1762: "Moving Boxes"
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 Soupspoon
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Re: 1762: "Moving Boxes"
I've had cause to think of my house contents recently. By weight, I think most of my boxes would just be labelled "Books". Bookshelves would be the dominant furniture type (but are themselves necessarily much lighter, and doesn't count the 'bookcase' that is the floor).
(And having thought I'd just mentally counted all the bookcases in my house, I've just realised that there are two more, just a small head turn away, that I'd totally forgotten.)
(And having thought I'd just mentally counted all the bookcases in my house, I've just realised that there are two more, just a small head turn away, that I'd totally forgotten.)
Re: 1762: "Moving Boxes"
Flumble wrote:addams wrote:uhoh....Is someone moving?
xkcd: Soon to be coming from Canada?
Given his express interest in the Netherlands in the past, maybe xkcd will be dutch soon?
..oh...Right. Good on him.
..Yes. That would be my first choice, too.
Life is, just, an exchange of electrons; It is up to us to give it meaning.
We are all in The Gutter.
Some of us see The Gutter.
Some of us see The Stars.
by mr. Oscar Wilde.
Those that want to Know; Know.
Those that do not Know; Don't tell them.
They do terrible things to people that Tell Them.
We are all in The Gutter.
Some of us see The Gutter.
Some of us see The Stars.
by mr. Oscar Wilde.
Those that want to Know; Know.
Those that do not Know; Don't tell them.
They do terrible things to people that Tell Them.
Re: 1762: "Moving Boxes"
orthogon wrote:ps.02 wrote: [...] CLOCKS (FAST) [...]
If you mean that the clocks gain time, a good GRaware removal company will put them in the attic of the new house. (Or is it the basement?)
Attic, I think. It's really easy to confuse which direction the effect goes, for you and me both.
(Note though: maximum gravity isn't the basement, but ground level  it goes to zero as you approach the center of the earth. So, do put CLOCKS (SLOW) in the basement, but only in the expectation that they can be hung high on the walls.)

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Re: 1762: "Moving Boxes"
azule wrote:Is a box labeled Pornography more or less likely to be "mishandled" by paid movers?
Well, apparently the ruse worked, so I guess the movers were either too embarrassed or too scared to look.
 Wee Red Bird
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Re: 1762: "Moving Boxes"
In my last move, I had a box marked The decapitated heads of my enemies Garage Stuff. Was left in the middle of the garage.
Re: 1762: "Moving Boxes"
Perhaps I've been watching too much Only Connect: my instinct on seeing an image like this is that there's some cryptic pattern to the box labels, usually involving the letters in the words and not the words themselves. Words ending with AM on the left, NZ on the right, or the number of unique vowels increases as you go down levels (so garam masala goes on the top and sequoia seeds on the bottom) and I have to figure out the pattern for 5 marks on my nonverbal reasoning exam.
Is there any such pattern discernible here or are the labels just fanciful box contents?
Also I agree that the box labeled
"DARK
MATTER"
contains both dark and matter  the only other obvious multiword phrase is "field lines" and that appears on a single line.
Is there any such pattern discernible here or are the labels just fanciful box contents?
Also I agree that the box labeled
"DARK
MATTER"
contains both dark and matter  the only other obvious multiword phrase is "field lines" and that appears on a single line.
Re: 1762: "Moving Boxes"
I've moved houses quite often. I've never labelled a box. Is it a normal thing to do? Do most people do it?
Re: 1762: "Moving Boxes"
Plutarch wrote:I've moved houses quite often. I've never labelled a box. Is it a normal thing to do? Do most people do it?
Last time we moved, we used boxes labelled using an Xnn schema  X being a prefix indicating which room (or at least region) of the house to move it to; nn being a number. We also had inventory sheets summarising the contents of each numbered box.
So the removal men could be told "boxes starting with K go in the Kitchen; anything starting D can stay downstairs; anything else goes upstairs" or similar.
 azule
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Re: 1762: "Moving Boxes"
Plutarch wrote:I've moved houses quite often. I've never labelled a box. Is it a normal thing to do? Do most people do it?
I used to label them. But, eh, don't anymore. If you don't label them it's like Xmas morning everyday until you're done unpacking.
