EPT1-5
12-15
23-25
34-35
44-45
123, 124, 125, 134, 135, 145
234, 235, 245
345
1234, 1235
1345
12345
So that’s 5+4+3+2+6+3+1+2+1+1 = 28 possibilities.
28^5= 17,210,368 combinations.
Ask and answer stupid questions here! - Page 737
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KwarK
United States43545 Posts
1-5 12-15 23-25 34-35 44-45 123, 124, 125, 134, 135, 145 234, 235, 245 345 1234, 1235 1345 12345 So that’s 5+4+3+2+6+3+1+2+1+1 = 28 possibilities. 28^5= 17,210,368 combinations. | ||
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DarkPlasmaBall
United States45258 Posts
On August 27 2019 04:17 Dangermousecatdog wrote: 182 EZ. You all suck. Did this in my head in under 2 min. Pls thank me for doing your homework for you. Unless there are additional restrictions to the original problem, 182 is too small of a total. How did you get that? | ||
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Simberto
Germany11739 Posts
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DarkPlasmaBall
United States45258 Posts
On August 27 2019 05:39 Simberto wrote: I kind of already explained how to solve this above. Yeah, with permutations rather than combinations  | ||
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GreenHorizons
United States23621 Posts
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DarkPlasmaBall
United States45258 Posts
On August 27 2019 07:58 GreenHorizons wrote: Dyscalculia or Acalculia, real or not? I know from working with students with the disability that Dyscalculia is real, nuanced, and not just a generic "I'm bad at math" excuse. Some basic info on the legitimate disability: https://en.wikipedia.org/wiki/Dyscalculia From that link: "Mathematical disabilities can occur as the result of some types of brain injury, in which case the proper term, acalculia, is to distinguish it from dyscalculia which is of innate, genetic or developmental origin." | ||
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FiWiFaKi
Canada9859 Posts
On August 27 2019 05:18 KwarK wrote: Each number can be 1-5 12-15 23-25 34-35 44-45 123, 124, 125, 134, 135, 145 234, 235, 245 345 1234, 1235 1345 12345 So that’s 5+4+3+2+6+3+1+2+1+1 = 28 possibilities. 28^5= 17,210,368 combinations. That's be wrong too, as you might have 28 possibilities of a number, but if you have a combined number, you don't have 5 numbers any more, for example doing 12, now you only have 3 numbers remaining. Also, this isn't a homework problem. I was installing a mechanical lock, and accidentally changed the combination. Without opening it all up, which is a pretty complex locksmith procedure, there is no other way that brute force. So I want to guess every combination until it opens (it's legit a real life application) . Doing the nPk thing, for k=1, 5 k=2, 20 k=3, 60 k=4, 120 k=5, 120 So for single numbers I'm up to 325. Now where it gets weird, let's do combined number combinations All 2 digit numbers: 1) 5c2 = 10 All 3 digit combinations: 1) All 3 digits pressed at once would be 5c3 = 20 2) All 2 digit combinations + single, so 5c2 =10, now treat it as 2 numbers so 4p2 =12, 12x10 =120 All 4 digit combinations: 1) All 4 digits pressed at once is 5c4 = 5 2) 3 digits + single is 5c3 = 20, then with that 3p2 = 6, 6x20 = 120 3) 2 digits + 2 singles 5c2 =10, then 4p3 = 24, so 240 4) 2 digits + 2 digits 5c2 =10, then 3c2 = 3, here I'm not sure but I want to say it's 2p2 = 2, so 10x3x2 = 60 All 5 digit combinations: 1) All 5 numbers pressed at once is 5c5 = 1 2) 4 digits + single is 5c4 = 5, then 2p2 = 2, 10 3) 3 digits + 2 digits is 5c3 = 20, then 2c2 = 1, then 2p2 = 2, 40 4) 3 digit + 2 singles is 5c3 =20, then 3p3 = 6, 120 5) 2 digits + 2 digits + single is 5c2 = 10, 3c2 = 3, 3p3=6, 180 6) 2 digits + 3 singles is 5c2=10, then 4c4 = 24, 240 That sounds be all of them, so: 365+10+140+325+591 = 1431 Anything that seems like it doesn't check out? Initially I expected it to be less, and I was just thinking what a shitty lock, but 1431 isn't so bad if it's correct. Also I'd love if someone had a more mathematically beautiful solution lol. And it'd be cool to be able to extrapolate the result for a lock like this but say it had 6 numbers with the same properties. And for those who care, I ended up guessing what the combination I entered was. I was mashing fairly predictable numbers, so I guessed the right one in around 10 minutes of trying haha. | ||
