Dgdrxrt

Why is the design of haulage companies so “special”?

Are tax years 2016 & 2017 back taxes deductible for tax year 2018?

Why are 150k or 200k jobs considered good when there are 300k+ births a month?

Pronouncing Dictionary.com's W.O.D "vade mecum" in English

Do airline pilots ever risk not hearing communication directed to them specifically, from traffic controllers?

What typically incentivizes a professor to change jobs to a lower ranking university?

What is the command to reset a PC without deleting any files

N.B. ligature in Latex

Email Account under attack (really) - anything I can do?

Is it possible to make sharp wind that can cut stuff from afar?

Is there really no realistic way for a skeleton monster to move around without magic?

Patience, young "Padovan"

What are these boxed doors outside store fronts in New York?

GPS Rollover on Android Smartphones

A Journey Through Space and Time

Continuity at a point in terms of closure

How can I hide my bitcoin transactions to protect anonymity from others?

A function which translates a sentence to title-case

whey we use polarized capacitor?

Is it possible to do 50 km distance without any previous training?

How long does it take to type this?

Can an x86 CPU running in real mode be considered to be basically an 8086 CPU?

DOS, create pipe for stdin/stdout of command.com(or 4dos.com) in C or Batch?

How can I fix this gap between bookcases I made?

Why do we divide by $2^2$ for $(6, 6, 5, 5)$ to find the permutations? [closed]

What is the proof of permutations of similar objects?Why counting $C(n,r)$ needs to divide $k$?permutations and the binomial coefficientNumber of subset-permutations for a setWhy do we divide values to count permutations with repeated letters?How many permutations are there for the letters in the word “meеt”?Formula for counting distinct n- letter long array permutationsPermutations with Repetition Formulacircular r-permutations of nWhy do we divide Permutations to get to Combinations?

-2

$begingroup$

Why do we divide by $2^2$
and are we using the permutations formula?

$(6, 6, 5, 5)$

$frac4!2^2= 6$

combinatorics permutations

share|cite|improve this question

edited Mar 29 at 20:55

N. F. Taussig

45.1k103358

asked Mar 29 at 20:48

Ebrahim A HajiEbrahim A Haji

42

$endgroup$

closed as unclear what you're asking by anomaly, Andrei, John Omielan, Cesareo, YiFan Mar 30 at 1:03

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It's completely unclear what is your final goal. What do you want to find? If you answer that, maybe we can help
  $endgroup$
  – Andrei
  Mar 29 at 20:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You should clarify that you wish to count the number of permutations of the sequence $(6, 6, 5, 5)$.
  $endgroup$
  – N. F. Taussig
  Mar 29 at 20:57
  

  
  
  
  

add a comment | 

-2

$begingroup$

Why do we divide by $2^2$
and are we using the permutations formula?

$(6, 6, 5, 5)$

$frac4!2^2= 6$

combinatorics permutations

share|cite|improve this question

edited Mar 29 at 20:55

N. F. Taussig

45.1k103358

asked Mar 29 at 20:48

Ebrahim A HajiEbrahim A Haji

42

$endgroup$

closed as unclear what you're asking by anomaly, Andrei, John Omielan, Cesareo, YiFan Mar 30 at 1:03

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It's completely unclear what is your final goal. What do you want to find? If you answer that, maybe we can help
  $endgroup$
  – Andrei
  Mar 29 at 20:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You should clarify that you wish to count the number of permutations of the sequence $(6, 6, 5, 5)$.
  $endgroup$
  – N. F. Taussig
  Mar 29 at 20:57
  

  
  
  
  

add a comment | 

$begingroup$

Why do we divide by $2^2$
and are we using the permutations formula?

$(6, 6, 5, 5)$

$frac4!2^2= 6$

combinatorics permutations

share|cite|improve this question

edited Mar 29 at 20:55

N. F. Taussig

45.1k103358

asked Mar 29 at 20:48

Ebrahim A HajiEbrahim A Haji

42

$endgroup$

share|cite|improve this question

edited Mar 29 at 20:55

N. F. Taussig

45.1k103358

asked Mar 29 at 20:48

Ebrahim A HajiEbrahim A Haji

42

share|cite|improve this question

edited Mar 29 at 20:55

N. F. Taussig

45.1k103358

asked Mar 29 at 20:48

Ebrahim A HajiEbrahim A Haji

42

edited Mar 29 at 20:55

N. F. Taussig

45.1k103358

edited Mar 29 at 20:55

N. F. Taussig

45.1k103358

asked Mar 29 at 20:48

Ebrahim A HajiEbrahim A Haji

42

asked Mar 29 at 20:48

Ebrahim A HajiEbrahim A Haji

42

3 Answers
3

active

oldest

votes

1

$begingroup$

Rather than a "division by symmetry" argument, I find it far more convincing to avoid division entirely.

