Statistics question: Out of 16 people, 12 are chosen to go home, how many such groups are possible.?

Question

I have several questions like this and just need someone to point me in the right direction...thanks!

Answer 1

You have a case of Combinations (not pemutations as the order chosen doesn't matter)

n!
---------- = n_Choose_k
k!(n - k)!

for you 16=n 12=k

The ! means factorial is the and for that you do
4! = 4*3*2*1
8! = 8*7*6*5*4*3*2*1

Answer 2

You start choosing people, you have 16 ways you can choose the first person. Once this person is chosen, there are only 15 people left, so there are only 15 ways to choose the second person. And so forth. But the key thing to remember is that those 15 possibilities for a second person exist FOR EACH OF the 16 possibilities for the first person. So for a two-person go-home group, there are 16*15 possible pairings. For a 12 person group, it would be:
16*15*14*13*12*11etc until you have 12 numbers.

Answer 3

That's a combination problem, as I recall. That's provided there are no repeats. My stats book is at home. I should remember, I just took the final a couple months ago. I can give you the answer around 8:00 central time, if no one else does. The equation should look like 12 C 16, but don't hold me to that just yet, I need to refresh my memory.