questions is irrelevant?

Denis
02-11-2011, 01:22 AM
Good question!

02-11-2011, 12:57 PM
Can anyone help me with this? Thanks

02-11-2011, 10:04 PM
This is the answer I came up with.

n C r = n!/(r!(n-r)!)

In this case, n = 12 and r = 8, so

n C r = 12 C 8 = (12!)/(8!(12-8)!) = (12*11*10*9*8!)/(8!*4!) = (12*11*10*9)/(4*3*2*1) = 11880/24 = 495

So there are 495 ways to answer 8 questions from the 12 given.

Mrspi
02-11-2011, 10:10 PM
This is the answer I came up with.

n C r = n!/(r!(n-r)!)

In this case, n = 12 and r = 8, so

n C r = 12 C 8 = (12!)/(8!(12-8)!) = (12*11*10*9*8!)/(8!*4!) = (12*11*10*9)/(4*3*2*1) = 11880/24 = 495

So there are 495 ways to answer 8 questions from the 12 given.

I agree! Since the order in which the 8 questions are chosen does not matter, you are looking at "how many ways can I choose a group of 8 questions out of 12 questions?" OR, how many combinations are there of 12 items chosen 8 at a time.

You've used the correct formula, and done the arithmetic correctly as well. Good job!