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Permutations and Combinations( Miscellaneous) [on hold]
Restricted Permutations and Combinationsmixed permutations and combinationsDistinguishing between combinations and permutationsPermutations and combinations - distributing objects into groupsPermutations and Combinations - Disc101Seating Problem ( Permutations and Combinations )Combinations and permutations helpPermutations and Combinations with lettersPermutations or combinations?permutations/combinations with repeated symbols
$begingroup$
In how many distinguishable ways can the symbols a,a,a,b,c,d,e,f be permuted in groups of 4?
combinatorics permutations
asked Mar 10 at 20:40
Sjava MabuzaSjava Mabuza
82
New contributor
Sjava Mabuza is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
put on hold as off-topic by mrtaurho, Rafa Budría, Eric Wofsey, YiFan, Eevee Trainer Mar 10 at 22:25
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." – mrtaurho, Rafa Budría, Eric Wofsey, YiFan, Eevee Trainer
add a comment |
$begingroup$
In how many distinguishable ways can the symbols a,a,a,b,c,d,e,f be permuted in groups of 4?
combinatorics permutations
asked Mar 10 at 20:40
Sjava MabuzaSjava Mabuza
82
New contributor
Sjava Mabuza is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
put on hold as off-topic by mrtaurho, Rafa Budría, Eric Wofsey, YiFan, Eevee Trainer Mar 10 at 22:25
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." – mrtaurho, Rafa Budría, Eric Wofsey, YiFan, Eevee Trainer
1
$begingroup$
What do you mean by "permuted in groups of 4"?
$endgroup$
– glowstonetrees
Mar 10 at 20:43
$begingroup$
That's how is the question asked.
$endgroup$
– Sjava Mabuza
Mar 10 at 20:45
1
$begingroup$
I expect that this is either lost in translation, or the question author is bastardizing the words being used. I expect that what you truly mean to ask is how many sequences of length four can be created using characters from $a,a,a,b,c,d,e,f$, noting that no character can be used more than once with the exception of $a$ which can be used up to three times and the different $a$'s are otherwise considered identical.
$endgroup$
– JMoravitz
Mar 10 at 20:47
$begingroup$
Ok. If that's the case, what is the solution?
$endgroup$
– Sjava Mabuza
Mar 10 at 20:51
add a comment |
$begingroup$
In how many distinguishable ways can the symbols a,a,a,b,c,d,e,f be permuted in groups of 4?
combinatorics permutations
asked Mar 10 at 20:40
Sjava MabuzaSjava Mabuza
82
New contributor
Sjava Mabuza is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
In how many distinguishable ways can the symbols a,a,a,b,c,d,e,f be permuted in groups of 4?
combinatorics permutations
combinatorics permutations
asked Mar 10 at 20:40
Sjava MabuzaSjava Mabuza
82
New contributor
Sjava Mabuza is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 10 at 20:40
Sjava MabuzaSjava Mabuza
82
New contributor
Sjava Mabuza is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 10 at 20:40
Sjava MabuzaSjava Mabuza
82
New contributor
Sjava Mabuza is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 10 at 20:40
Sjava MabuzaSjava Mabuza
82
asked Mar 10 at 20:40
Sjava MabuzaSjava Mabuza
82
82
New contributor
Sjava Mabuza is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Sjava Mabuza is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Sjava Mabuza is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
put on hold as off-topic by mrtaurho, Rafa Budría, Eric Wofsey, YiFan, Eevee Trainer Mar 10 at 22:25
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." – mrtaurho, Rafa Budría, Eric Wofsey, YiFan, Eevee Trainer
put on hold as off-topic by mrtaurho, Rafa Budría, Eric Wofsey, YiFan, Eevee Trainer Mar 10 at 22:25
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." – mrtaurho, Rafa Budría, Eric Wofsey, YiFan, Eevee Trainer
1
$begingroup$
What do you mean by "permuted in groups of 4"?
$endgroup$
– glowstonetrees
Mar 10 at 20:43
$begingroup$
That's how is the question asked.
$endgroup$
– Sjava Mabuza
Mar 10 at 20:45
1
$begingroup$
I expect that this is either lost in translation, or the question author is bastardizing the words being used. I expect that what you truly mean to ask is how many sequences of length four can be created using characters from $a,a,a,b,c,d,e,f$, noting that no character can be used more than once with the exception of $a$ which can be used up to three times and the different $a$'s are otherwise considered identical.
$endgroup$
– JMoravitz
Mar 10 at 20:47
$begingroup$
Ok. If that's the case, what is the solution?
$endgroup$
– Sjava Mabuza
Mar 10 at 20:51
add a comment |
1
$begingroup$
What do you mean by "permuted in groups of 4"?
