Easy permutations formula/assistance req'd!
Discussion
Hi all, would like to determine how many different combinations that could be achieved from a range of, say Lego parts.
Let's say I have 10 diferent parts, labelled A, B, C >> J
If A can be used with one other A, then that would be the first combination. Then x2 A's etc etc up to x10, for example.
Then x1 A could work with x1 B, x2 B, x3 B etc etc.
Does that make sense - can anyone help determine the number of all other permutations/combinations from x10 different Lego parts with a quantity of x1 each, please?
Let's say I have 10 diferent parts, labelled A, B, C >> J
If A can be used with one other A, then that would be the first combination. Then x2 A's etc etc up to x10, for example.
Then x1 A could work with x1 B, x2 B, x3 B etc etc.
Does that make sense - can anyone help determine the number of all other permutations/combinations from x10 different Lego parts with a quantity of x1 each, please?
)ReaderScars said:
can anyone help determine the number of all other permutations/combinations from x10 different Lego parts with a quantity of x1 each, please?
Excluding the empty set it is 2^10 - 1 = 1023.ETA - that is just the number of combinations. Do you want permutations as in AB is different to BA?
Edited by Jabbah on Thursday 1st December 15:24
Sorry Jabbah, that last x1 in my OP should be x10 - or to be specific, up to x10.
So I'd like to be able to state how many combinations there could be when using ten parts, which have ten different shapes, in as many different combinations as possible (don't need to calculate what those actual combinations are, just the final number).
Hopefully that makes it a little simpler?
So I'd like to be able to state how many combinations there could be when using ten parts, which have ten different shapes, in as many different combinations as possible (don't need to calculate what those actual combinations are, just the final number).
Hopefully that makes it a little simpler?
It depends on how many of each type of part you have, and how many you use. If you have 10 different types, have at least 10 of each type, and want to make a structure using 10 bricks, then yes, it's 10^10.
If you have only 10 different parts then the number of combinations isn't 10^10.
Assuming the parts can only be bolted end-on-end, the number of permutations for one brick out of 10 is 10.
The number of permutations for two bricks out of 10 is 10 x 9. You have a choice of 10 bricks for the first and 9 for the second.
The number of permutations for three bricks from 10 is 10 x 9 x 8.
The number of permutations for r bricks from a set of n bricks is n!/(n-r)!
If using all 10, the number of different permutations is 10! = 3 628 800
If you have only 10 different parts then the number of combinations isn't 10^10.
Assuming the parts can only be bolted end-on-end, the number of permutations for one brick out of 10 is 10.
The number of permutations for two bricks out of 10 is 10 x 9. You have a choice of 10 bricks for the first and 9 for the second.
The number of permutations for three bricks from 10 is 10 x 9 x 8.
The number of permutations for r bricks from a set of n bricks is n!/(n-r)!
If using all 10, the number of different permutations is 10! = 3 628 800
Edited by V8LM on Thursday 1st December 20:18
Mother nature has similar problems. Think about this:
Amino acids are the Lego building blocks of proteins, and proteins are the building blocks of life.
There are 20 different types of naturally-occurring amino acid (brick).
These bricks are linked end-on-end to form a string, and this string curls up to form a protein.
The average length of the string making up a protein in the human genome is around 400 amino acids.
So, how many combinations of 400 amino acid protein sequences are there?
If you were to go into a lab and make one of each, how big a test-tube would you need and what would it weigh if containing only one of each?
Amino acids are the Lego building blocks of proteins, and proteins are the building blocks of life.
There are 20 different types of naturally-occurring amino acid (brick).
These bricks are linked end-on-end to form a string, and this string curls up to form a protein.
The average length of the string making up a protein in the human genome is around 400 amino acids.
So, how many combinations of 400 amino acid protein sequences are there?
If you were to go into a lab and make one of each, how big a test-tube would you need and what would it weigh if containing only one of each?
Some extra food for thought then:
A protein is a chain of amino acids linked into a string. A very simple model of the links is that there are three different configurations for each link. A di-peptide (two amino acids joined together) can adopt three conformations; a tri-peptide can adopt nine conformations; etc. The average length of a protein in the human genome is 400 amino acids. How many conformations can each protein adopt in this simple model?
If one were to try each conformation, and could do so as fast as possible (one conformation per Planck time) then how long would it take, on average, to find the most stable structure?
Mother Nature folds the string into the same structure (almost) each and every time in a fraction of a second!
A protein is a chain of amino acids linked into a string. A very simple model of the links is that there are three different configurations for each link. A di-peptide (two amino acids joined together) can adopt three conformations; a tri-peptide can adopt nine conformations; etc. The average length of a protein in the human genome is 400 amino acids. How many conformations can each protein adopt in this simple model?