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Re: 1762: "Moving Boxes"
I think I used to label them, but it's been too long. However, some time ago I got a two line phone that iets me forward an incoming call on one line to a different number via a different line. It's a really cool feature and is the main reason I got the phone. As it turns out, the one time it would have come in handy, I didn't set it up, and missed a very important call (about a house I was negotiating over). Ironically, I was actually just upstairs, in the real estate office of my landlord and agent (who wasn't the one making the call though).
Anyway, we got the house, I packed the phone in with all my other stuff in boxes that were probably labeled well enough (it would have been my style), and we moved in. It's been over thirty years and the phone has not turned up. In fairness, not all boxes got opened either, perhaps because of what the labels said. And actually, I don't even know where the unopened boxes are at this point.
As an aside, I got some cheap padlocks for free. They are of the type that has ten buttons, and you depress five of them (they stay depressed; the order doesn't matter) and you can open the lock. I wrote the combination on the lock in magic marker, but in a nonobvious but easy to remember code.
I have no idea what the code is anymore, can't figure out the combination, and can't open the lock. I guess the code was good enough.
Jose
Anyway, we got the house, I packed the phone in with all my other stuff in boxes that were probably labeled well enough (it would have been my style), and we moved in. It's been over thirty years and the phone has not turned up. In fairness, not all boxes got opened either, perhaps because of what the labels said. And actually, I don't even know where the unopened boxes are at this point.
As an aside, I got some cheap padlocks for free. They are of the type that has ten buttons, and you depress five of them (they stay depressed; the order doesn't matter) and you can open the lock. I wrote the combination on the lock in magic marker, but in a nonobvious but easy to remember code.
I have no idea what the code is anymore, can't figure out the combination, and can't open the lock. I guess the code was good enough.
Jose
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 Soupspoon
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Re: 1762: "Moving Boxes"
ucim wrote:As an aside, I got some cheap padlocks for free. They are of the type that has ten buttons, and you depress five of them (they stay depressed; the order doesn't matter) and you can open the lock. I wrote the combination on the lock in magic marker, but in a nonobvious but easy to remember code.
I have no idea what the code is anymore, can't figure out the combination, and can't open the lock. I guess the code was good enough.
Just 252 combinations, easily permuted through, if I've worked it out correctly... Give it a go, and you'll probably get the first open within twenty minutes, the second within ten (having got into the pattern, through the practice on the first), and the rest in not much over five, if even that. Especially if you find yourself able to backcalculate your cluecode.
Quicker, if its the old type that has a subtle 'feel' when you spoil the code. (Like the early door locks with 1..9,0,X,Y,Z and the C for the reset (was that maybe 6 of 13 keys pressed, for 1716 combos?) , like I remembered playing with back in the '80s each time we visited a campsite with one of those on the pedestrian access gate/doorintheboundarywall. Having the current code, on each visit, but also having plenty of time on my hand... )
Re: 1762: "Moving Boxes"
Yeah, not worth it. They were really junk locks anyway and I don't even know if I have them any more.Soupspoon wrote:Just 252 combinations, easily permuted through, if I've worked it out correctly... Give it a go, and you'll probably get the first open within twenty minutes...
There was another I lost the combination to  it was one of those brass ones with four dials numbered 09, so 10000 combinations. You had to push the hasp in to get it to spring out, and you could change the combination. 0000 push 0001 push 0002 push.... I got halfway through before giving up. Took forever. Even odds I passed the combination and it didn't open because, well, sometimes they stick. Dunno where that one is either.
Jose
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 Sableagle
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Re: 1762: "Moving Boxes"
Most "cheap and crappy little" locks can be opened with 55645 and a few symbols.
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Re: 1762: "Moving Boxes"
Sableagle wrote:Most "cheap and crappy little" locks can be opened with 55645 and a few symbols.
It can probably be opened with a screwdriver. The buttons are all plastic.
However, the combination is not 55645. There are ten buttons, and each one is either pushed or not pushed. Exactly five are pushed. 12467 is a possible combination, but 12255 (equivalent to 125) is not. I don't even know if I have them, but if I do, I'll post the code and see if any of you can figure it out.
Jose
Order of the Sillies, Honoris Causam  bestowed by charlie_grumbles on NP 859 * OTTscar winner: Wordsmith  bestowed by yappobiscuts and the OTT on NP 1832 * Ecclesiastical Calendar of the Order of the Holy Contradiction * Heartfelt thanks from addams and from me  you really made a difference.
