How many $ntimes n$ binary matrices are there, such that at least one row is filled by only $0'$s? [closed]
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How many $ntimes n$ binary matrices (values are only $0'$s and $1'$s) are there, such that at least one row is filled by only $0'$s?
How to approach this problem?
combinatorics matrices
edited Dec 7 '18 at 11:40
Namaste
1
asked Dec 7 '18 at 11:28
lellerleller
715
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closed as off-topic by Did, mrtaurho, Saad, user91500, Key Flex Dec 26 '18 at 8:22
This question appears to be off-topic. The users who voted to close gave this specific reason:
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How many $ntimes n$ binary matrices (values are only $0'$s and $1'$s) are there, such that at least one row is filled by only $0'$s?
How to approach this problem?
combinatorics matrices
edited Dec 7 '18 at 11:40
Namaste
1
asked Dec 7 '18 at 11:28
lellerleller
715
$endgroup$
closed as off-topic by Did, mrtaurho, Saad, user91500, Key Flex Dec 26 '18 at 8:22
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Did, mrtaurho, Saad, user91500, Key Flex
If this question can be reworded to fit the rules in the help center, please edit the question.
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This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc.
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– Did
Dec 23 '18 at 21:35
add a comment |
$begingroup$
How many $ntimes n$ binary matrices (values are only $0'$s and $1'$s) are there, such that at least one row is filled by only $0'$s?
How to approach this problem?
combinatorics matrices
edited Dec 7 '18 at 11:40
Namaste
1
asked Dec 7 '18 at 11:28
lellerleller
715
$endgroup$
How many $ntimes n$ binary matrices (values are only $0'$s and $1'$s) are there, such that at least one row is filled by only $0'$s?
How to approach this problem?
combinatorics matrices
combinatorics matrices
edited Dec 7 '18 at 11:40
Namaste
1
asked Dec 7 '18 at 11:28
lellerleller
715
edited Dec 7 '18 at 11:40
Namaste
1
asked Dec 7 '18 at 11:28
lellerleller
715
edited Dec 7 '18 at 11:40
Namaste
1
edited Dec 7 '18 at 11:40
Namaste
1
edited Dec 7 '18 at 11:40
Namaste
1
1
asked Dec 7 '18 at 11:28
lellerleller
715
asked Dec 7 '18 at 11:28
lellerleller
715
asked Dec 7 '18 at 11:28
lellerleller
715
715
closed as off-topic by Did, mrtaurho, Saad, user91500, Key Flex Dec 26 '18 at 8:22
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Did, mrtaurho, Saad, user91500, Key Flex
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Did, mrtaurho, Saad, user91500, Key Flex Dec 26 '18 at 8:22
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Did, mrtaurho, Saad, user91500, Key Flex
If this question can be reworded to fit the rules in the help center, please edit the question.
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This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc.
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– Did
Dec 23 '18 at 21:35
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– Did
Dec 23 '18 at 21:35
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– Did
Dec 23 '18 at 21:35
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1 Answer
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The number of binary matrices with at-least one "zero" row $=$ Total no. of binary matrices $-$ No. of binary matrices with no "zero" row
Total no. of binary matrices $=2^{n^2}because$ the matrix contains $n^2$ entries each of which can be $0,1$.
For the second part, observe each row of the binary matrix can be set up in $2^n$ ways by virtue of having $n$ entries that may be $0,1$. Each row is a non-"zero" row, so out of the $2^n$ possibilities for a given row, all but one (that is the the "zero" row) are admissible. So, each row can be set up in $2^n-1$ ways. Since there are $n$ rows, no. of binary matrices with no "zero" row $=(2^n-1)^n$.
This answer is equal to $2^{n^2}-(2^n-1)^n$.
answered Dec 7 '18 at 11:45
Shubham JohriShubham Johri
5,204718
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The number of binary matrices with at-least one "zero" row $=$ Total no. of binary matrices $-$ No. of binary matrices with no "zero" row
Total no. of binary matrices $=2^{n^2}because$ the matrix contains $n^2$ entries each of which can be $0,1$.
For the second part, observe each row of the binary matrix can be set up in $2^n$ ways by virtue of having $n$ entries that may be $0,1$. Each row is a non-"zero" row, so out of the $2^n$ possibilities for a given row, all but one (that is the the "zero" row) are admissible. So, each row can be set up in $2^n-1$ ways. Since there are $n$ rows, no. of binary matrices with no "zero" row $=(2^n-1)^n$.
This answer is equal to $2^{n^2}-(2^n-1)^n$.
answered Dec 7 '18 at 11:45
Shubham JohriShubham Johri
5,204718
$endgroup$
add a comment |
$begingroup$
The number of binary matrices with at-least one "zero" row $=$ Total no. of binary matrices $-$ No. of binary matrices with no "zero" row
Total no. of binary matrices $=2^{n^2}because$ the matrix contains $n^2$ entries each of which can be $0,1$.
For the second part, observe each row of the binary matrix can be set up in $2^n$ ways by virtue of having $n$ entries that may be $0,1$. Each row is a non-"zero" row, so out of the $2^n$ possibilities for a given row, all but one (that is the the "zero" row) are admissible. So, each row can be set up in $2^n-1$ ways. Since there are $n$ rows, no. of binary matrices with no "zero" row $=(2^n-1)^n$.
This answer is equal to $2^{n^2}-(2^n-1)^n$.
answered Dec 7 '18 at 11:45
Shubham JohriShubham Johri
5,204718
$endgroup$
add a comment |
$begingroup$
The number of binary matrices with at-least one "zero" row $=$ Total no. of binary matrices $-$ No. of binary matrices with no "zero" row
Total no. of binary matrices $=2^{n^2}because$ the matrix contains $n^2$ entries each of which can be $0,1$.
For the second part, observe each row of the binary matrix can be set up in $2^n$ ways by virtue of having $n$ entries that may be $0,1$. Each row is a non-"zero" row, so out of the $2^n$ possibilities for a given row, all but one (that is the the "zero" row) are admissible. So, each row can be set up in $2^n-1$ ways. Since there are $n$ rows, no. of binary matrices with no "zero" row $=(2^n-1)^n$.
This answer is equal to $2^{n^2}-(2^n-1)^n$.
answered Dec 7 '18 at 11:45
Shubham JohriShubham Johri
5,204718
$endgroup$
The number of binary matrices with at-least one "zero" row $=$ Total no. of binary matrices $-$ No. of binary matrices with no "zero" row
Total no. of binary matrices $=2^{n^2}because$ the matrix contains $n^2$ entries each of which can be $0,1$.
For the second part, observe each row of the binary matrix can be set up in $2^n$ ways by virtue of having $n$ entries that may be $0,1$. Each row is a non-"zero" row, so out of the $2^n$ possibilities for a given row, all but one (that is the the "zero" row) are admissible. So, each row can be set up in $2^n-1$ ways. Since there are $n$ rows, no. of binary matrices with no "zero" row $=(2^n-1)^n$.
This answer is equal to $2^{n^2}-(2^n-1)^n$.
answered Dec 7 '18 at 11:45
Shubham JohriShubham Johri
5,204718
answered Dec 7 '18 at 11:45
Shubham JohriShubham Johri
5,204718
answered Dec 7 '18 at 11:45
Shubham JohriShubham Johri
5,204718
answered Dec 7 '18 at 11:45
Shubham JohriShubham Johri
5,204718
5,204718
add a comment |
add a comment |
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This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc.
$endgroup$
– Did
Dec 23 '18 at 21:35