

Maths Question




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So, say you have 1000 types of beer... and you can mix/match those beers to make a six pack. How many variations of a six pack could you make?
I'm guessing 1000*1000*1000*1000*1000*1000=10 to the 18th (or... ~ quintillion variations)
Is that correct? Also, how many Billions would that be.









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Its too many for the OS X scientific calculator apparently. I think the equation is 1005! / (6! x 999!).

I have plenty of more important things to do, if only I could bring myself to do them....








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Yes but since it's a six pack the order doesn't matter so it combinations not permutations. Though the combination does include all the cases were there's more than one of each kind in the six pack.









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That depends on whether you allow for having several bottles of one kind of beer or if you insist on having 6 different types of beer in your six pack.
For the first case you have:
1000^6/6! ≈ 10^(183) = 10^15
Each time you pick a bottle, you have 1000 choices, but you have to divide by 6! since the order in which the beer bottles are picked does not matter.
In the second case, the number is slightly lower, but practically the same:
1000*999*998*997*996*995/6! = 1000!/(10006)! 6! ≈ 10^16
The first time, you have 1000 beers to choose from, then 999 (since one type of beer has already been taken, etc.).

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Originally Posted by OreoCookie
That depends on whether you allow for having several bottles of one kind of beer or if you insist on having 6 different types of beer in your six pack.
For the first case you have:
1000^6/6! ≈ 10^(183) = 10^15
Each time you pick a bottle, you have 1000 choices, but you have to divide by 6! since the order in which the beer bottles are picked does not matter.
In the second case, the number is slightly lower, but practically the same:
1000*999*998*997*996*995/6! = 1000!/(10006)! 6! ≈ 10^16
The first time, you have 1000 beers to choose from, then 999 (since one type of beer has already been taken, etc.).
To be more specific, you can have doubles in the six pack.
So, that would be
100,000,000,000,000,000 options?
I just don't want to be out in left field.
Thanks for all of the help!









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10^16 or 10^15, your still gonna die from liver failure first.









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If you're allowing doubles, wouldn't it be 1000*1000*999*999*998?









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There is no answer to the original question because he never stated how many bottles you have. He said you have 1000 "types", which would set a floor on the number of bottles at 1 per type. But you could have 10,000 bottles and still have just 1000 types.









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The correct answer involves the choose function C(n,k).
C(n,k) is the number of ways to pick k objects from a group of n objects (if order doesn't matter).
C(n,k) = n! / (k! (nk)!)
For example, the number of 5card hands from a standard 52card deck is
C(52, 5) = 52! /(5! x 47!) = 2,598,960
So for 1000 beers and picking 6,
C(1000, 6) = 1,368,173,298,991,500 (from wolfram alpha)









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But the original question didn't give us the value of k.
He didn't say we had 1000 beers. He said we had 1000 types of beer. It's not the same thing.









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Originally Posted by asd
So for 1000 beers and picking 6,
C(1000, 6) = 1,368,173,298,991,500 (from wolfram alpha)
That's what I posted if you look, but if it's only singles and pairs then it's 1000!  994! for the singles plus 1000!  997! for the pairs.









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well, i think from his post he meant 1000 beers, and how many different 6packs could u make.









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What if one should happen to fall?

"…I contend that we are both atheists. I just believe in one fewer god than
you do. When you understand why you dismiss all the other possible gods,
you will understand why I dismiss yours."  Stephen F. Roberts








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Ooooooops I didn't read carefully through the replies. BLAZE_MklV got it right. my bad.









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Originally Posted by olePigeon
What if one should happen to fall?
You've had enough. Call a cab.









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Hmm ... I wonder if this math problem is intended to placate the wife.
Yeth dear, 'M solvving a maths problem. Chek this thread for de tales. hic!









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Ok, ok... 1000 unique bottles. Sorry.
So... 1000*1000*1000*1000*1000*1000, right?









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You'll get repeats, though. You'll have duplicate flavors in a six packs, but the bottles will just be in different places.

"…I contend that we are both atheists. I just believe in one fewer god than
you do. When you understand why you dismiss all the other possible gods,
you will understand why I dismiss yours."  Stephen F. Roberts








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Originally Posted by olePigeon
You'll get repeats, though. You'll have duplicate flavors in a six packs, but the bottles will just be in different places.
Ahhhhh, now I get it... Awesome, thanks!









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Originally Posted by mitchell_pgh
Ok, ok... 1000 unique bottles. Sorry.
So... 1000*1000*1000*1000*1000*1000, right?
No.
It's easier to visualize if you consider a »three pack«: for simplicity, assume you have 10 different beers to choose (one bottle each!) from and you have drawn beers 1, 2 and 3. Then you have 3! = 1 x 2 x 3 = 6 »threepacks« of beer which contain the same choices:
Code:
1 2 3 1 3 2 2 1 3 2 3 1 3 2 1 3 1 2
So overall, you have 120 = 10 x 9 x 8 / 3! = 10! / ((103)! x 3!) = »3 out of 10« different six packs.
If you have 10 different kinds, but more than 3 bottles of each, you have to do a bit more math: you have to add the number of combinations you have where 3 beers are different, 2 beers are different and all the beers are the same.
The first number, »3 out of 10« = 10 x 9 x 8 / 3! has been computed above. To compute the number of combinations where you have two bottles of the same kind, we repeat the procedure: you can only choose two types of beer, so you have 10 x 9 = 90 different combinations (remembering the order in which the bottles were drawn). But 3! / 2! = 3 of these combinations yield the same »threepack«:
So you have 30 = 10 x 9 / 3 different »threepacks«.
Lastly, there are only 10 »three packs« where all three beers are identical and there is only one such »threepack«.
Overall, this means, you have 120 + 30 + 10 = 150 different »threepacks«. Note that there are less combinations if you assume that one or several bottles are identical. Note that the combinatorics in your example are much more complicated if you allow up to two bottles of the same beer, because you have four different cases: six different beers, one pair (5 different), two pairs (4 different) and three pairs. However, if you have 1000 beers to choose from,
1000^6/6! ≈ 1000^5 will give you a good approximation. And it also means, your proposed solution was off by a factor of 1000.

