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This is a really basic question, and I should have paid more attention in discrete math class. I have $3$ booleans, and each can be represented by "yes" or "no". So that's kind of like $6$ possible options. How many possible combinations can I have out of these $3$ booleans?

Doing it manually, I only come up with $8$ possibilites. Would this be a $\binom 6 3$ problem, because the result of that is $20$, so I'm not sure if that would be correct. Or would it be $3 \times 3$? Or what would be the general equation?

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    That is not like 6 possible options.2012-07-06
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    There are two ways to choose the first value. For each of these, there are two ways to choose the second. For each of the $2\cdot2=4$ ways to choose the first two values, there are two ways to choose the third. So there are $2\cdot2\cdot2=8$ ways to choose all three.2012-07-06
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    Ahhh I see..it was that basic all along, I was overthinking it..2012-07-06
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    Why is this different from 6 possible options?2012-07-06
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    @mohabitar: Because you are not choosing 1 thing (or 3 things) from a set of 6 things, you are choosing 1 thing each from 3 sets of 2 things. Based on your guess of $6\choose 3$, you would have allowed things like choosing both yes and no in the first boolean, choosing yes in the second, and not choosing anything for the third.2012-07-06

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