 Sableagle
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Re: 1762: "Moving Boxes"
The symbols are a decimal point, a multiplication sign and a couple of lowercase "m"s. Get the angle right and have a bucket of water there.
Zohar wrote:You don't know what you're talking about. Please spare me your quote sniping and general obliviousness.
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 Soupspoon
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Re: 1762: "Moving Boxes"
ucim wrote:Sableagle wrote:Most "cheap and crappy little" locks can be opened with 55645 and a few symbols.
It can probably be opened with a screwdriver. The buttons are all plastic.
However, the combination is not 55645. There are ten buttons, and each one is either pushed or not pushed. Exactly five are pushed. 12467 is a possible combination, but 12255 (equivalent to 125) is not. I don't even know if I have them, but if I do, I'll post the code and see if any of you can figure it out.
Jose
You're thinking of the combinations wrongly.
They range from 0000011111 to 1111100000, 'binary' pressed/not pressed by each 1 or 0, but the transition is the next 'binary' number of exactly five set (and five unset) bits.
One transition is:
0000011111
0000101111 (most significant bit shifted)
0000110111 (each nextmost
0000111011 significant bit now
0000111101 shifted into the
0000111110 gap opened up)
0001001111 (shift the MSB again)
0001010111 (move the nMSB into the gap)
0001011011 (into whose own
0001011101 gap you shuffle
0001011110 the remainder)
0001100111 (then reset tyem aw you move the nMSB [ifurther][/i] into the gap)
... Repeat with the dregs to get the nextnMSB to the gap ceiling of the doublegap, then the nnnMSB, until the entire gap is atbthe bottom, then reset the (n+)MSBs whilst moving the MSB up to create a triplegap, which you then shuffle into with similar methods.
(This order just handily follows the subset of all binary numbers that fulfil the 'five set bits' rule, there are better transitions that work like a bidirectional bubblesort that would minimise switchflicking in a 'permute until it works' way, but as you probably have to press an eleventh 'reset' button after each failed attempt, that's reset+five+try on each attempt after the initial one, regardless, so a good sequence that you can keep track of probably is best.)
That's the 'physical layer' model, but you can also look at the sequences, in abstraction, as a(n inferior) form of RunLength Encoding. The number of 0s between each 1, or if you prefer) the number of 1s between each zero. Taking the latter (counting the trailing digitsofnote as counted until the endofsequence with the above pattern, you have 00005, 00014, 00023, 00032, 00041, 00050 then 00104, 00113, 00122, etc, which (with less digits than ten) you again get a consistent an ascending list of numbers in an arbitrary base that is actually also limited to a sumofalldigits being exactly five.
...and I want to go onto how you can find the number of combinations through that variablyvariablebase counting system, but I'm about to sit down to a meal with relatives, and I'll be getting funny looks if Imdon't engage with their conversation. I'll leave it there, and maybe come back to it later...
 Soupspoon
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Re: 1762: "Moving Boxes"
...tell you what, I won't go into the number of combinations, because that rushed (typoladen) message probably belongs more in the Mathematics forum.
But I'll start you off, if you want to work it out yourself. The full (01)x10 space is of course 2^{10} combinations. But you can remove the single combination that is 0000000000 and the opposite one that is 1111111111, obviously. And the ten from 0111111111 to 1111111110 with a single zero taking each position, ditto allbutonearezero with a travelling 1. And for two notliketherest, it gets complicated but calculable (hint: 1andoneinnine plus 0and2innine, the latter of which is 1andoneineight plus..., etc), which leads you on to the (two complimentary) threeunliketherest and onward yet until fourunlikethe rest, all paired. Subtract those totals from the 2^{10}, as what is left is fiveandfive total.
(Fourandsix was easier for me to work out, through reduction, than the fiveandfive, so going the long way round, later confirmed with brute force, was my method. But maybe there's a better direct method, after all, based upon a single recursive value.)