$endgroup$
– glowstonetrees
Mar 10 at 20:43
$begingroup$
That's how is the question asked.
$endgroup$
– Sjava Mabuza
Mar 10 at 20:45
1
$begingroup$
I expect that this is either lost in translation, or the question author is bastardizing the words being used. I expect that what you truly mean to ask is how many sequences of length four can be created using characters from $a,a,a,b,c,d,e,f$, noting that no character can be used more than once with the exception of $a$ which can be used up to three times and the different $a$'s are otherwise considered identical.
$endgroup$
– JMoravitz
Mar 10 at 20:47
$begingroup$
Ok. If that's the case, what is the solution?
$endgroup$
– Sjava Mabuza
Mar 10 at 20:51
1
1
$begingroup$
What do you mean by "permuted in groups of 4"?
$endgroup$
– glowstonetrees
Mar 10 at 20:43
$begingroup$
What do you mean by "permuted in groups of 4"?
$endgroup$
– glowstonetrees
Mar 10 at 20:43
$begingroup$
That's how is the question asked.
$endgroup$
– Sjava Mabuza
Mar 10 at 20:45
$begingroup$
That's how is the question asked.
$endgroup$
– Sjava Mabuza
Mar 10 at 20:45
1
1
$begingroup$
I expect that this is either lost in translation, or the question author is bastardizing the words being used. I expect that what you truly mean to ask is how many sequences of length four can be created using characters from $a,a,a,b,c,d,e,f$, noting that no character can be used more than once with the exception of $a$ which can be used up to three times and the different $a$'s are otherwise considered identical.
$endgroup$
– JMoravitz
Mar 10 at 20:47
$begingroup$
I expect that this is either lost in translation, or the question author is bastardizing the words being used. I expect that what you truly mean to ask is how many sequences of length four can be created using characters from $a,a,a,b,c,d,e,f$, noting that no character can be used more than once with the exception of $a$ which can be used up to three times and the different $a$'s are otherwise considered identical.
$endgroup$
– JMoravitz
Mar 10 at 20:47
$begingroup$
Ok. If that's the case, what is the solution?
$endgroup$
– Sjava Mabuza
Mar 10 at 20:51
$begingroup$
Ok. If that's the case, what is the solution?
$endgroup$
– Sjava Mabuza
Mar 10 at 20:51
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Setup:
Break into cases based on the number of $a$'s being used.
If exactly $n$ copies of $a$ are used in the string, pick which locations in the sequence they occupy in $binom4n$ ways. Then, from left to right, pick which character occupies each remaining space.
For example, if exactly two $a$'s are used, pick where they go in $binom42$ ways. Then pick the left-most non-$a$ character in the string in $5$ ways (as there are five non-$a$ characters available to choose from) and then pick the remaining character in $4$ ways (as after picking the earlier character it is no longer available, leaving only $4$ left to choose from). This gives a total of $binom42cdot 5cdot 4$ different ways where exactly two $a$'s were used.
Sum over all possible values of $n$ to get the final count.
answered Mar 10 at 20:52
JMoravitzJMoravitz
48.6k43988
$endgroup$
$begingroup$
A downvote? Just because I'm not willing to give the OP a final numerical answer without forcing them to put in some effort to try to understand? Surely, its not because my answer is incorrect or doesn't lead to an easy to understand solution...
$endgroup$
– JMoravitz
Mar 10 at 20:54
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Setup:
Break into cases based on the number of $a$'s being used.
If exactly $n$ copies of $a$ are used in the string, pick which locations in the sequence they occupy in $binom4n$ ways. Then, from left to right, pick which character occupies each remaining space.
For example, if exactly two $a$'s are used, pick where they go in $binom42$ ways. Then pick the left-most non-$a$ character in the string in $5$ ways (as there are five non-$a$ characters available to choose from) and then pick the remaining character in $4$ ways (as after picking the earlier character it is no longer available, leaving only $4$ left to choose from). This gives a total of $binom42cdot 5cdot 4$ different ways where exactly two $a$'s were used.
Sum over all possible values of $n$ to get the final count.
answered Mar 10 at 20:52
JMoravitzJMoravitz
48.6k43988
$endgroup$
$begingroup$
A downvote? Just because I'm not willing to give the OP a final numerical answer without forcing them to put in some effort to try to understand? Surely, its not because my answer is incorrect or doesn't lead to an easy to understand solution...
$endgroup$
– JMoravitz
Mar 10 at 20:54
add a comment |
$begingroup$
Setup:
Break into cases based on the number of $a$'s being used.
If exactly $n$ copies of $a$ are used in the string, pick which locations in the sequence they occupy in $binom4n$ ways. Then, from left to right, pick which character occupies each remaining space.