If one were to try each conformation, and could do so as fast as possible (one conformation per Planck time) then how long would it take, on average, to find the most stable structure?
Mother Nature folds the string into the same structure (almost) each and every time in a fraction of a second!
Edited by V8LM on Friday 2nd December 05:45
V8LM said:
Some extra food for thought then:
A protein is a chain of amino acids linked into a string. A very simple model of the links is that there are three different configurations for each link. A di-peptide (two amino acids joined together) can adopt three conformations; a tri-peptide can adopt nine conformations; etc. The average length of a protein in the human genome is 400 amino acids. How many conformations can each protein adopt in this simple model?
If one were to try each conformation, and could do so as fast as possible (one conformation per Planck time) then how long would it take, on average, to find the most stable structure?
Mother Nature folds the string into the same structure (almost) each and every time in a fraction of a second!
And why we have folding@home A protein is a chain of amino acids linked into a string. A very simple model of the links is that there are three different configurations for each link. A di-peptide (two amino acids joined together) can adopt three conformations; a tri-peptide can adopt nine conformations; etc. The average length of a protein in the human genome is 400 amino acids. How many conformations can each protein adopt in this simple model?
If one were to try each conformation, and could do so as fast as possible (one conformation per Planck time) then how long would it take, on average, to find the most stable structure?
Mother Nature folds the string into the same structure (almost) each and every time in a fraction of a second!
I saw a "Brickumentary" recently and there's been a study on how many different ways you can connect 6 2x4 Lego Blocks. Some mathematician came back with close to a million different combinations just from those 6 bricks, add another brick and the number goes through the roof (has yet to be accurately calculated I believe).
AshVX220 said:
I saw a "Brickumentary" recently and there's been a study on how many different ways you can connect 6 2x4 Lego Blocks. Some mathematician came back with close to a million different combinations just from those 6 bricks, add another brick and the number goes through the roof (has yet to be accurately calculated I believe).
I think there are 46 ways to add a 2x4 brick to a 2x4 brick, so with six bricks thats 46^5 = 205 962 976. Many of these are symmetric though, but even if we discount half of the 46 it's way more than 1 million (23^5 = 6 436 343) (I think).ETA: As there is only one plane of symmetry there are probably 205 962 976 / 2 = 102 981 488 different shapes, but happy to be corrected.
ETATA: In fact, there are far more than that. 102 981 488 is the number of different shapes if you add them one on top of the next. Of course, with Lego, you can add the third to either the top or the bottom of the second, so there are more than 46 combinations for each bricks > 2. And you can add blocks to any of the previous. Then we have to calculate the excluded volume - not all of the combinations are possible as the bricks would overlap.
ETATATA: And this assumes the blocks connected by a single lug form right angles. There are an infinite number of shapes that can be formed with just two blocks if they connect at their corner lug and are allowed to adopt any rotational angle.
ETATATATA: The number also depends on whether one is allowed to loosen the structure to add a new brick assuming building one brick at a time.
ETATATATATA: I'm off.
Edited by V8LM on Monday 5th December 22:46
Surely the answer depends on the type of part as well as the number.
If you had a two 2x2 bricks that are different colours (say red and blue) - then there are 6 unique ways they can be connected (ignoring rotation around the connection point).
You could stack the red directly on top of the blue using all 4 connection points - and vice versa.
You could stack the red on top of the blue using only two connection points (to make a step) - and vice versa.
You could stack the red on top of the blue using only one connection point (i.e. a corner step) - and vice versa.
As the size of the bricks goes up (and hence the number of connection points on each brick) - the number of possible permutations increases massively - even for two bricks.
If you had a two 2x2 bricks that are different colours (say red and blue) - then there are 6 unique ways they can be connected (ignoring rotation around the connection point).
You could stack the red directly on top of the blue using all 4 connection points - and vice versa.
You could stack the red on top of the blue using only two connection points (to make a step) - and vice versa.
You could stack the red on top of the blue using only one connection point (i.e. a corner step) - and vice versa.
As the size of the bricks goes up (and hence the number of connection points on each brick) - the number of possible permutations increases massively - even for two bricks.
Sorry to drag this up again, but seeing some results which I don't understand, for ten each of x25 different shaped blocks.
1.551121e+25<this was the result of googling "25!"
Can anyone put me right again please? Just to clarify, trying to determine the number of combinations of shapes that could be made with 25 different blocks, x10 of each.
1.551121e+25<this was the result of googling "25!"
Can anyone put me right again please? Just to clarify, trying to determine the number of combinations of shapes that could be made with 25 different blocks, x10 of each.
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