Mixing
Possible combinations of placing 14 different colors in a matrix 8x8
$begingroup$
I have a matrix *8*x*8* where i can place color bricks. I can place bricks everywhere except on all four edges - So I have 60 different places where I can place bricks. I have 14 different colors of bricks and do not need to fill all 60 places with bricks (So I can leave all places empty or I can fill them all).
I want to calculate all different combinations I can create. I know that with 1 color I can create 60! different combinations.
But what happens when I start to mix colors on a matix? How can I calculate the right number of possibilities?
Thanks
matrices permutations combinations
asked Jan 21 at 17:30
Hugo ZupanHugo Zupan
31
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add a comment |
$begingroup$
I have a matrix *8*x*8* where i can place color bricks. I can place bricks everywhere except on all four edges - So I have 60 different places where I can place bricks. I have 14 different colors of bricks and do not need to fill all 60 places with bricks (So I can leave all places empty or I can fill them all).
I want to calculate all different combinations I can create. I know that with 1 color I can create 60! different combinations.
But what happens when I start to mix colors on a matix? How can I calculate the right number of possibilities?
Thanks
matrices permutations combinations
asked Jan 21 at 17:30
Hugo ZupanHugo Zupan
31
$endgroup$
add a comment |
$begingroup$
I have a matrix *8*x*8* where i can place color bricks. I can place bricks everywhere except on all four edges - So I have 60 different places where I can place bricks. I have 14 different colors of bricks and do not need to fill all 60 places with bricks (So I can leave all places empty or I can fill them all).
I want to calculate all different combinations I can create. I know that with 1 color I can create 60! different combinations.
But what happens when I start to mix colors on a matix? How can I calculate the right number of possibilities?
Thanks
matrices permutations combinations
asked Jan 21 at 17:30
Hugo ZupanHugo Zupan
31
$endgroup$
I have a matrix *8*x*8* where i can place color bricks. I can place bricks everywhere except on all four edges - So I have 60 different places where I can place bricks. I have 14 different colors of bricks and do not need to fill all 60 places with bricks (So I can leave all places empty or I can fill them all).
I want to calculate all different combinations I can create. I know that with 1 color I can create 60! different combinations.
But what happens when I start to mix colors on a matix? How can I calculate the right number of possibilities?
Thanks
matrices permutations combinations
matrices permutations combinations
asked Jan 21 at 17:30
Hugo ZupanHugo Zupan
31
asked Jan 21 at 17:30
Hugo ZupanHugo Zupan
31
asked Jan 21 at 17:30
Hugo ZupanHugo Zupan
31
asked Jan 21 at 17:30
Hugo ZupanHugo Zupan
31
asked Jan 21 at 17:30
Hugo ZupanHugo Zupan
31
31
add a comment |
add a comment |
1 Answer
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The answer is $(k+1)^{60}$, where $k$ is the number of colours you want to use. To see why this is so, consider a single cell: it may take $k$ different values for the different colours and an additional value if the cell is empty. Since the choice of each cell is independent of the others we get a multiplication. A single combination can be written as a vector: $v in {0,...,k}^{60}$. Note that the size of the set is exactly $(k+1)^{60}$.
Note that when you said edges you meant corners.
answered Jan 21 at 17:48
lightxbulblightxbulb
1,040311
$endgroup$
$begingroup$
Thank you very much. Just one more question. If I would have just one color I could create 60! (8.3209871e+81) different combinations. With your formula I can create with all colors 3.6768469e+70 different combinations. So I get less combinations with 14 colors than with 1?
$endgroup$
– Hugo Zupan
Jan 22 at 9:23
$begingroup$
No you are right. Because some combinations repeat and it is same combination. Thank you again.
$endgroup$
– Hugo Zupan
Jan 22 at 9:31
$begingroup$
@Hugo Zupan You can mark it as answered if it answers your question.
$endgroup$
– lightxbulb
Jan 22 at 12:03
add a comment |
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
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oldest
votes
$begingroup$
The answer is $(k+1)^{60}$, where $k$ is the number of colours you want to use. To see why this is so, consider a single cell: it may take $k$ different values for the different colours and an additional value if the cell is empty. Since the choice of each cell is independent of the others we get a multiplication. A single combination can be written as a vector: $v in {0,...,k}^{60}$. Note that the size of the set is exactly $(k+1)^{60}$.
Note that when you said edges you meant corners.
answered Jan 21 at 17:48
lightxbulblightxbulb
1,040311
$endgroup$
$begingroup$
Thank you very much. Just one more question. If I would have just one color I could create 60! (8.3209871e+81) different combinations. With your formula I can create with all colors 3.6768469e+70 different combinations. So I get less combinations with 14 colors than with 1?
$endgroup$
– Hugo Zupan
Jan 22 at 9:23
$begingroup$
No you are right. Because some combinations repeat and it is same combination. Thank you again.
$endgroup$
– Hugo Zupan
Jan 22 at 9:31
$begingroup$
@Hugo Zupan You can mark it as answered if it answers your question.
