I wonder how they arrived at that figure?
I make it 1,307,674,368,000 which is a much sexier number.
Nice work by the way. 
Chat GPT, where else? I canāt figure out the math
2āæ - 1 where n = number of plugins
In a recent thread one of the SketchUp Team mentioned a debugging method but I canāt find the thread.
But an internet search suggests that the method is binary search debugging, and asking Gemini about it, binary search debugging might find a single culprit plugin in 4ā5 testsā¦
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I despair - I guess you still have to ask the right question.
āInsanity is doing the same thing over and over again and expecting different results.ā
turn off half of your plugins.
if crash, then turn off half of the plugins that were on.
if no crash, then switch.
rinse and repeat.
this works well to find a problematic plugin. if itās a combination, you can usually rule out some very simple ones, compatibility issues tend to arise with complex plugins not playing nice together.
yup
for future reference,
each plugin has 2 states. on or off.
so the math starts by a 2
then if you have 15 plugins, each having 2 modes, itās 2x2x2x2x2x2ā¦x2 so in the end, 2^15
if you had 6 things with 3 possible modes, it would be 3^6
and so on.
chat gpt is a linguistic predictive model. it answers what it believes people are most likely to want to read, based on other texts it read.
it can do basic math. google search engine tooā¦
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I may be wrong, but as its early on a Saturday morning and I have a few minutes let me put this out there.
The OP wrote:
I would say if you have just 1 plugin thereās 1 combination, and with 2 there are 2 combinations, but with 3 plugins there are 6 (on, on, off) (on, off, on) (off, off, on) (off, off, off) (off, on, on) (off, on, off) and with 4 there are 24, so I think its n! i.e. (n Factorial) rather than n^x, where n = the number of plugins
so:
1! = 1 x 1 = 1
2! = 1 x 2 = 2
3! = 1 x 2 x 3 = 6
4! = 1 x 2 x 3 x 4 = 24
5! = 1 x 2 x 3 x 4 x 5 = 120
15! = 1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 x 9 x 10 x 11 x 12 x13 x 14 x 15 = 1,307,674,368,000
What do you think?
you forgot (on, on, on) an (on, off, off)
so thatās 8. aka 2^3 
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No!
one: 0 or 1
two: 00 01 10 11
tre: 000 001 011 010 100 101 111 110
The right way as @ateliernab say.
sataes^number of plugin >> 2^15 = 32768
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Sorry - I still think I am correct 
If Iām not mistaken, the factorial thing would be use to find specific orders.
Say you have 3 plugins, A B and C
you can sort them as
ABC
ACB
BAC
BCA
CAB
CBA
thatās 3! ways of sorting. and we donāt care whether theyāre on or off.
but here, the sorting order is irrelevant, they are all here, regardless of their order, and we want to know the ones that are turned on or off. therefore its 2^n 
edit:
I mean, we just proved that you missed 2 combinations, making your demonstration false but sure
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I was thinking more about the number of ways the plugins can be switched on as that could effect whether they cause a problem or not.
For instance, plugin 3 may work fine with plugin 7, but only if its switched on after plugin 8 is activated. If its switched on before plugin 8 then it messes with plugin 7. You get the idea. Then the order they are switched on becomes relevant and the number of combinations is 15!
Just a thought?
that would be a combo.
first, 2^n to find out how many different on/off combos there are.
then, for a specific combo, a m! where m is the number of āonā
you have your 3 plugins. thatās 8 combos of on/off (2^3)
then,
000 ā 0! so 1 chance to find the right order to turn it on (it works or it doesnāt)
001 / 010 / 100 ā 1! (same, itās all on off, and you flip one on.)
011 / 101 / 110 ā 2!. you can turn one on first, then the other. or switch.
111 ā 3! ways of doing it.
so with 3 plugins, you can have 8 on/off combination, and a total of 16 orders to turn them on
so thatās orders of magnitude more than both n! and 2^3
so if you isolate one specific solution, and say āok I have 15 plugins, all turned on, and I want to see in what order to do itā then itās indeed 15!
but if you have 15 plugins, some on some off, and you want to know for each on/off combo in what order to turn them on, itās the more complicated math up there.
Some seriously bored people in here.
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I only got the idea because I am really keen on the Rubikās cube, and I was recently studying the maths involved.
The statistic that got me was how long a line of standard size cubes would be you had one for every combination of possible colour scrambles placed side by side. I asked a few people to guess the answer, and offered them Ā£50 if they got it within a factor of 10.
Have a guess without looking it up - I think you will be surprised
I didnāt have much friend growing up and it turns out maths were on my spectrum card. statistics, probabilities. oh man, geometry, so sweet.
idk, more than 10 ? am I right ? am I technically right ?
Personally I am unashamedly procrastinating I am trying to draw a curved building and Iām not finding it easy.
Its something like 260 light years.