Question 1180043: There are 24 students in an MA 206 class at FCC. How many ways are there to select 5 students for a group project?
Answer by math_tutor2020(3816)   (Show Source):
You can put this solution on YOUR website!
Let's say for a moment that order mattered (it doesn't but let's just go with it for now). If order mattered, then we'd have 24*23*22*21*20 = 5,100,480 different permutations. We start with 24 and count our way down one at a time until we fill up five slots.
So again there are 5,100,480 different ways to select a group of five people if order mattered. However, order does not matter. Any group is the same no matter how you order that particular group around. It's assumed that there are no special roles or seats for any of the projects. Think of it like a committee where no member outranks another.
Consider one group of five people. There are 5*4*3*2*1 = 120 different ways to arrange this group. For instance, we could have ABCDE contrast with ABCED. This figure of 120 is the factor we're overcounting by when we got that initial value of 5,100,480
So that means we need to divide 5,100,480 by 120 to get the proper number of combinations. Doing so gets us
(5,100,480)/120 = 42,504
There are 42,504 different combinations here. Order does not matter.
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We could use the nCr combination formula as an alternative method.
We use n = 24 and r = 5
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 The factorial terms "19!" cancel out.
 This showed up earlier in the previous section.
 We get the same answer as before.
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Answer: 42,504
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