$endgroup$
– lightxbulb
Jan 22 at 12:03
add a comment |
$begingroup$
The answer is $(k+1)^{60}$, where $k$ is the number of colours you want to use. To see why this is so, consider a single cell: it may take $k$ different values for the different colours and an additional value if the cell is empty. Since the choice of each cell is independent of the others we get a multiplication. A single combination can be written as a vector: $v in {0,...,k}^{60}$. Note that the size of the set is exactly $(k+1)^{60}$.
Note that when you said edges you meant corners.
answered Jan 21 at 17:48
lightxbulblightxbulb
1,040311
$endgroup$
$begingroup$
Thank you very much. Just one more question. If I would have just one color I could create 60! (8.3209871e+81) different combinations. With your formula I can create with all colors 3.6768469e+70 different combinations. So I get less combinations with 14 colors than with 1?
$endgroup$
– Hugo Zupan
Jan 22 at 9:23
$begingroup$
No you are right. Because some combinations repeat and it is same combination. Thank you again.
$endgroup$
– Hugo Zupan
Jan 22 at 9:31
$begingroup$
@Hugo Zupan You can mark it as answered if it answers your question.
$endgroup$
– lightxbulb
Jan 22 at 12:03
add a comment |
$begingroup$
The answer is $(k+1)^{60}$, where $k$ is the number of colours you want to use. To see why this is so, consider a single cell: it may take $k$ different values for the different colours and an additional value if the cell is empty. Since the choice of each cell is independent of the others we get a multiplication. A single combination can be written as a vector: $v in {0,...,k}^{60}$. Note that the size of the set is exactly $(k+1)^{60}$.
Note that when you said edges you meant corners.
answered Jan 21 at 17:48
lightxbulblightxbulb
1,040311
$endgroup$
The answer is $(k+1)^{60}$, where $k$ is the number of colours you want to use. To see why this is so, consider a single cell: it may take $k$ different values for the different colours and an additional value if the cell is empty. Since the choice of each cell is independent of the others we get a multiplication. A single combination can be written as a vector: $v in {0,...,k}^{60}$. Note that the size of the set is exactly $(k+1)^{60}$.
Note that when you said edges you meant corners.
answered Jan 21 at 17:48
lightxbulblightxbulb
1,040311
answered Jan 21 at 17:48
lightxbulblightxbulb
1,040311
answered Jan 21 at 17:48
lightxbulblightxbulb
1,040311
answered Jan 21 at 17:48
lightxbulblightxbulb
1,040311
1,040311
$begingroup$
Thank you very much. Just one more question. If I would have just one color I could create 60! (8.3209871e+81) different combinations. With your formula I can create with all colors 3.6768469e+70 different combinations. So I get less combinations with 14 colors than with 1?
$endgroup$
– Hugo Zupan
Jan 22 at 9:23
$begingroup$
No you are right. Because some combinations repeat and it is same combination. Thank you again.
$endgroup$
– Hugo Zupan
Jan 22 at 9:31
$begingroup$
@Hugo Zupan You can mark it as answered if it answers your question.
$endgroup$
– lightxbulb
Jan 22 at 12:03
add a comment |
$begingroup$
Thank you very much. Just one more question. If I would have just one color I could create 60! (8.3209871e+81) different combinations. With your formula I can create with all colors 3.6768469e+70 different combinations. So I get less combinations with 14 colors than with 1?
$endgroup$
– Hugo Zupan
Jan 22 at 9:23
$begingroup$
No you are right. Because some combinations repeat and it is same combination. Thank you again.
$endgroup$
– Hugo Zupan
Jan 22 at 9:31
$begingroup$
@Hugo Zupan You can mark it as answered if it answers your question.
$endgroup$
– lightxbulb
Jan 22 at 12:03
$begingroup$
Thank you very much. Just one more question. If I would have just one color I could create 60! (8.3209871e+81) different combinations. With your formula I can create with all colors 3.6768469e+70 different combinations. So I get less combinations with 14 colors than with 1?
$endgroup$
– Hugo Zupan
Jan 22 at 9:23
$begingroup$
Thank you very much. Just one more question. If I would have just one color I could create 60! (8.3209871e+81) different combinations. With your formula I can create with all colors 3.6768469e+70 different combinations. So I get less combinations with 14 colors than with 1?
$endgroup$
– Hugo Zupan
Jan 22 at 9:23
$begingroup$
No you are right. Because some combinations repeat and it is same combination. Thank you again.
$endgroup$
– Hugo Zupan
Jan 22 at 9:31
$begingroup$
No you are right. Because some combinations repeat and it is same combination. Thank you again.
$endgroup$
– Hugo Zupan
Jan 22 at 9:31
$begingroup$
@Hugo Zupan You can mark it as answered if it answers your question.
$endgroup$
– lightxbulb
Jan 22 at 12:03
$begingroup$
@Hugo Zupan You can mark it as answered if it answers your question.
$endgroup$
– lightxbulb
Jan 22 at 12:03
add a comment |
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