How many sequences of 11 ones and 7 zeroes without consecutive zeroes exist? [closed]
$begingroup$
I've been looking for solutions to such a problem, but only found solutions for any amount of zeroes and ones (without consecutive zeroes).
How many such sequences exist?
combinatorics
asked Jan 1 at 15:07
E.R.E.R.
81
$endgroup$
closed as off-topic by NCh, Namaste, hardmath, Jyrki Lahtonen, Paul Frost Jan 1 at 21:55
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." – NCh, Namaste, hardmath, Jyrki Lahtonen, Paul Frost
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
I've been looking for solutions to such a problem, but only found solutions for any amount of zeroes and ones (without consecutive zeroes).
How many such sequences exist?
combinatorics
asked Jan 1 at 15:07
E.R.E.R.
81
$endgroup$
closed as off-topic by NCh, Namaste, hardmath, Jyrki Lahtonen, Paul Frost Jan 1 at 21:55
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." – NCh, Namaste, hardmath, Jyrki Lahtonen, Paul Frost
If this question can be reworded to fit the rules in the help center, please edit the question.
4
$begingroup$
Well, it's more or less a Stars and Bars problem. Put spaces before, after, and between the seven $0's$. The between ones must have at least one $1$ so put one in each. Now you have a standard problem. Can you finish from here?
$endgroup$
– lulu
Jan 1 at 15:09
$begingroup$
One way to add context to a "bare problem statement" is by working out some small examples. If eleven ones and seven zeros is too hard, how about cutting the size of the problem roughly in half? Often the process of working out an example will suggest methods to attack the general problem. In any event it will give your Readers a clear idea of the extent to which you have digested the problem statement. Other ways of adding context are also possible...
$endgroup$
– hardmath
Jan 1 at 19:04
add a comment |
$begingroup$
I've been looking for solutions to such a problem, but only found solutions for any amount of zeroes and ones (without consecutive zeroes).
How many such sequences exist?
combinatorics
asked Jan 1 at 15:07
E.R.E.R.
81
$endgroup$
I've been looking for solutions to such a problem, but only found solutions for any amount of zeroes and ones (without consecutive zeroes).
How many such sequences exist?
combinatorics
combinatorics
asked Jan 1 at 15:07
E.R.E.R.
81
asked Jan 1 at 15:07
E.R.E.R.
81
asked Jan 1 at 15:07
E.R.E.R.
81
asked Jan 1 at 15:07
E.R.E.R.
81
asked Jan 1 at 15:07
E.R.E.R.
81
81
closed as off-topic by NCh, Namaste, hardmath, Jyrki Lahtonen, Paul Frost Jan 1 at 21:55
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." – NCh, Namaste, hardmath, Jyrki Lahtonen, Paul Frost
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by NCh, Namaste, hardmath, Jyrki Lahtonen, Paul Frost Jan 1 at 21:55
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." – NCh, Namaste, hardmath, Jyrki Lahtonen, Paul Frost
If this question can be reworded to fit the rules in the help center, please edit the question.
4
$begingroup$
Well, it's more or less a Stars and Bars problem. Put spaces before, after, and between the seven $0's$. The between ones must have at least one $1$ so put one in each. Now you have a standard problem. Can you finish from here?
$endgroup$
– lulu
Jan 1 at 15:09
$begingroup$
One way to add context to a "bare problem statement" is by working out some small examples. If eleven ones and seven zeros is too hard, how about cutting the size of the problem roughly in half? Often the process of working out an example will suggest methods to attack the general problem. In any event it will give your Readers a clear idea of the extent to which you have digested the problem statement. Other ways of adding context are also possible...
$endgroup$
– hardmath
Jan 1 at 19:04
add a comment |
4
$begingroup$
Well, it's more or less a Stars and Bars problem. Put spaces before, after, and between the seven $0's$. The between ones must have at least one $1$ so put one in each. Now you have a standard problem. Can you finish from here?
$endgroup$
– lulu
Jan 1 at 15:09
$begingroup$
One way to add context to a "bare problem statement" is by working out some small examples. If eleven ones and seven zeros is too hard, how about cutting the size of the problem roughly in half? Often the process of working out an example will suggest methods to attack the general problem. In any event it will give your Readers a clear idea of the extent to which you have digested the problem statement. Other ways of adding context are also possible...
$endgroup$
– hardmath
Jan 1 at 19:04
4
4
$begingroup$
Well, it's more or less a Stars and Bars problem. Put spaces before, after, and between the seven $0's$. The between ones must have at least one $1$ so put one in each. Now you have a standard problem. Can you finish from here?
$endgroup$
– lulu
Jan 1 at 15:09
$begingroup$
Well, it's more or less a Stars and Bars problem. Put spaces before, after, and between the seven $0's$. The between ones must have at least one $1$ so put one in each. Now you have a standard problem. Can you finish from here?
