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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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    @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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Assume you have $n$ events that are each independent of each other, which I take to mean the outcome of any event will not affect the outcome of any other event. If there are $o_i$ outcomes for event $i$, then the total number of outcomes will be $(o_1)(o_2) \cdots (o_n)$.

In this case specifically, you have 3 events and each has 2 possible outcomes, so there are $2 \cdot 2 \cdot 2 = 8$ total possible outcomes.