To count the number of arrangements of the digits $5,5,6,6$ you can count this directly by deciding in which two of the four available positions the $5$'s are occupying. This can be done in $binom42=6$ ways. The $6$'s will then occupy whatever spaces are left.

Even more generally, the number of ways of arranging $k_1$ copies of $1$, $k_2$ copies of $2$, on up to $k_n$ copies of $n$ where $k_1+k_2+dots+k_n = N$ can be counted direction using the multinomial coefficient $binomNk_1,k_2,dots,k_n=fracN!k_1!k_2!cdots k_n!$ and this is seen to be equal to having chosen where the $1$'s are placed, multiplying by the number of ways of where the $2$'s are placed from those still available spots, etc... to give:

$$binomNk_1,k_2,dots,k_n = binomNk_1binomN-k_1k_2binomN-k_1-k_2k_3cdotsbinomN-k_1-dots-k_n-1k_n$$

Of course, the algebra is made much simpler to just take note of the final result and use $binomNk_1,k_2,dots,k_n = fracN!k_1!k_2!dots k_n!$ directly.

share|cite|improve this answer

answered Mar 29 at 21:14

JMoravitzJMoravitz

48.8k43988

$endgroup$

add a comment | 

0

$begingroup$

Because the four permutations
$$
(colorred6, colorblue6, colorred5, colorblue5)\
(colorred6, colorblue6, colorblue5, colorred5)\
(colorblue6, colorred6, colorred5, colorblue5)\
(colorblue6, colorred6, colorblue5, colorred5)
$$
are considered to be the same.

share|cite|improve this answer

answered Mar 29 at 20:57

ArthurArthur

122k7122211

$endgroup$

add a comment | 

0

$begingroup$

The correct way to write it is
$$
frac4!2!2!
$$

The terms in denominator reflect the fact that permutation of identical elements does not matter. Since the number in numerator counts all permutations as if all the elements were distinct, it should be divided by numbers of all possible permutations in every group of identical elements (counted as if the group elements were distinct).

For $(1,2,2,3,3,3)$ it would be
$$
frac 6!1!2!3!$$
and so on.

share|cite|improve this answer

edited Mar 29 at 21:49

answered Mar 29 at 20:56

useruser

6,32411031

$endgroup$

add a comment | 

1

$begingroup$

Rather than a "division by symmetry" argument, I find it far more convincing to avoid division entirely.

To count the number of arrangements of the digits $5,5,6,6$ you can count this directly by deciding in which two of the four available positions the $5$'s are occupying. This can be done in $binom42=6$ ways. The $6$'s will then occupy whatever spaces are left.

Even more generally, the number of ways of arranging $k_1$ copies of $1$, $k_2$ copies of $2$, on up to $k_n$ copies of $n$ where $k_1+k_2+dots+k_n = N$ can be counted direction using the multinomial coefficient $binomNk_1,k_2,dots,k_n=fracN!k_1!k_2!cdots k_n!$ and this is seen to be equal to having chosen where the $1$'s are placed, multiplying by the number of ways of where the $2$'s are placed from those still available spots, etc... to give:

$$binomNk_1,k_2,dots,k_n = binomNk_1binomN-k_1k_2binomN-k_1-k_2k_3cdotsbinomN-k_1-dots-k_n-1k_n$$

Of course, the algebra is made much simpler to just take note of the final result and use $binomNk_1,k_2,dots,k_n = fracN!k_1!k_2!dots k_n!$ directly.

share|cite|improve this answer

answered Mar 29 at 21:14

JMoravitzJMoravitz

48.8k43988

$endgroup$

add a comment | 

1

$begingroup$

Rather than a "division by symmetry" argument, I find it far more convincing to avoid division entirely.

To count the number of arrangements of the digits $5,5,6,6$ you can count this directly by deciding in which two of the four available positions the $5$'s are occupying. This can be done in $binom42=6$ ways. The $6$'s will then occupy whatever spaces are left.

Even more generally, the number of ways of arranging $k_1$ copies of $1$, $k_2$ copies of $2$, on up to $k_n$ copies of $n$ where $k_1+k_2+dots+k_n = N$ can be counted direction using the multinomial coefficient $binomNk_1,k_2,dots,k_n=fracN!k_1!k_2!cdots k_n!$ and this is seen to be equal to having chosen where the $1$'s are placed, multiplying by the number of ways of where the $2$'s are placed from those still available spots, etc... to give:

$$binomNk_1,k_2,dots,k_n = binomNk_1binomN-k_1k_2binomN-k_1-k_2k_3cdotsbinomN-k_1-dots-k_n-1k_n$$

Of course, the algebra is made much simpler to just take note of the final result and use $binomNk_1,k_2,dots,k_n = fracN!k_1!k_2!dots k_n!$ directly.

share|cite|improve this answer

answered Mar 29 at 21:14

JMoravitzJMoravitz

48.8k43988

$endgroup$

add a comment | 

$begingroup$

Rather than a "division by symmetry" argument, I find it far more convincing to avoid division entirely.