$endgroup$
– lulu
Jan 1 at 15:09
$begingroup$
One way to add context to a "bare problem statement" is by working out some small examples. If eleven ones and seven zeros is too hard, how about cutting the size of the problem roughly in half? Often the process of working out an example will suggest methods to attack the general problem. In any event it will give your Readers a clear idea of the extent to which you have digested the problem statement. Other ways of adding context are also possible...
$endgroup$
– hardmath
Jan 1 at 19:04
$begingroup$
One way to add context to a "bare problem statement" is by working out some small examples. If eleven ones and seven zeros is too hard, how about cutting the size of the problem roughly in half? Often the process of working out an example will suggest methods to attack the general problem. In any event it will give your Readers a clear idea of the extent to which you have digested the problem statement. Other ways of adding context are also possible...
$endgroup$
– hardmath
Jan 1 at 19:04
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Place the eleven ones out first and leave a bit of space to either side of the ones.
$$underbrace{~}_{a}1underbrace{~}_{b}1underbrace{~}_{c}1underbrace{~}_{d}1underbrace{~}_{e}1underbrace{~}_{f}1underbrace{~}_{g}1underbrace{~}_{h}1underbrace{~}_{i}1underbrace{~}_{j}1underbrace{~}_{k}1underbrace{~}_{ell}$$
Now, pick which seven of the spaces are used by zeroes and close all the remaining unused spaces. Choose the spaces simultaneously. To count this, use binomial coefficients. We have twelve spaces available (11 + 1) and we want to choose seven of them to be used by zeroes.
The answer then is $$binom{12}{7}$$
By choosing in this fashion, either zero or one $0$ will be in each space and as a result no $0$ will be next to another $0$.
answered Jan 1 at 15:30
JMoravitzJMoravitz
49.4k44091
$endgroup$
add a comment |
$begingroup$
Take such a sequence and stick a $1$ at the end. You have a sequence
of $12$ ones and $7$ zeroes with each zero followed by a $1$. So you have
seven subwords $01$. Replace each of those by a $2$. You then have a sequence
of $5$ ones and seven twos. There are $binom{12}5$ of these.
You can reverse the process. So there are $binom{12}5$ of your sort of sequence.
answered Jan 1 at 15:26
Lord Shark the UnknownLord Shark the Unknown
109k1163136
$endgroup$
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Place the eleven ones out first and leave a bit of space to either side of the ones.
$$underbrace{~}_{a}1underbrace{~}_{b}1underbrace{~}_{c}1underbrace{~}_{d}1underbrace{~}_{e}1underbrace{~}_{f}1underbrace{~}_{g}1underbrace{~}_{h}1underbrace{~}_{i}1underbrace{~}_{j}1underbrace{~}_{k}1underbrace{~}_{ell}$$
Now, pick which seven of the spaces are used by zeroes and close all the remaining unused spaces. Choose the spaces simultaneously. To count this, use binomial coefficients. We have twelve spaces available (11 + 1) and we want to choose seven of them to be used by zeroes.
The answer then is $$binom{12}{7}$$
By choosing in this fashion, either zero or one $0$ will be in each space and as a result no $0$ will be next to another $0$.
answered Jan 1 at 15:30
JMoravitzJMoravitz
49.4k44091
$endgroup$
add a comment |
$begingroup$
Place the eleven ones out first and leave a bit of space to either side of the ones.
$$underbrace{~}_{a}1underbrace{~}_{b}1underbrace{~}_{c}1underbrace{~}_{d}1underbrace{~}_{e}1underbrace{~}_{f}1underbrace{~}_{g}1underbrace{~}_{h}1underbrace{~}_{i}1underbrace{~}_{j}1underbrace{~}_{k}1underbrace{~}_{ell}$$
Now, pick which seven of the spaces are used by zeroes and close all the remaining unused spaces. Choose the spaces simultaneously. To count this, use binomial coefficients. We have twelve spaces available (11 + 1) and we want to choose seven of them to be used by zeroes.
The answer then is $$binom{12}{7}$$
By choosing in this fashion, either zero or one $0$ will be in each space and as a result no $0$ will be next to another $0$.
answered Jan 1 at 15:30
JMoravitzJMoravitz
49.4k44091
$endgroup$
add a comment |
$begingroup$
Place the eleven ones out first and leave a bit of space to either side of the ones.
$$underbrace{~}_{a}1underbrace{~}_{b}1underbrace{~}_{c}1underbrace{~}_{d}1underbrace{~}_{e}1underbrace{~}_{f}1underbrace{~}_{g}1underbrace{~}_{h}1underbrace{~}_{i}1underbrace{~}_{j}1underbrace{~}_{k}1underbrace{~}_{ell}$$
Now, pick which seven of the spaces are used by zeroes and close all the remaining unused spaces. Choose the spaces simultaneously. To count this, use binomial coefficients. We have twelve spaces available (11 + 1) and we want to choose seven of them to be used by zeroes.