But I'll start you off, if you want to work it out yourself. The full (01)x10 space is of course 2^{10} combinations. But you can remove the single combination that is 0000000000 and the opposite one that is 1111111111, obviously. And the ten from 0111111111 to 1111111110 with a single zero taking each position, ditto allbutonearezero with a travelling 1. And for two notliketherest, it gets complicated but calculable (hint: 1andoneinnine plus 0and2innine, the latter of which is 1andoneineight plus..., etc), which leads you on to the (two complimentary) threeunliketherest and onward yet until fourunlikethe rest, all paired. Subtract those totals from the 2^{10}, as what is left is fiveandfive total.
(Fourandsix was easier for me to work out, through reduction, than the fiveandfive, so going the long way round, later confirmed with brute force, was my method. But maybe there's a better direct method, after all, based upon a single recursive value.)
 Sableagle
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Re: 1762: "Moving Boxes"
While you were working that out I walked to the hardware store, bought a pair of bolt croppers, walked back and cut the lock off.
Zohar wrote:You don't know what you're talking about. Please spare me your quote sniping and general obliviousness.
CorruptUser wrote:Just admit that you were wrong ... and your entire life, cyberspace and meatspace both, would be orders of magnitude more enjoyable for you and others around you.
Re: 1762: "Moving Boxes"
The number of combinations is just ten choose five. That is, 10! / (5!*5!) which equals 10*9*8*7*6 / (5*4*3*2)
This equals (2*9*2*7*2) / 2 which is 7*9*4 = 252
Jose
This equals (2*9*2*7*2) / 2 which is 7*9*4 = 252
That's actually the basis of my code. Each lock had some letters on it. Consider a lock that had "ADE" written on it.Soupspoon wrote:One transition is:
0000011111
0000101111 (most significant bit shifted)
0000110111 (each nextmost
While you were at the store, I used a bic pen to poke the plastic buttons out (each of them actually goes completely through the lock).Sableagle wrote:While you were working that out I walked to the hardware store, bought a pair of bolt croppers, walked back and cut the lock off.
Jose
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Re: 1762: "Moving Boxes"
You were all talking past each other.
I was really confused about the conversation between Soupspoon, ucim and Sableagle until I realized that.
Apart from ucim's locks being either 10 buttons or a 4digit combination, so it's not a response to ucim, I have no clue what kind of lock can open with '55645 and a few symbols'.
@Soupspoon, since the whole space of 2^10 is only 4 times as large, just trying all binary codes is less errorprone and might be just as fast if you're familiar with binary adders (or snakes). Then again, if you have to reset the buttons after every try, it's faster and less errorprone to write a script that shows you visually which buttons to press for every try.
I'd rather do a knownplaintext attack.
I was really confused about the conversation between Soupspoon, ucim and Sableagle until I realized that.
Sableagle wrote:Most "cheap and crappy little" locks can be opened with 55645 and a few symbols.
Apart from ucim's locks being either 10 buttons or a 4digit combination, so it's not a response to ucim, I have no clue what kind of lock can open with '55645 and a few symbols'.
@Soupspoon, since the whole space of 2^10 is only 4 times as large, just trying all binary codes is less errorprone and might be just as fast if you're familiar with binary adders (or snakes). Then again, if you have to reset the buttons after every try, it's faster and less errorprone to write a script that shows you visually which buttons to press for every try.
ucim wrote:Consider a lock that had "ADE" written on it.
I'd rather do a knownplaintext attack.
 Soupspoon
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Re: 1762: "Moving Boxes"
ucim wrote:The number of combinations is just ten choose five. That is, 10! / (5!*5!) which equals 10*9*8*7*6 / (5*4*3*2)
This equals (2*9*2*7*2) / 2 which is 7*9*4 = 252
I gotta think through that "m choose n == m!/(n!²)" thang, there. Doesn't sound right, to me, but you got the right result anyway. (I departed from the statistical mathematics route at the Further Education level, so even though I've worked alongside statisticians later on I'm perhaps lacking some of the standard nonselfderived formulae and understandings.)
Anyway, at "10*9*8*7*6 / (5*4*3*2)" I'd go further than you to reduce that by (10/5=2), (8/4=2), (9/3=3) and (6/2=3) to become "2*3*2*7*3" over unity, to the same result.
