Rhythm combinations in a 16 notes bar.
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How many combinations are possible between a note-event and a silence-event (i.e. a played note and a pause) over 16 consecutive beats? In other words, if one musical bar is divided into 16 parts (16th notes) of equal duration, filled with either a sound or a silence, how many combinations of the two are possible within the given structure of 16 events?
I hope this is clear enough...sorry, musician here, therefore almost completely unprepared to approach the question. I am not even sure if I am asking about combinations or permutations.
An example: (1= sound 0= pause) .
1000 0010 0011 0011
1111 0000 0111 1010
combinatorics
edited 1 hour ago
Hans Lundmark
34.3k564110
asked 2 hours ago
Ale
211
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up vote
4
down vote
favorite
How many combinations are possible between a note-event and a silence-event (i.e. a played note and a pause) over 16 consecutive beats? In other words, if one musical bar is divided into 16 parts (16th notes) of equal duration, filled with either a sound or a silence, how many combinations of the two are possible within the given structure of 16 events?
I hope this is clear enough...sorry, musician here, therefore almost completely unprepared to approach the question. I am not even sure if I am asking about combinations or permutations.
An example: (1= sound 0= pause) .
1000 0010 0011 0011
1111 0000 0111 1010
combinatorics
edited 1 hour ago
Hans Lundmark
34.3k564110
asked 2 hours ago
Ale
211
New contributor
Ale is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Possible duplicate of Musical and combinatorial proof
– Micah
2 hours ago
3
The linked question applies under the musically reasonable condition that multiple adjacent rests are identical to a longer rest, but multiple adjacent notes are not (i.e., two quarter rests are the same as a half rest, but two quarter notes are different from a half note). If that's what you want, then there are $F(2n+1)$ possible $n$-beat sequences, where $F(k)$ is the $k$th Fibonacci number. So in your case there are $F(33)=3524578$ possible 16-beat sequences.
– Micah
2 hours ago
add a comment |
up vote
4
down vote
favorite
up vote
4
down vote
favorite
How many combinations are possible between a note-event and a silence-event (i.e. a played note and a pause) over 16 consecutive beats? In other words, if one musical bar is divided into 16 parts (16th notes) of equal duration, filled with either a sound or a silence, how many combinations of the two are possible within the given structure of 16 events?
I hope this is clear enough...sorry, musician here, therefore almost completely unprepared to approach the question. I am not even sure if I am asking about combinations or permutations.
An example: (1= sound 0= pause) .
1000 0010 0011 0011
1111 0000 0111 1010
combinatorics
edited 1 hour ago
Hans Lundmark
34.3k564110
asked 2 hours ago
Ale
211
New contributor
Ale is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
How many combinations are possible between a note-event and a silence-event (i.e. a played note and a pause) over 16 consecutive beats? In other words, if one musical bar is divided into 16 parts (16th notes) of equal duration, filled with either a sound or a silence, how many combinations of the two are possible within the given structure of 16 events?
I hope this is clear enough...sorry, musician here, therefore almost completely unprepared to approach the question. I am not even sure if I am asking about combinations or permutations.
An example: (1= sound 0= pause) .
1000 0010 0011 0011
1111 0000 0111 1010
combinatorics
combinatorics
edited 1 hour ago
Hans Lundmark
34.3k564110
asked 2 hours ago
Ale
211
New contributor
Ale is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 1 hour ago
Hans Lundmark
34.3k564110
asked 2 hours ago
Ale
211
New contributor
Ale is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 1 hour ago
Hans Lundmark
34.3k564110
edited 1 hour ago
Hans Lundmark
34.3k564110
edited 1 hour ago
Hans Lundmark
34.3k564110
34.3k564110
asked 2 hours ago
Ale
211
New contributor
Ale is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 hours ago
Ale
211
asked 2 hours ago
Ale
211
211
New contributor
Ale is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Ale is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Ale is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Possible duplicate of Musical and combinatorial proof
– Micah
2 hours ago
3
The linked question applies under the musically reasonable condition that multiple adjacent rests are identical to a longer rest, but multiple adjacent notes are not (i.e., two quarter rests are the same as a half rest, but two quarter notes are different from a half note). If that's what you want, then there are $F(2n+1)$ possible $n$-beat sequences, where $F(k)$ is the $k$th Fibonacci number. So in your case there are $F(33)=3524578$ possible 16-beat sequences.