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Simberto
Germany11739 Posts
6) 2 digits + 3 singles is 5c2=10, then 4c4 = 24, 240 you are once again using c in part two, where you shouldn't, because order is once again important. 534(12) is not the same combination as 53(12)4. So you need 5c2 * 4p4 2) All 2 digit combinations + single, so 5c2 =10, now treat it as 2 numbers so 4p2 =12, 12x10 =120 This isn't correct either. First you need to choose which numbers you use, then how many combinations of those exist. So you are at 5c2 for the two digits, and then at 3c1 for the single additional digit to get all combinations of doubles and singles. Then, you only have to do 2p2 for the order of doubles + single. So you get a total of 5c2 * 3c1 * 2p2 = 10*3*2 = 60 for combinations which look like this (12)5 or (34)1 I think some of this may come from a misunderstanding of what i wrote earlier. But that is okay, because what i wrote earlier wouldn't work either, because you will count some things double. Your way works, if you do it correctly. | ||
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Acrofales
Spain18208 Posts
On August 27 2019 17:36 Simberto wrote: I don't think you have all of them. I see multiple problems with what you are doing. you are once again using c in part two, where you shouldn't, because order is once again important. 534(12) is not the same combination as 53(12)4. So you need 5c2 * 4p4 This isn't correct either. First you need to choose which numbers you use, then how many combinations of those exist. So you are at 5c2 for the two digits, and then at 3c1 for the single additional digit to get all combinations of doubles and singles. Then, you only have to do 2p2 for the order of doubles + single. So you get a total of 5c2 * 3c1 * 2p2 = 10*3*2 = 60 for combinations which look like this (12)5 or (34)1 I think some of this may come from a misunderstanding of what i wrote earlier. But that is okay, because what i wrote earlier wouldn't work either, because you will count some things double. Your way works, if you do it correctly. No. I think the second part is wrong. You do 5c2 to find all 2-key-press combos, and then you get it as a keypad with 4 keys where order is important again, so 5c2 * (4p1 + 4p2 + 4p3 + 4p4) = 10 * (4+12+24+24) = 640 Then you do the same for 5c3 * (3p1 + 3p2 + 3p3) = 150 , 5c4 * (2p1 + 2p2) = 20 and 5c5 = 1 So that's 811 combinations for all the combined key presses. Assuming fiwi mathed it right for the single key presses, a combined total of 811+325=1136 | ||
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Simberto
Germany11739 Posts
Now, we could try to figure out how many lock combinations we count doubly each time, and subtract them from our total result. Or we go the route that FiWiFaKi went, and split it up even further. So we have things like permutations involving a pair and two singles, or permutations involving two pairs and one single, or permutations involving a triplet and a single. Those are the things he is counting, but he is not counting them correctly. | ||
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Acrofales
Spain18208 Posts
Also, is it possible to make double combinations? So (12)(34)? Or (123)(45)? There aren't that many, but still, if that's a possibility, they need to be accounted for. | ||
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Simberto
Germany11739 Posts
3) 3 digits + 2 digits is 5c3 = 20, then 2c2 = 1, then 2p2 = 2, 40 In this case, they are actually counted correctly, too. | ||
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Splax
Sweden51 Posts
+ Show Spoiler +
Hope I'm not missing something obvious here.. A simple google gives me wolfram alpha that says if all buttons are required to be pressed, we're looking at 541 possible combinations for n=5, so if we exclude our counts for less than 5 buttons pressed total, we should get 541. | ||
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Zambrah
United States7393 Posts
If a Battlecruiser warps in at lightspeed, does it not kill all the inhabitants on the Battlecruiser? | ||
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Simberto
Germany11739 Posts