Well straight hex is no good, 1010 1101 1110 has overflow (whichever way) on ten bits, assuming no deliberate 'junk bits' (and it's 8v4, not 7v5 with the 5 being important). Even with bigendian/littleendian switching at either bit or demibyte level, it doesn't. And you reply to the binary depictions, not the RLEish ones, so it aintn't based upon that, even if ADE = 1,4,5. (1,5,4 could be 1000001111, but that means another layer of code transposition.) Maybe you're encoding initial parity? But in a(n unsolved) sample of one there's really too few clues. If you'd care to expand the sample in either dimension (or ideally both..?) then maybe we could do better than those people you had originally intended the locks to intellectually defeat... If we could do it with one such example, any halfdetermined couriering professional with multiple examples (and/or a plastic pen!) could surely have cracked your code (if not locks!)....Consider a lock that had "ADE" written on it.
 Soupspoon
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Re: 1762: "Moving Boxes"
Reply to self, deliberately, to record my thoughts on the matter...
I set about thinking it through and soon realised that it wasn't "m!/(n!²)", it was "m!/(n! * (mn)!)"... n! being all possible combinations of (for any given five bits) you can select those bits to mark as (say) 1. And the (mn)! being all possible combinations of selecting the (other!) bits and marking them as (say) 0. (It just happens that mn=n, of course.)
And then it took a few more minutes (more than it should have) to realise that the m! is therefore all the possible orders to choose "set a first 1 (of ten), second 1 (of nine), third 1 (of eight), fourth 1 (of seven), fifth 1 (of six), first 0 (of five), ... fifth 0 (of one)". The 10! and the 5! approaches I had considered myself in working out the original combination number, the other day, then excluded because of needing to irreleventise the order. But with all orders of 1111100000 above the divisor line and then (all orders of 11111) and (all orders of 00000) below the line, naturally it cancels out the unwanted ordering multiplicities in each case and gives us the orderless result.
I liked doing that. It is just past midnight, local time, and already I have just learnt a completely new thing, today. Thank you.
Soupspoon wrote:ucim wrote:The number of combinations is just ten choose five. That is, 10! / (5!*5!) which equals 10*9*8*7*6 / (5*4*3*2)
This equals (2*9*2*7*2) / 2 which is 7*9*4 = 252
I gotta think through that "m choose n == m!/(n!²)" thang, there. Doesn't sound right, to me, but you got the right result anyway.
I set about thinking it through and soon realised that it wasn't "m!/(n!²)", it was "m!/(n! * (mn)!)"... n! being all possible combinations of (for any given five bits) you can select those bits to mark as (say) 1. And the (mn)! being all possible combinations of selecting the (other!) bits and marking them as (say) 0. (It just happens that mn=n, of course.)
And then it took a few more minutes (more than it should have) to realise that the m! is therefore all the possible orders to choose "set a first 1 (of ten), second 1 (of nine), third 1 (of eight), fourth 1 (of seven), fifth 1 (of six), first 0 (of five), ... fifth 0 (of one)". The 10! and the 5! approaches I had considered myself in working out the original combination number, the other day, then excluded because of needing to irreleventise the order. But with all orders of 1111100000 above the divisor line and then (all orders of 11111) and (all orders of 00000) below the line, naturally it cancels out the unwanted ordering multiplicities in each case and gives us the orderless result.
I liked doing that. It is just past midnight, local time, and already I have just learnt a completely new thing, today. Thank you.
 Sableagle
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Re: 1762: "Moving Boxes"
Yep. That bit I remember: mCn = m! / n! (mn)! and mPn = m! / (mn)!, numbers of combinations and permutations of n from m.
5.56x45mm versus good locks.
5.56x45mm versus good locks.
Zohar wrote:You don't know what you're talking about. Please spare me your quote sniping and general obliviousness.
CorruptUser wrote:Just admit that you were wrong ... and your entire life, cyberspace and meatspace both, would be orders of magnitude more enjoyable for you and others around you.
Re: 1762: "Moving Boxes"
The whole m!/(mn)! thing always bothered me, since it's just a notational trick for writing "the product of the integers between (mn+1) and n" (actually you could use bigpi notation, which would be better). It's not as though calculating the factorials and dividing them is more efficient than calculating the product you actually need, and it obfuscates the derivation (pick one of n, then for the next choice you've got n1 to choose from, etc; repeat m times), which you can easily do in your head.
xtifr wrote:... and orthogon merely sounds undecided.
