Switches and Number of orders

Hello,

So here is the question: A special-purpose computer has 2 switches, each of which can be set in 3 different positions and 1 switch that can be set in 2 positions. In how many ways can the computer's switches be set?

What i tried: is 2 * 2 * 2 * 2 (because in every position there can be 2 switches, so 2^3) and for the 1 switch that can be in 2 positions so multiply by another 2

Is this correct?

Thanks

Re: Switches and Number of orders

Hey Ahasueros.

If each switch is independent from the other (i.e. changing one switch doesn't impact the others in any way), then you multiply the combinations.

Since you have three positions for two switches the combinations for this are 3*3 = 9. Now you have another switch which has two choices which is independent from the others so multiplying gives 9*2 = 18.

I'm not quite sure why you did 2^3 if you know that each switch has three choices (i.e. the positions).

Intuitively another way to do think about this multiplication is to think of creating one "super switch" with all possibilities. So for a single 2-state 2-switch system you have {Off,Off}, {Off,On}, {On, Off], {On, On} so you could replace this set of two switches with one "super switch" with four possibilities.

Re: Switches and Number of orders

Oh the way i understood was for every positions we have two choices on or off, which leads to 2^3?

So what i understand now is that since there are 2 switches each can be placed in three different positions that's why it is 3x3 and not 2^3? and since the 1 switch can be placed in 2 different spots we multiply by 2