But in general, speed does not kill things. Acceleration does. So the question is not how fast it moves, but how fast it decelerates afterwards, and if there are any SciFiTech "Inertial Dampeners" to stop everyone from being a stain on the wall. Humans are usually pretty fine at up to 5g acceleration, fighter pilots can take up to 10g with special equipment, and there have been occurences of humans surviving higher accelerations. But even if we say that with futuretech, humans can survive 20g of acceleration. That is 200m/s². Which means to get from near lightspeed to zero, it would take you 1.5 million seconds, or about 17 days to decelerate from near lightspeed to zero at an acceleration that is probably still too high for humans to survive. I think we can safely say that that is not the amount of time Battlecruisers take to decelerate. There are ways around this, of course. The basic classic SciFi tropes are that FTL travel isn't actually achieved by accelerating past the speed of light, but by doing something qualitatively different, like warp, hyperspace, or whatever. At this point, we can't really use normal science to estimate anything, because it is basically magic. And/or we have inertial dampeners which mean that the inside of the ship doesn't "feel" the acceleration. Which once again is basically magic, and we cannot make any reasonable estimates of what that would lead to. If you have none of these, and do what battlecruisers do, it doesn't actually matter if you are a human inside that battlecruiser, or a steel part of the hull of it, or basically anything else. Nothing can take the kind of force necessary for that kind of acceleration. If you go from near light to zero in one second, you need an acceleration of 300 million m/s², which equal 30 million N /kg of mass. That will nicely rip everything to shreds, and depending on how we apply the force, either compact it into a nice neutronium ball, or spread the battlecruiser out as a quickly expanding gas cloud. | ||
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_fool
Netherlands681 Posts
How do you clean your chopsticks. Dishwasher? Just rinse them in the sink? Just leave them for the next day? Do Asian dishwashers have a dedicated chopsticks area? | ||
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Uldridge
Belgium5041 Posts
But for like 99.99% of the time just rinsing with hot water is all you need. And they're sticks, so it takes like literally 5s after your water is hot. In the mean time, if your water takes a while to heat up, just use that to fill a drinking bottle or water your plants with. | ||
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_fool
Netherlands681 Posts
On August 29 2019 03:20 Uldridge wrote: In the mean time, if your water takes a while to heat up, just use that to fill a drinking bottle or water your plants with. It's like you know my kitchen inside out It takes ages before the water gets hot! Tnx dude. | ||
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Uldridge
Belgium5041 Posts
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Dangermousecatdog
United Kingdom7084 Posts
-Has 5 numbers, 1 through 5. -Each number can be pressed at most once. and seperately -Can press as many numbers as wanted simultaneously. For example pressing 2 and 4 at same time is a different combination than pressing 2 then 4, or pressing 4 then 2 So For 1=1 For 2=2 For 3=6 For 4=24 For 5=120 So for each number pressed once at most once is 1+2+6+24+120=153 Then for sumbers pressed simultaneously, draw 5 dots in a pentagram and join all the lines together. For 1=5 For 2=10 For 3=10 For 4=5 For 5=1 So 31 combinations 153+31=184 but both have pressing 1 twice so -1 to that so its 183. If it isn't separate it's a bit more complicated and I'll just leave out of embarrassment. On August 29 2019 03:14 _fool wrote: I occasionally eat my dinner with (reusable) chopsticks. But they're a pain to put in the dishwasher! I can't put them in the cutlery basket with the forks and knives, because they fall through. I can't put them in the top because they are pushed around by the water jets, ending up on the bottom and/or blocking the rotating thingy. How do you clean your chopsticks. Dishwasher? Just rinse them in the sink? Just leave them for the next day? Do Asian dishwashers have a dedicated chopsticks area? My cutlery basket have holes small enough that chopsticks can't fall through. | ||
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