 Soupspoon
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Re: 1762: "Moving Boxes"
It is handier to notate, though, than doing a sum (for nm+1 to m) of ln(x) and then raising that back up as a power of e, or somesuch alternate workaround.
 Sableagle
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Re: 1762: "Moving Boxes"
orthogon wrote:The whole m!/(mn)! thing always bothered me, since it's just a notational trick for writing "the product of the integers between (mn+1) and n" (actually you could use bigpi notation, which would be better). It's not as though calculating the factorials and dividing them is more efficient than calculating the product you actually need, and it obfuscates the derivation (pick one of n, then for the next choice you've got n1 to choose from, etc; repeat m times), which you can easily do in your head.
That's what I lined up in my head to make sure I had it right before I clicked "Submit," yes. With a calculator, the ! button is the way to go. Without it, doing only 11+ stuff, cancelling out the numbers first is the way forward.
Number of possible combinations of 6 balls from the 59 now used in our National Lottery:
59*58*57*56*55*54*53*52*51*...*2*1
53*52*52*...*2*1 * 6*5*4*3*2*1
59*58*57*56*55*54
6*5*4*3*2*1
59*
45057474
Fortyfive million to one against you hitting the jackpot, and even if you do you have to share it, with probability 1  (1  1/45057474)^{"15 to 45 million"}. 71.68% probability that nobody else hits it, assuming no bias in number selection among other players. Nonzero probability that 132 other people hit the jackpot, as apparently happened back in '95. They got £122,510 each.
That would be almost enough to buy a 3bedroom endterrace here, so it's totally relevant to moving boxes.
Zohar wrote:You don't know what you're talking about. Please spare me your quote sniping and general obliviousness.
CorruptUser wrote:Just admit that you were wrong ... and your entire life, cyberspace and meatspace both, would be orders of magnitude more enjoyable for you and others around you.
Re: 1762: "Moving Boxes"
Well, pretty much all notation is a trick. m*n is just a tricky way of writing m+m+m+m+m+m+m+ (n times). I'd propose overloading the factorial operator to make it binary:orthogon wrote:The whole m!/(mn)! thing always bothered me, since it's just a notational trick...
m!n would be m*(m1)*(m2)...*(n+2)*(n+1)*n. Thus m! would be a shorthand ("trick") for m!1, m!m would equal m, and m!n would only be defined for positive integers where m>=n (although the definition could be expanded to make m!n=n!m since multiplication itself is commutative).
Jose
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Re: 1762: "Moving Boxes"
Sableagle wrote:Most "cheap and crappy little" locks can be opened with 55645 and a few symbols.
I find they can all be opened rapidly with a dremel. Far more rapid than the above math.
Re: 1762: "Moving Boxes"
orthogon wrote:The whole m!/(mn)! thing always bothered me, since it's just a notational trick for writing "the product of the integers between (mn+1) and n" (actually you could use bigpi notation, which would be better).
...or is it? You can also derive it as: get all the permutations of m items, then, because we only care about the first n items picked, we don't care about the order of the mn other items so we divide by (mn)!. Likewise, if you also don't care about the order of the first n items, you divide by n!.
The "choose one of m, choose one of m1, ..., choose one of mn" approach is more 'constructive', whereas the "put all m items in all orders, discard all permutations of mn items" is easier to generalize.
Re: 1762: "Moving Boxes"
The representation using a ratio of factorials is nice as far as definitions go, but fortunately they came up with the specialized notations like C with superscript n and subscript k, or the stacked n and k inside a long pair of parentheses, for convenience in further use.
Re: 1762: "Moving Boxes"
Flumble wrote:orthogon wrote:The whole m!/(mn)! thing always bothered me, since it's just a notational trick for writing "the product of the integers between (mn+1) and n" (actually you could use bigpi notation, which would be better).
...or is it? You can also derive it as: get all the permutations of m items, then, because we only care about the first n items picked, we don't care about the order of the mn other items so we divide by (mn)!. Likewise, if you also don't care about the order of the first n items, you divide by n!
That's brilliant, thanks. One thing that I always loved about maths and the harder sciences is the way that you can think a problem through in different ways and they all give the same answer.
xtifr wrote:... and orthogon merely sounds undecided.
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