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P(6 2) how do you evaluate this? help please!!
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P(6 2)?
PERMUTATIONS Yes, P(6,2)
\[P=\frac{n!}{(n-k)!}\]
Using 6 and 2, \[P=\frac{6!}{(6-2)!}\]
Are you familiar with factorials?
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yes.
So what do you get?
6 factorial means \[6!=6 \times 5 \times 4 \times 3 \times 2 \times 1\]
i did the equation and i get 180, what am i doing wrong?
12 30 360 these are the answer choices.
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I'm not sure but \[\frac{6!}{(6-2)!}=\frac{6 \times 5 \times 4 \times 3 \times 2 \times 1}{4!}=\frac{6 \times 5 \times \cancel{4 \times 3 \times 2 \times 1}}{\cancel{4 \times 3 \times 2 \times 1}}=6 \times 5=30\]
Do you get it now?
yes, thank you! :D
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