Permutations: How does limiting the orientation of objects change the permutation formula? Notatation? Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)What is the proof of permutations of similar objects?Circular permutation; trouble understandingFinding the number of objects in permutationNumber of permutations of the word “PERMUTATION”How to intuitively understand the formula for permutation on non-distinct objects?Permutations with repetitions, when you're not picking all the objects?are there any elementary formula of $r'_3(n)$?How is the number of permutations of alike objects different from from the normal number of permutations?Proving the context of the formula $frack!k_1!cdot k_2!…k_n!$ - Need help to understand the contextDoes limiting iteration count when performing permutation on a set change the time complexity?
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Permutations: How does limiting the orientation of objects change the permutation formula? Notatation?
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)What is the proof of permutations of similar objects?Circular permutation; trouble understandingFinding the number of objects in permutationNumber of permutations of the word “PERMUTATION”How to intuitively understand the formula for permutation on non-distinct objects?Permutations with repetitions, when you're not picking all the objects?are there any elementary formula of $r'_3(n)$?How is the number of permutations of alike objects different from from the normal number of permutations?Proving the context of the formula $frack!k_1!cdot k_2!…k_n!$ - Need help to understand the contextDoes limiting iteration count when performing permutation on a set change the time complexity?
$begingroup$
Hello MathStackExchange Community,
Confused about how to change the notation-formula for permutations when you have more complex processes. i.e. multiple independent action choices of different types of objects and orientation limitations.
- The generic formula is : $P(n,r)= n!/(n-r)!$
$n!$ = being the total number of objects in an option set
$(n-r)!$ = the limitations caused by the choices made.
but this only works for very simple permutation-processes where you make one single choice-type.
If I have 15 particles, 10 neutral and 5 positive particles
and I care about the arrangement
he formula is $P(n,r)= 15!/10!cdot5!$
What I don't get is how does 10! 5! translate to $(n-r)!$
the second part of the problem asks, if like charges repel, and positive charges can't sit next to each other- number of possibilities then?
What is the way to think about the second portion? Could draw it out and try to find a pattern, but if I had avagadro's number of particles that would be hard.
Thanks,
ThermoRestart
probability combinatorics permutations
asked Mar 20 at 2:39
ThermoRestartThermoRestart
33
$endgroup$
add a comment |
$begingroup$
Hello MathStackExchange Community,
Confused about how to change the notation-formula for permutations when you have more complex processes. i.e. multiple independent action choices of different types of objects and orientation limitations.
- The generic formula is : $P(n,r)= n!/(n-r)!$
$n!$ = being the total number of objects in an option set
$(n-r)!$ = the limitations caused by the choices made.
but this only works for very simple permutation-processes where you make one single choice-type.
If I have 15 particles, 10 neutral and 5 positive particles
and I care about the arrangement
he formula is $P(n,r)= 15!/10!cdot5!$
What I don't get is how does 10! 5! translate to $(n-r)!$
the second part of the problem asks, if like charges repel, and positive charges can't sit next to each other- number of possibilities then?
What is the way to think about the second portion? Could draw it out and try to find a pattern, but if I had avagadro's number of particles that would be hard.
Thanks,
ThermoRestart
probability combinatorics permutations
asked Mar 20 at 2:39
ThermoRestartThermoRestart
33
$endgroup$
add a comment |
$begingroup$
Hello MathStackExchange Community,
Confused about how to change the notation-formula for permutations when you have more complex processes. i.e. multiple independent action choices of different types of objects and orientation limitations.
- The generic formula is : $P(n,r)= n!/(n-r)!$
$n!$ = being the total number of objects in an option set
$(n-r)!$ = the limitations caused by the choices made.
but this only works for very simple permutation-processes where you make one single choice-type.
If I have 15 particles, 10 neutral and 5 positive particles
and I care about the arrangement
he formula is $P(n,r)= 15!/10!cdot5!$
What I don't get is how does 10! 5! translate to $(n-r)!$
the second part of the problem asks, if like charges repel, and positive charges can't sit next to each other- number of possibilities then?
What is the way to think about the second portion? Could draw it out and try to find a pattern, but if I had avagadro's number of particles that would be hard.
Thanks,
ThermoRestart
probability combinatorics permutations
asked Mar 20 at 2:39
ThermoRestartThermoRestart
33
$endgroup$
Hello MathStackExchange Community,
Confused about how to change the notation-formula for permutations when you have more complex processes. i.e. multiple independent action choices of different types of objects and orientation limitations.
- The generic formula is : $P(n,r)= n!/(n-r)!$
$n!$ = being the total number of objects in an option set
$(n-r)!$ = the limitations caused by the choices made.
but this only works for very simple permutation-processes where you make one single choice-type.
