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Mathematics 19 Online
OpenStudy (anonymous):

Find C(100, 2)

OpenStudy (anonymous):

C(n,k) is the number of combinations of k elements you can make from a set of n elements. \[C(n,k) = \frac{n!}{k!(n-k)!}\]\[\implies C(100,2) = \frac{100!}{2!(100-2)!} = \frac{100!}{2!\ 98!} = \frac{100\cdot 99}{2}\]

OpenStudy (anonymous):

is that the answer?

OpenStudy (anonymous):

Well you can certainly simplify it further

OpenStudy (anonymous):

can u please give me the final answer plz

OpenStudy (anonymous):

Nah. I think I'm good. You can figure it out. Or if you can't, you can ask questions.

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