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I was trying to explain this to my nine year old. I blame the wine but I ran into trouble.
Our open plan living room has 40 bulbs/ lamps
However the important bit is they are controlled by 8 switches. All are on or off apart one that switches a circuit of 5 independent lights.
My maths says there are 126 million lighting permutations.
Am I mad? |
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By (user no longer on site)
over a year ago
|
"I was trying to explain this to my nine year old. I blame the wine but I ran into trouble.
Our open plan living room has 40 bulbs/ lamps
However the important bit is they are controlled by 8 switches. All are on or off apart one that switches a circuit of 5 independent lights.
My maths says there are 126 million lighting permutations.
Am I mad?"
I've read this 3 times and I think the answer is Faraday Cage |
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"I was trying to explain this to my nine year old. I blame the wine but I ran into trouble.
Our open plan living room has 40 bulbs/ lamps
However the important bit is they are controlled by 8 switches. All are on or off apart one that switches a circuit of 5 independent lights.
My maths says there are 126 million lighting permutations.
Am I mad?"
Ok
It seems you have
2 to the power 9
2 to the 8 covers the 8 switches and most lights
With a doubling of permutations for the independent 5
2to the 8 is 256
Thus 512 combinations
The number of lamps is irrelevant as you suggest they go on and off as one unit
So for simply
You have 8 binary switches that control one lamp with 40 bulbs
And one switch that controls one lamp with 5 bulbs
If so 512 positions |
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"I was trying to explain this to my nine year old. I blame the wine but I ran into trouble.
Our open plan living room has 40 bulbs/ lamps
However the important bit is they are controlled by 8 switches. All are on or off apart one that switches a circuit of 5 independent lights.
My maths says there are 126 million lighting permutations.
Am I mad?"
Only if you’ve left them on for ever then get a big leckie bill .  |
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You can think of the configuration of 8 switches as an 8 digit binary word where 00000000 represents all switches off and 11111111 represents all switches on. All states can be represented. For instance, 01000000 would be just the second switch on.
The number of combinations is given by the maximum number that can be described in an 8 bit word, which is 2 to the power 8 or 256.
If you have another light and switch this is doubled to 512 combinations. |
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"I was trying to explain this to my nine year old. I blame the wine but I ran into trouble.
Our open plan living room has 40 bulbs/ lamps
However the important bit is they are controlled by 8 switches. All are on or off apart one that switches a circuit of 5 independent lights.
My maths says there are 126 million lighting permutations.
Am I mad?
Ok
It seems you have
2 to the power 9
2 to the 8 covers the 8 switches and most lights
With a doubling of permutations for the independent 5
2to the 8 is 256
Thus 512 combinations
The number of lamps is irrelevant as you suggest they go on and off as one unit
So for simply
You have 8 binary switches that control one lamp with 40 bulbs
And one switch that controls one lamp with 5 bulbs
If so 512 positions"
Thanks
Obvious really.
Although I think the answer is 2 to the power of 12 as one of the 8 switches turns on the five freestanding lamps and each has its own independent inline switch
So 4096 options |
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"I was trying to explain this to my nine year old. I blame the wine but I ran into trouble.
Our open plan living room has 40 bulbs/ lamps
However the important bit is they are controlled by 8 switches. All are on or off apart one that switches a circuit of 5 independent lights.
My maths says there are 126 million lighting permutations.
Am I mad?"
Yes for employing such a shite electrician  |
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By (user no longer on site)
over a year ago
|
"I was trying to explain this to my nine year old. I blame the wine but I ran into trouble.
Our open plan living room has 40 bulbs/ lamps
However the important bit is they are controlled by 8 switches. All are on or off apart one that switches a circuit of 5 independent lights.
My maths says there are 126 million lighting permutations.
Am I mad?
Ok
It seems you have
2 to the power 9
2 to the 8 covers the 8 switches and most lights
With a doubling of permutations for the independent 5
2to the 8 is 256
Thus 512 combinations
The number of lamps is irrelevant as you suggest they go on and off as one unit
So for simply
You have 8 binary switches that control one lamp with 40 bulbs
And one switch that controls one lamp with 5 bulbs
If so 512 positions
Thanks
Obvious really.