If I have 15 particles, 10 neutral and 5 positive particles
and I care about the arrangement
he formula is $P(n,r)= 15!/10!cdot5!$
What I don't get is how does 10! 5! translate to $(n-r)!$
the second part of the problem asks, if like charges repel, and positive charges can't sit next to each other- number of possibilities then?
What is the way to think about the second portion? Could draw it out and try to find a pattern, but if I had avagadro's number of particles that would be hard.
Thanks,
ThermoRestart
probability combinatorics permutations
probability combinatorics permutations
asked Mar 20 at 2:39
ThermoRestartThermoRestart
33
asked Mar 20 at 2:39
ThermoRestartThermoRestart
33
edited Apr 1 at 12:07
ThermoRestart
asked Mar 20 at 2:39
ThermoRestartThermoRestart
33
asked Mar 20 at 2:39
ThermoRestartThermoRestart
33
asked Mar 20 at 2:39
ThermoRestartThermoRestart
33
33
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
What I don't get is how does 10! 5! translate to (n-r)!
You cannot see how the formula is applied because ... it is not being applied. A slightly different formula is.
This is an application of the binomial coeffient, sometimes written $^nmathrm C_r$, $mathrm C(n,r)$, or as $binom nr$. It counts ways to select $r$ elements from a set of $n$ distinct elements.
$$^nmathrm C_r = dfracn!r!~(n-r)!$$
In this case, we are counting ways to select places to put the $10$ neutral particles from the $15$ distinct places available.
$$^15mathrm C_10=dfrac15!10!~5!$$
the second part of the problem asks, if like charges repel, and positive charges can't sit next to each other- number of possibilities then?
For this we have to know what shape the collections of fifteen particals has, such that "next to" has meaning.
Let us say we're stringing the particals in a chain. In that case, line the neutral particals up, and place each positive partical between two neutral particals, or at an end of the string (ie there are eleven posible places). $$_circ_circ_circ_circ_circ_circ_circ_circ_circ_circ_$$ Count the ways.
answered Mar 20 at 5:03
Graham KempGraham Kemp
88k43579
$endgroup$
$begingroup$
Doesn't that end up being the same value wise as considering the permutations of if it didn't matter if cations were next to neutral particles? Shouldn't choosing limitations limit the number of arrangements possible?
$endgroup$
– ThermoRestart
Mar 20 at 7:05
add a comment |
$begingroup$
Limiting the orientation of particles limits the number of arrangements possible and makes the permutation counting concerned about the blocks of spaces that are cation or neutral instead of discrete spaces.
The correct number of permutations is 462.
If cation particles can't sit next to cation particles, that redefines the problem to consider spaces available instead of discrete particles.
This is true because, the arrangements possible are completely limited by the number of cation particle spaces.
_ C _ C _ C_ C_C_ <--- describes the set up, where C= cation particle
this leaves 6 spaces for neutral particles.
The number of discrete neutral particles aren't considered because they don't limit the arrangement.
Apply the permutation formula for multiple choices.
N! = total objects available to choose from
n1! ... nt!= each a subset of indistinguishable objects, which were chosen from N
N_total_available_spaces!/n_neutral_spaces! n_cation_spaces!
11!/5!6!= 462
answered Mar 31 at 19:12
ThermoRestartThermoRestart
33
$endgroup$
add a comment |
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2 Answers
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$begingroup$
What I don't get is how does 10! 5! translate to (n-r)!
You cannot see how the formula is applied because ... it is not being applied. A slightly different formula is.
This is an application of the binomial coeffient, sometimes written $^nmathrm C_r$, $mathrm C(n,r)$, or as $binom nr$. It counts ways to select $r$ elements from a set of $n$ distinct elements.
$$^nmathrm C_r = dfracn!r!~(n-r)!$$
In this case, we are counting ways to select places to put the $10$ neutral particles from the $15$ distinct places available.
$$^15mathrm C_10=dfrac15!10!~5!$$
the second part of the problem asks, if like charges repel, and positive charges can't sit next to each other- number of possibilities then?
For this we have to know what shape the collections of fifteen particals has, such that "next to" has meaning.
Let us say we're stringing the particals in a chain. In that case, line the neutral particals up, and place each positive partical between two neutral particals, or at an end of the string (ie there are eleven posible places). $$_circ_circ_circ_circ_circ_circ_circ_circ_circ_circ_$$ Count the ways.
answered Mar 20 at 5:03
Graham KempGraham Kemp
88k43579
$endgroup$
$begingroup$
Doesn't that end up being the same value wise as considering the permutations of if it didn't matter if cations were next to neutral particles? Shouldn't choosing limitations limit the number of arrangements possible?