– Micah
2 hours ago
add a comment |
Possible duplicate of Musical and combinatorial proof
– Micah
2 hours ago
3
The linked question applies under the musically reasonable condition that multiple adjacent rests are identical to a longer rest, but multiple adjacent notes are not (i.e., two quarter rests are the same as a half rest, but two quarter notes are different from a half note). If that's what you want, then there are $F(2n+1)$ possible $n$-beat sequences, where $F(k)$ is the $k$th Fibonacci number. So in your case there are $F(33)=3524578$ possible 16-beat sequences.
– Micah
2 hours ago
Possible duplicate of Musical and combinatorial proof
– Micah
2 hours ago
Possible duplicate of Musical and combinatorial proof
– Micah
2 hours ago
3
3
The linked question applies under the musically reasonable condition that multiple adjacent rests are identical to a longer rest, but multiple adjacent notes are not (i.e., two quarter rests are the same as a half rest, but two quarter notes are different from a half note). If that's what you want, then there are $F(2n+1)$ possible $n$-beat sequences, where $F(k)$ is the $k$th Fibonacci number. So in your case there are $F(33)=3524578$ possible 16-beat sequences.
– Micah
2 hours ago
The linked question applies under the musically reasonable condition that multiple adjacent rests are identical to a longer rest, but multiple adjacent notes are not (i.e., two quarter rests are the same as a half rest, but two quarter notes are different from a half note). If that's what you want, then there are $F(2n+1)$ possible $n$-beat sequences, where $F(k)$ is the $k$th Fibonacci number. So in your case there are $F(33)=3524578$ possible 16-beat sequences.
– Micah
2 hours ago
add a comment |
1 Answer
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up vote
3
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The number of $16$ long sequences of $0$ and $1$ is $2^{16} = 65536$. I think that's what you are asking for.
answered 2 hours ago
Ethan Bolker
38.8k543102
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
The number of $16$ long sequences of $0$ and $1$ is $2^{16} = 65536$. I think that's what you are asking for.
answered 2 hours ago
Ethan Bolker
38.8k543102
add a comment |
up vote
3
down vote
The number of $16$ long sequences of $0$ and $1$ is $2^{16} = 65536$. I think that's what you are asking for.
answered 2 hours ago
Ethan Bolker
38.8k543102
add a comment |
up vote
3
down vote
up vote
3
down vote
The number of $16$ long sequences of $0$ and $1$ is $2^{16} = 65536$. I think that's what you are asking for.
answered 2 hours ago
Ethan Bolker
38.8k543102
The number of $16$ long sequences of $0$ and $1$ is $2^{16} = 65536$. I think that's what you are asking for.
answered 2 hours ago
Ethan Bolker
38.8k543102
answered 2 hours ago
Ethan Bolker
38.8k543102
answered 2 hours ago
Ethan Bolker
38.8k543102
answered 2 hours ago
Ethan Bolker
38.8k543102
38.8k543102
add a comment |
add a comment |
Ale is a new contributor. Be nice, and check out our Code of Conduct.
Ale is a new contributor. Be nice, and check out our Code of Conduct.
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2XeHxSjzcYV6sIAwT2vxMmDd7oA9yxC yKUie
Possible duplicate of Musical and combinatorial proof
– Micah
2 hours ago
3
The linked question applies under the musically reasonable condition that multiple adjacent rests are identical to a longer rest, but multiple adjacent notes are not (i.e., two quarter rests are the same as a half rest, but two quarter notes are different from a half note). If that's what you want, then there are $F(2n+1)$ possible $n$-beat sequences, where $F(k)$ is the $k$th Fibonacci number. So in your case there are $F(33)=3524578$ possible 16-beat sequences.
– Micah
2 hours ago