Although I think the answer is 2 to the power of 12 as one of the 8 switches turns on the five freestanding lamps and each has its own independent inline switch
So 4096 options "
Last one out turn the god damn lights out he's used enough electricity to power Blackpool lights |
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"I was trying to explain this to my nine year old. I blame the wine but I ran into trouble.
Our open plan living room has 40 bulbs/ lamps
However the important bit is they are controlled by 8 switches. All are on or off apart one that switches a circuit of 5 independent lights.
My maths says there are 126 million lighting permutations.
Am I mad?
Ok
It seems you have
2 to the power 9
2 to the 8 covers the 8 switches and most lights
With a doubling of permutations for the independent 5
2to the 8 is 256
Thus 512 combinations
The number of lamps is irrelevant as you suggest they go on and off as one unit
So for simply
You have 8 binary switches that control one lamp with 40 bulbs
And one switch that controls one lamp with 5 bulbs
If so 512 positions
Thanks
Obvious really.
Although I think the answer is 2 to the power of 12 as one of the 8 switches turns on the five freestanding lamps and each has its own independent inline switch
So 4096 options
Last one out turn the god damn lights out he's used enough electricity to power Blackpool lights "
All low energy LEDs
They are not all used at the same time. The whole point of the bank of switches is that lots of different moods can be created.
Just one switch is used normally.
The room is quite large 80m2 and 5m high in the centre. Think oak barn. |
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"I was trying to explain this to my nine year old. I blame the wine but I ran into trouble.
Our open plan living room has 40 bulbs/ lamps
However the important bit is they are controlled by 8 switches. All are on or off apart one that switches a circuit of 5 independent lights.
My maths says there are 126 million lighting permutations.
Am I mad?
Ok
It seems you have
2 to the power 9
2 to the 8 covers the 8 switches and most lights
With a doubling of permutations for the independent 5
2to the 8 is 256
Thus 512 combinations
The number of lamps is irrelevant as you suggest they go on and off as one unit
So for simply
You have 8 binary switches that control one lamp with 40 bulbs
And one switch that controls one lamp with 5 bulbs
If so 512 positions
Thanks
Obvious really.
Although I think the answer is 2 to the power of 12 as one of the 8 switches turns on the five freestanding lamps and each has its own independent inline switch
So 4096 options
Last one out turn the god damn lights out he's used enough electricity to power Blackpool lights
All low energy LEDs
They are not all used at the same time. The whole point of the bank of switches is that lots of different moods can be created.
Just one switch is used normally.
The room is quite large 80m2 and 5m high in the centre. Think oak barn."
Infact I've decided
The bulbs are pure red herring
Switch has two states
Regardless of bulb configuration it's always going to be
2 to the power of switches or less |
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"I was trying to explain this to my nine year old. I blame the wine but I ran into trouble.
Our open plan living room has 40 bulbs/ lamps
However the important bit is they are controlled by 8 switches. All are on or off apart one that switches a circuit of 5 independent lights.
My maths says there are 126 million lighting permutations.
Am I mad?
Ok
It seems you have
2 to the power 9
2 to the 8 covers the 8 switches and most lights
With a doubling of permutations for the independent 5
2to the 8 is 256
Thus 512 combinations
The number of lamps is irrelevant as you suggest they go on and off as one unit
So for simply
You have 8 binary switches that control one lamp with 40 bulbs
And one switch that controls one lamp with 5 bulbs
If so 512 positions
Thanks
Obvious really.
Although I think the answer is 2 to the power of 12 as one of the 8 switches turns on the five freestanding lamps and each has its own independent inline switch
So 4096 options "
Not being funny but that’s the sort of question my brother would pose as a riddle, and managing to omit the vital information- in this case that there’s another switch for each of the five lights, ie 14 switches in total....  |
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"Sounds like some confused thinking going on.
The possible combinations of the 40 lights is factorial 40.
"
Not with 11 14 15 8 switches it isnt
And factorials work when numbers are not repeated
So simply youre just completely wrong x  |
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"Sounds like some confused thinking going on.
The possible combinations of the 40 lights is factorial 40.
"
One switch
Regardless equals two possibly
Regardless of light number
2 switches gives maximum of four regardless of configuration however less could be wired configured
3 8 maximum yet two switches giving a bank a binary option would reduce the combination
We need to differentiate also between max switch position combinations and the actual lights lit combinations which as I say although can be wired to match the switches they cannot exceed 2 to power switches and can be less xx |
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