$endgroup$
– ThermoRestart
Mar 20 at 7:05
add a comment |
$begingroup$
What I don't get is how does 10! 5! translate to (n-r)!
You cannot see how the formula is applied because ... it is not being applied. A slightly different formula is.
This is an application of the binomial coeffient, sometimes written $^nmathrm C_r$, $mathrm C(n,r)$, or as $binom nr$. It counts ways to select $r$ elements from a set of $n$ distinct elements.
$$^nmathrm C_r = dfracn!r!~(n-r)!$$
In this case, we are counting ways to select places to put the $10$ neutral particles from the $15$ distinct places available.
$$^15mathrm C_10=dfrac15!10!~5!$$
the second part of the problem asks, if like charges repel, and positive charges can't sit next to each other- number of possibilities then?
For this we have to know what shape the collections of fifteen particals has, such that "next to" has meaning.
Let us say we're stringing the particals in a chain. In that case, line the neutral particals up, and place each positive partical between two neutral particals, or at an end of the string (ie there are eleven posible places). $$_circ_circ_circ_circ_circ_circ_circ_circ_circ_circ_$$ Count the ways.
answered Mar 20 at 5:03
Graham KempGraham Kemp
88k43579
$endgroup$
$begingroup$
Doesn't that end up being the same value wise as considering the permutations of if it didn't matter if cations were next to neutral particles? Shouldn't choosing limitations limit the number of arrangements possible?
$endgroup$
– ThermoRestart
Mar 20 at 7:05
add a comment |
$begingroup$
What I don't get is how does 10! 5! translate to (n-r)!
You cannot see how the formula is applied because ... it is not being applied. A slightly different formula is.
This is an application of the binomial coeffient, sometimes written $^nmathrm C_r$, $mathrm C(n,r)$, or as $binom nr$. It counts ways to select $r$ elements from a set of $n$ distinct elements.
$$^nmathrm C_r = dfracn!r!~(n-r)!$$
In this case, we are counting ways to select places to put the $10$ neutral particles from the $15$ distinct places available.
$$^15mathrm C_10=dfrac15!10!~5!$$
the second part of the problem asks, if like charges repel, and positive charges can't sit next to each other- number of possibilities then?
For this we have to know what shape the collections of fifteen particals has, such that "next to" has meaning.
Let us say we're stringing the particals in a chain. In that case, line the neutral particals up, and place each positive partical between two neutral particals, or at an end of the string (ie there are eleven posible places). $$_circ_circ_circ_circ_circ_circ_circ_circ_circ_circ_$$ Count the ways.
answered Mar 20 at 5:03
Graham KempGraham Kemp
88k43579
$endgroup$
What I don't get is how does 10! 5! translate to (n-r)!
You cannot see how the formula is applied because ... it is not being applied. A slightly different formula is.
This is an application of the binomial coeffient, sometimes written $^nmathrm C_r$, $mathrm C(n,r)$, or as $binom nr$. It counts ways to select $r$ elements from a set of $n$ distinct elements.
$$^nmathrm C_r = dfracn!r!~(n-r)!$$
In this case, we are counting ways to select places to put the $10$ neutral particles from the $15$ distinct places available.
$$^15mathrm C_10=dfrac15!10!~5!$$
the second part of the problem asks, if like charges repel, and positive charges can't sit next to each other- number of possibilities then?
For this we have to know what shape the collections of fifteen particals has, such that "next to" has meaning.
Let us say we're stringing the particals in a chain. In that case, line the neutral particals up, and place each positive partical between two neutral particals, or at an end of the string (ie there are eleven posible places). $$_circ_circ_circ_circ_circ_circ_circ_circ_circ_circ_$$ Count the ways.
answered Mar 20 at 5:03
Graham KempGraham Kemp
88k43579
edited Mar 20 at 8:21
answered Mar 20 at 5:03
Graham KempGraham Kemp
88k43579
answered Mar 20 at 5:03
Graham KempGraham Kemp
88k43579
answered Mar 20 at 5:03
Graham KempGraham Kemp
88k43579
88k43579
$begingroup$
Doesn't that end up being the same value wise as considering the permutations of if it didn't matter if cations were next to neutral particles? Shouldn't choosing limitations limit the number of arrangements possible?
$endgroup$
– ThermoRestart
Mar 20 at 7:05
add a comment |
$begingroup$
Doesn't that end up being the same value wise as considering the permutations of if it didn't matter if cations were next to neutral particles? Shouldn't choosing limitations limit the number of arrangements possible?