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I'd like to toss quantum entanglement into the mix.
If you drink one of the beers, how will its unseen and whereabouts unknown twin respond? Will it too become devoid of alcoholic nourishment so that the next smart ass that tries this gets an empty bottle? Or will the Universe implode into the newly opened empty bottle? Would Doofy's cats suddenly appear from the depths of the empty bottle? Would the bottle even be empty?
I'll never sleep tonight.

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Allowing for doubles (or more) implies doubling (trioling, etc.) the initial population of available beers, basically increasing the complexity of the problem by an order of magnitude for each possible multiple.
Just sayin'....

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If I may, I have a math question too.
Imagine a very simple Euclidean universe in which the largest object is a circle 10 centimeters in diameter and the smallest object a circle 1 centimeter in diameter.
A circle occupying the midpoint between these two sizes has a diameter of 5 cm.
In a comic novel by Thomas Mann, a professor wonders airily whether the average dimensions of the human body occupy the midpoint between the largest and smallest possible objects thus restoring man's centrality in the centerless Copernician universe.
This supposition may be unprovable, even laughable, but it raises a question that has been bothering me ever since.
If the largest object in the universe is infinitely large and the smallest object infinitely small, what size is an object occupying the midpoint between the two?
And, prior to that, can an infinitely small object exist, even in theory? Or is its infinite smallness tantamount to nonexistence?









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(I know from reading Borges that the midpoint of a line of infinite length is, counterintuitively, any point on that line. A circle of infinite circumference is an infinite straight line. An infinite sphere has no circumference and any point within it can be regarded as its center; furthermore, the infinite sphere itself can be regarded as an infinite plane. I'm guessing that the answer to the above question, like most problems that invoke infinity, is similarly vertiginous and paradoxical.)









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Unless you are talking about this universe, in which case there are (very probably) no infinitely large objects.
You might run into problems with your definition of an object too. Does a galaxy count as an object given all the empty space in it? How about an atom? Those are more space than not too.
If you don't count galaxies, then you are left with black holes which technically are very (in fact infinitely) small, so then you get left with big stars. VY Canis Majoris is the biggest we know of.
Its not infinite, but it really is very, very big.
Scale of Earth, Sun, Rigel, and VY Canis Majoris. [full zoom at the end]  YouTube
If you want to stick to a theoretical universe where you have infinitely large objects, then I see no reason why your answer would be any different to the string analogy. Any point between infinitely large and infinitely small is the midpoint.

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^ Good points.
I had already assumed that, applying the question to a hypothetical universe in which we allow for the existence of an infinitely large and small object, the answer was that every object of finite size could be regarded as occupying the midpoint between the two.
Infinity seems to defy, and almost corrupt, logic. Perhaps that is why the Pythagoreans regarded it as evil.









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The shortest possible length is believed to be the Planck length (~1.6 × 10^−35 meters) below which spacetime becomes inconsistent (but this isn't proven). Black hole singularities are predicted by relativity, which does not take quantum mechanics into account, so it's not safe to use their zerosize in a scale.
The visible universe has a radius of ~46 billion LY, or a diameter of ~9.2 x 10^16 meters. The actual size of the universe is unknown. It might be smaller than the observable size, or substantially larger.
Using the observable universe size gives a possible size range of 5.7 x 10^51. Using the log scale, the midpoint might be somewhere around 1 x 10^9 meters or 1 nanometer. That puts the "centered species" somewhat smaller than a virus, in the size range of molecules. Perhaps a simple protein. Humans are too big to be centered, unless the Universe is incredibly larger than the visible size.









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The smallest measurable time is the delay between the light turning green and the guy behind you honking. True fact....

Glenn OTR/L, MOT, Tx








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Originally Posted by Tiresias
If the largest object in the universe is infinitely large and the smallest object infinitely small, what size is an object occupying the midpoint between the two?
Logically, the midpoint between infinitely large and infinitely small would be infinitely average. Thus the size of said object would be approximately 5'3", 250 lbs and sits on her huge ass bitching about things on Facebook all day.

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Originally Posted by osiris
I'd like to toss quantum entanglement into the mix.
If you drink one of the beers, how will its unseen and whereabouts unknown twin respond?
If you drink one beer, the other is not drunk. If you don't drink the beer, then the other is drunk. However, until you drink either beer, they're both drunk and not drunk simultaneously.

"…I contend that we are both atheists. I just believe in one fewer god than
you do. When you understand why you dismiss all the other possible gods,
you will understand why I dismiss yours."  Stephen F. Roberts















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