For example, if exactly two $a$'s are used, pick where they go in $binom42$ ways. Then pick the left-most non-$a$ character in the string in $5$ ways (as there are five non-$a$ characters available to choose from) and then pick the remaining character in $4$ ways (as after picking the earlier character it is no longer available, leaving only $4$ left to choose from). This gives a total of $binom42cdot 5cdot 4$ different ways where exactly two $a$'s were used.
Sum over all possible values of $n$ to get the final count.
answered Mar 10 at 20:52
JMoravitzJMoravitz
48.6k43988
$endgroup$
$begingroup$
A downvote? Just because I'm not willing to give the OP a final numerical answer without forcing them to put in some effort to try to understand? Surely, its not because my answer is incorrect or doesn't lead to an easy to understand solution...
$endgroup$
– JMoravitz
Mar 10 at 20:54
add a comment |
$begingroup$
Setup:
Break into cases based on the number of $a$'s being used.
If exactly $n$ copies of $a$ are used in the string, pick which locations in the sequence they occupy in $binom4n$ ways. Then, from left to right, pick which character occupies each remaining space.
For example, if exactly two $a$'s are used, pick where they go in $binom42$ ways. Then pick the left-most non-$a$ character in the string in $5$ ways (as there are five non-$a$ characters available to choose from) and then pick the remaining character in $4$ ways (as after picking the earlier character it is no longer available, leaving only $4$ left to choose from). This gives a total of $binom42cdot 5cdot 4$ different ways where exactly two $a$'s were used.
Sum over all possible values of $n$ to get the final count.
answered Mar 10 at 20:52
JMoravitzJMoravitz
48.6k43988
$endgroup$
Setup:
Break into cases based on the number of $a$'s being used.
If exactly $n$ copies of $a$ are used in the string, pick which locations in the sequence they occupy in $binom4n$ ways. Then, from left to right, pick which character occupies each remaining space.
For example, if exactly two $a$'s are used, pick where they go in $binom42$ ways. Then pick the left-most non-$a$ character in the string in $5$ ways (as there are five non-$a$ characters available to choose from) and then pick the remaining character in $4$ ways (as after picking the earlier character it is no longer available, leaving only $4$ left to choose from). This gives a total of $binom42cdot 5cdot 4$ different ways where exactly two $a$'s were used.
Sum over all possible values of $n$ to get the final count.
answered Mar 10 at 20:52
JMoravitzJMoravitz
48.6k43988
answered Mar 10 at 20:52
JMoravitzJMoravitz
48.6k43988
answered Mar 10 at 20:52
JMoravitzJMoravitz
48.6k43988
answered Mar 10 at 20:52
JMoravitzJMoravitz
48.6k43988
48.6k43988
$begingroup$
A downvote? Just because I'm not willing to give the OP a final numerical answer without forcing them to put in some effort to try to understand? Surely, its not because my answer is incorrect or doesn't lead to an easy to understand solution...
$endgroup$
– JMoravitz
Mar 10 at 20:54
add a comment |
$begingroup$
A downvote? Just because I'm not willing to give the OP a final numerical answer without forcing them to put in some effort to try to understand? Surely, its not because my answer is incorrect or doesn't lead to an easy to understand solution...
$endgroup$
– JMoravitz
Mar 10 at 20:54
$begingroup$
A downvote? Just because I'm not willing to give the OP a final numerical answer without forcing them to put in some effort to try to understand? Surely, its not because my answer is incorrect or doesn't lead to an easy to understand solution...
$endgroup$
– JMoravitz
Mar 10 at 20:54
$begingroup$
A downvote? Just because I'm not willing to give the OP a final numerical answer without forcing them to put in some effort to try to understand? Surely, its not because my answer is incorrect or doesn't lead to an easy to understand solution...
$endgroup$
– JMoravitz
Mar 10 at 20:54
add a comment |
1
$begingroup$
What do you mean by "permuted in groups of 4"?
$endgroup$
– glowstonetrees
Mar 10 at 20:43
$begingroup$
That's how is the question asked.
$endgroup$
– Sjava Mabuza
Mar 10 at 20:45
1
$begingroup$
I expect that this is either lost in translation, or the question author is bastardizing the words being used. I expect that what you truly mean to ask is how many sequences of length four can be created using characters from $a,a,a,b,c,d,e,f$, noting that no character can be used more than once with the exception of $a$ which can be used up to three times and the different $a$'s are otherwise considered identical.
$endgroup$
– JMoravitz
Mar 10 at 20:47
$begingroup$
Ok. If that's the case, what is the solution?
$endgroup$
– Sjava Mabuza
Mar 10 at 20:51