# Possible ways to choose from $n$ different numbers

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. – Jonas Meyer Jul 6 '12 at 16:59
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. – David Mitra Jul 6 '12 at 16:59
Ahhh I see..it was that basic all along, I was overthinking it.. – maq Jul 6 '12 at 17:01
Why is this different from 6 possible options? – maq Jul 6 '12 at 17:01
@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. – Jonas Meyer Jul 6 '12 at 17:04

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.