It has been quite some time since I've done permutations and combinations, and I'm attempting to remember the proper way to go about solving this issue (not a homework assignment, more of a thought exercise).

Say that I have 10 objects, each of which can have 1 of 6 possible operators applied, and furthermore each operator can be designated as ON or OFF.

To me, this would start out as a basic combination where n = 10 and r = 6. The binary aspect is where I get a bit fuzzy.

Would this become 2^nCr? So, 2^10C6? Or is this logic incorrect?

For example, there are 10 closed boxes. Each box can only have a single item inside, however that item is drawn from a set of 6. It is fine for multiple boxes to to have the same item. The binary aspect may be if the box is open or closed.

So, Box 1 has an orange, Box 2 has an apple, Box 3 has a grape, and so on. Box 4 also has an orange, however that box is closed (or OFF, whereas the rest are ON). Basically, each box will be assigned an item, and that item may be either visible or not visible. I need to find the total number of combinations of boxes to items to items that are visible.

Here is a secondary example:

Say I have 10 flashlights. Each flashlight can have a different color lightbulb, and the flashlight can be turned on or off. So, a green flashlight that is off is a different state than a green flashlight that is on, and similarly, a green flashlight that is off is different than a red flashlight in any state.

I'm trying to coerce my example into something non-software related. If I were to sketch out a small bit of pseudo-code, it'd look like this:

object[1].operation = OPERATION_1
object[1].operation_status = FALSE

object[2].operation = OPERATION_5
object[2].operation_status = TRUE

  • $\begingroup$ if I understood the problem, each of the 10 objects has 6 options for an operator and that operator could be on or off.. so isn't it $10*6*2$? $\endgroup$
    – benji
    Jan 14, 2015 at 4:04
  • $\begingroup$ Not entirely clear what you are asking but I would have said that there are $12$ possible states for each of $10$ objects, and so the total number of possibilities is $12^{10}$. $\endgroup$
    – David
    Jan 14, 2015 at 4:04
  • 2
    $\begingroup$ Suggestion: in a counting problem, the first thing you need to know is exactly what you are counting. If you could give two or three examples of possible outcomes, that would help to clarify the problem. $\endgroup$
    – David
    Jan 14, 2015 at 4:06
  • $\begingroup$ Sorry, I will add an example $\endgroup$
    – erik
    Jan 14, 2015 at 4:10
  • $\begingroup$ Frankly, it's still a little unclear. If you have one closed box with an apple inside then it counts as indistinguishable from a closed box with an orange inside? And if either of these aforementioned boxes (one with apple and one with orange inside) is open then they count as different? $\endgroup$
    – Fizz
    Jan 14, 2015 at 4:21

1 Answer 1


One object can have $12$ states-$6$ for the operator times $2$ for on/off. As you have $10$ objects and the state of each object is independent, the number of total states is $12^{10}$

  • $\begingroup$ Ah, that makes sense. $\endgroup$
    – erik
    Jan 14, 2015 at 4:47
  • $\begingroup$ @espais: It would have made sense a lot sooner if you didn't use open/closed boxes with apples/oranges inside as analogy, but something that would have made it clear that you considered them different states in all combinations, e.g. peeled/intact apples/oranges. $\endgroup$
    – Fizz
    Jan 14, 2015 at 4:56
  • $\begingroup$ Sorry, I was trying to come up with a very generic example. $\endgroup$
    – erik
    Jan 14, 2015 at 5:29

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