The answer then is $$binom{12}{7}$$
By choosing in this fashion, either zero or one $0$ will be in each space and as a result no $0$ will be next to another $0$.
answered Jan 1 at 15:30
JMoravitzJMoravitz
49.4k44091
$endgroup$
Place the eleven ones out first and leave a bit of space to either side of the ones.
$$underbrace{~}_{a}1underbrace{~}_{b}1underbrace{~}_{c}1underbrace{~}_{d}1underbrace{~}_{e}1underbrace{~}_{f}1underbrace{~}_{g}1underbrace{~}_{h}1underbrace{~}_{i}1underbrace{~}_{j}1underbrace{~}_{k}1underbrace{~}_{ell}$$
Now, pick which seven of the spaces are used by zeroes and close all the remaining unused spaces. Choose the spaces simultaneously. To count this, use binomial coefficients. We have twelve spaces available (11 + 1) and we want to choose seven of them to be used by zeroes.
The answer then is $$binom{12}{7}$$
By choosing in this fashion, either zero or one $0$ will be in each space and as a result no $0$ will be next to another $0$.
answered Jan 1 at 15:30
JMoravitzJMoravitz
49.4k44091
answered Jan 1 at 15:30
JMoravitzJMoravitz
49.4k44091
answered Jan 1 at 15:30
JMoravitzJMoravitz
49.4k44091
answered Jan 1 at 15:30
JMoravitzJMoravitz
49.4k44091
49.4k44091
add a comment |
add a comment |
$begingroup$
Take such a sequence and stick a $1$ at the end. You have a sequence
of $12$ ones and $7$ zeroes with each zero followed by a $1$. So you have
seven subwords $01$. Replace each of those by a $2$. You then have a sequence
of $5$ ones and seven twos. There are $binom{12}5$ of these.
You can reverse the process. So there are $binom{12}5$ of your sort of sequence.
answered Jan 1 at 15:26
Lord Shark the UnknownLord Shark the Unknown
109k1163136
$endgroup$
add a comment |
$begingroup$
Take such a sequence and stick a $1$ at the end. You have a sequence
of $12$ ones and $7$ zeroes with each zero followed by a $1$. So you have
seven subwords $01$. Replace each of those by a $2$. You then have a sequence
of $5$ ones and seven twos. There are $binom{12}5$ of these.
You can reverse the process. So there are $binom{12}5$ of your sort of sequence.
answered Jan 1 at 15:26
Lord Shark the UnknownLord Shark the Unknown
109k1163136
$endgroup$
add a comment |
$begingroup$
Take such a sequence and stick a $1$ at the end. You have a sequence
of $12$ ones and $7$ zeroes with each zero followed by a $1$. So you have
seven subwords $01$. Replace each of those by a $2$. You then have a sequence
of $5$ ones and seven twos. There are $binom{12}5$ of these.
You can reverse the process. So there are $binom{12}5$ of your sort of sequence.
answered Jan 1 at 15:26
Lord Shark the UnknownLord Shark the Unknown
109k1163136
$endgroup$
Take such a sequence and stick a $1$ at the end. You have a sequence
of $12$ ones and $7$ zeroes with each zero followed by a $1$. So you have
seven subwords $01$. Replace each of those by a $2$. You then have a sequence
of $5$ ones and seven twos. There are $binom{12}5$ of these.
You can reverse the process. So there are $binom{12}5$ of your sort of sequence.
answered Jan 1 at 15:26
Lord Shark the UnknownLord Shark the Unknown
109k1163136
answered Jan 1 at 15:26
Lord Shark the UnknownLord Shark the Unknown
109k1163136
answered Jan 1 at 15:26
Lord Shark the UnknownLord Shark the Unknown
109k1163136
answered Jan 1 at 15:26
Lord Shark the UnknownLord Shark the Unknown
109k1163136
109k1163136
add a comment |
add a comment |
4
$begingroup$
Well, it's more or less a Stars and Bars problem. Put spaces before, after, and between the seven $0's$. The between ones must have at least one $1$ so put one in each. Now you have a standard problem. Can you finish from here?
$endgroup$
– lulu
Jan 1 at 15:09
$begingroup$
One way to add context to a "bare problem statement" is by working out some small examples. If eleven ones and seven zeros is too hard, how about cutting the size of the problem roughly in half? Often the process of working out an example will suggest methods to attack the general problem. In any event it will give your Readers a clear idea of the extent to which you have digested the problem statement. Other ways of adding context are also possible...
$endgroup$
– hardmath
Jan 1 at 19:04