relatively simple probability question

A

aholmes

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Dec 12, 2014
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Hello there! This is my first time posting, so I hope this works out.
I have a question. It seems simple, but I can't figure it out for the life of me.
If I have 5 switches, and each switch has two positions (on or off), how many possible combinations are there?
Order matters, i.e. ON - OFF - OFF - OFF - OFF - OFF is different than OFF - ON - OFF - OFF - OFF - OFF even though they both have 4 OFF's and one ON.

I feel like this should be really simple...not sure if it's 5! = 120 (seems too small to me?)

Also, I would like to extend this to n switches (so 6 or 7 or 8 or 100 switches, instead of 5).

Any help would be really really greatly appreciated!! Thank you:)
 
aholmes said:
Hello there! This is my first time posting, so I hope this works out.
I have a question. It seems simple, but I can't figure it out for the life of me.
If I have 5 switches, and each switch has two positions (on or off), how many possible combinations are there?
Order matters, i.e. ON - OFF - OFF - OFF - OFF - OFF is different than OFF - ON - OFF - OFF - OFF - OFF even though they both have 4 OFF's and one ON.

I feel like this should be really simple...not sure if it's 5! = 120 (seems too small to me?)

Also, I would like to extend this to n switches (so 6 or 7 or 8 or 100 switches, instead of 5).

Any help would be really really greatly appreciated!! Thank you:)
Click to expand...

You have 2 choices for the first, two choices for the second, ..., 2 choices for the last. Multiply them together. It doesn't grow as fast as you apparently think.

Another way to think about it is, suppose you have n possibilities and you add another switch. You will have n choices if you choose on first and n choices if you choose off first, so you now have 2 n possibilities. Suppose you add another switch. You will have 2n choices ...
 
Ishuda said:
You have 2 choices for the first, two choices for the second, ..., 2 choices for the last. Multiply them together. It doesn't grow as fast as you apparently think.

Another way to think about it is, suppose you have n possibilities and you add another switch. You will have n choices if you choose on first and n choices if you choose off first, so you now have 2 n possibilities. Suppose you add another switch. You will have 2n choices ...
Click to expand...


So you're saying that it's 2(5) = 10 choices? ....But listing them out, I already found more than that:

here's 18 possible options:

12345
ononononon
ononononoff
onononoffoff
ononoffoffoff
onoffoffoffoff
offoffoffoffoff
offonononon
offoffononon
offoffoffonon
offoffoffoffon
offonoffonoff
onoffonoffon
onoffononon
ononoffonon
onononoffon
offonoffoffoff
offoffonoffoff
offoffoffonoff
 
aholmes said:
So you're saying that it's 2(5) = 10 choices? ....But listing them out, I already found more than that:

here's 18 possible options:

12345
ononononon
ononononoff
onononoffoff
ononoffoffoff
onoffoffoffoff
offoffoffoffoff
offonononon
offoffononon
offoffoffonon
offoffoffoffon
offonoffonoff
onoffonoffon
onoffononon
ononoffonon
onononoffon
offonoffoffoff
offoffonoffoff
offoffoffonoff
Click to expand...
Not 2 * 5 but 2 * 2 * 2 * 2 * 2 = 32. Let o be on and O be off, you start with 1 switch
o
O
Add a second swithc
o o
o O
O o
O O
add a third switch. Copy and paste the above twice, then add o to the beginning for the first 4 and and O to the beginning for the next 4
o o o
o o O
o O o
o O O
O o o
O o O
O O o
O O O
add a fourth switch. Copy and paste ...
 
aholmes said:
Hello there! This is my first time posting, so I hope this works out.
I have a question. It seems simple, but I can't figure it out for the life of me.
If I have 5 switches, and each switch has two positions (on or off), how many possible combinations are there?
Order matters, i.e. ON - OFF - OFF - OFF - OFF - OFF is different than OFF - ON - OFF - OFF - OFF - OFF even though they both have 4 OFF's and one ON.
I feel like this should be really simple...It is totally simple!
Click to expand...
\(\displaystyle 2^5=32\)
 
2^5

The first light switch has 2 possible positions (up and down).

For position A (up), the next light switch has 2 positions (A and B), and for position B (down), there are also 2; this makes 2*2 combinations.

We continue this for 5 switches, so 2*2*2*2*2(2^5). Since this account for EVERY possible combination, the answer is 32.

Dwell on it.
 
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