$endgroup$
– ThermoRestart
Mar 20 at 7:05
$begingroup$
Doesn't that end up being the same value wise as considering the permutations of if it didn't matter if cations were next to neutral particles? Shouldn't choosing limitations limit the number of arrangements possible?
$endgroup$
– ThermoRestart
Mar 20 at 7:05
$begingroup$
Doesn't that end up being the same value wise as considering the permutations of if it didn't matter if cations were next to neutral particles? Shouldn't choosing limitations limit the number of arrangements possible?
$endgroup$
– ThermoRestart
Mar 20 at 7:05
add a comment |
$begingroup$
Limiting the orientation of particles limits the number of arrangements possible and makes the permutation counting concerned about the blocks of spaces that are cation or neutral instead of discrete spaces.
The correct number of permutations is 462.
If cation particles can't sit next to cation particles, that redefines the problem to consider spaces available instead of discrete particles.
This is true because, the arrangements possible are completely limited by the number of cation particle spaces.
_ C _ C _ C_ C_C_ <--- describes the set up, where C= cation particle
this leaves 6 spaces for neutral particles.
The number of discrete neutral particles aren't considered because they don't limit the arrangement.
Apply the permutation formula for multiple choices.
N! = total objects available to choose from
n1! ... nt!= each a subset of indistinguishable objects, which were chosen from N
N_total_available_spaces!/n_neutral_spaces! n_cation_spaces!
11!/5!6!= 462
answered Mar 31 at 19:12
ThermoRestartThermoRestart
33
$endgroup$
add a comment |
$begingroup$
Limiting the orientation of particles limits the number of arrangements possible and makes the permutation counting concerned about the blocks of spaces that are cation or neutral instead of discrete spaces.
The correct number of permutations is 462.
If cation particles can't sit next to cation particles, that redefines the problem to consider spaces available instead of discrete particles.
This is true because, the arrangements possible are completely limited by the number of cation particle spaces.
_ C _ C _ C_ C_C_ <--- describes the set up, where C= cation particle
this leaves 6 spaces for neutral particles.
The number of discrete neutral particles aren't considered because they don't limit the arrangement.
Apply the permutation formula for multiple choices.
N! = total objects available to choose from
n1! ... nt!= each a subset of indistinguishable objects, which were chosen from N
N_total_available_spaces!/n_neutral_spaces! n_cation_spaces!
11!/5!6!= 462
answered Mar 31 at 19:12
ThermoRestartThermoRestart
33
$endgroup$
add a comment |
$begingroup$
Limiting the orientation of particles limits the number of arrangements possible and makes the permutation counting concerned about the blocks of spaces that are cation or neutral instead of discrete spaces.
The correct number of permutations is 462.
If cation particles can't sit next to cation particles, that redefines the problem to consider spaces available instead of discrete particles.
This is true because, the arrangements possible are completely limited by the number of cation particle spaces.
_ C _ C _ C_ C_C_ <--- describes the set up, where C= cation particle
this leaves 6 spaces for neutral particles.
The number of discrete neutral particles aren't considered because they don't limit the arrangement.
Apply the permutation formula for multiple choices.
N! = total objects available to choose from
n1! ... nt!= each a subset of indistinguishable objects, which were chosen from N
N_total_available_spaces!/n_neutral_spaces! n_cation_spaces!
11!/5!6!= 462
answered Mar 31 at 19:12
ThermoRestartThermoRestart
33
$endgroup$
Limiting the orientation of particles limits the number of arrangements possible and makes the permutation counting concerned about the blocks of spaces that are cation or neutral instead of discrete spaces.
The correct number of permutations is 462.
If cation particles can't sit next to cation particles, that redefines the problem to consider spaces available instead of discrete particles.
This is true because, the arrangements possible are completely limited by the number of cation particle spaces.
_ C _ C _ C_ C_C_ <--- describes the set up, where C= cation particle
this leaves 6 spaces for neutral particles.
The number of discrete neutral particles aren't considered because they don't limit the arrangement.
Apply the permutation formula for multiple choices.
N! = total objects available to choose from
n1! ... nt!= each a subset of indistinguishable objects, which were chosen from N
N_total_available_spaces!/n_neutral_spaces! n_cation_spaces!
11!/5!6!= 462
answered Mar 31 at 19:12
ThermoRestartThermoRestart
33
edited Apr 1 at 12:06
answered Mar 31 at 19:12
ThermoRestartThermoRestart
33
answered Mar 31 at 19:12
ThermoRestartThermoRestart
33
answered Mar 31 at 19:12
ThermoRestartThermoRestart
33
33
add a comment |
add a comment |
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