To count the number of arrangements of the digits $5,5,6,6$ you can count this directly by deciding in which two of the four available positions the $5$'s are occupying. This can be done in $binom42=6$ ways. The $6$'s will then occupy whatever spaces are left.

Even more generally, the number of ways of arranging $k_1$ copies of $1$, $k_2$ copies of $2$, on up to $k_n$ copies of $n$ where $k_1+k_2+dots+k_n = N$ can be counted direction using the multinomial coefficient $binomNk_1,k_2,dots,k_n=fracN!k_1!k_2!cdots k_n!$ and this is seen to be equal to having chosen where the $1$'s are placed, multiplying by the number of ways of where the $2$'s are placed from those still available spots, etc... to give:

$$binomNk_1,k_2,dots,k_n = binomNk_1binomN-k_1k_2binomN-k_1-k_2k_3cdotsbinomN-k_1-dots-k_n-1k_n$$

Of course, the algebra is made much simpler to just take note of the final result and use $binomNk_1,k_2,dots,k_n = fracN!k_1!k_2!dots k_n!$ directly.

share|cite|improve this answer

answered Mar 29 at 21:14

JMoravitzJMoravitz

48.8k43988

$endgroup$

share|cite|improve this answer

answered Mar 29 at 21:14

JMoravitzJMoravitz

48.8k43988

answered Mar 29 at 21:14

JMoravitzJMoravitz

48.8k43988

answered Mar 29 at 21:14

JMoravitzJMoravitz

48.8k43988

0

$begingroup$

Because the four permutations
$$
(colorred6, colorblue6, colorred5, colorblue5)\
(colorred6, colorblue6, colorblue5, colorred5)\
(colorblue6, colorred6, colorred5, colorblue5)\
(colorblue6, colorred6, colorblue5, colorred5)
$$
are considered to be the same.

share|cite|improve this answer

answered Mar 29 at 20:57

ArthurArthur

122k7122211

$endgroup$

add a comment | 

0

$begingroup$

Because the four permutations
$$
(colorred6, colorblue6, colorred5, colorblue5)\
(colorred6, colorblue6, colorblue5, colorred5)\
(colorblue6, colorred6, colorred5, colorblue5)\
(colorblue6, colorred6, colorblue5, colorred5)
$$
are considered to be the same.

share|cite|improve this answer

answered Mar 29 at 20:57

ArthurArthur

122k7122211

$endgroup$

add a comment | 

$begingroup$

Because the four permutations
$$
(colorred6, colorblue6, colorred5, colorblue5)\
(colorred6, colorblue6, colorblue5, colorred5)\
(colorblue6, colorred6, colorred5, colorblue5)\
(colorblue6, colorred6, colorblue5, colorred5)
$$
are considered to be the same.

share|cite|improve this answer

answered Mar 29 at 20:57

ArthurArthur

122k7122211

$endgroup$

share|cite|improve this answer

answered Mar 29 at 20:57

ArthurArthur

122k7122211

answered Mar 29 at 20:57

ArthurArthur

122k7122211

answered Mar 29 at 20:57

ArthurArthur

122k7122211

0

$begingroup$

The correct way to write it is
$$
frac4!2!2!
$$

The terms in denominator reflect the fact that permutation of identical elements does not matter. Since the number in numerator counts all permutations as if all the elements were distinct, it should be divided by numbers of all possible permutations in every group of identical elements (counted as if the group elements were distinct).

For $(1,2,2,3,3,3)$ it would be
$$
frac 6!1!2!3!$$
and so on.

share|cite|improve this answer

edited Mar 29 at 21:49

answered Mar 29 at 20:56

useruser

6,32411031

$endgroup$

add a comment | 

0

$begingroup$

The correct way to write it is
$$
frac4!2!2!
$$

The terms in denominator reflect the fact that permutation of identical elements does not matter. Since the number in numerator counts all permutations as if all the elements were distinct, it should be divided by numbers of all possible permutations in every group of identical elements (counted as if the group elements were distinct).

For $(1,2,2,3,3,3)$ it would be
$$
frac 6!1!2!3!$$
and so on.

share|cite|improve this answer

edited Mar 29 at 21:49

answered Mar 29 at 20:56

useruser

6,32411031

$endgroup$

add a comment | 

$begingroup$

The correct way to write it is
$$
frac4!2!2!
$$

The terms in denominator reflect the fact that permutation of identical elements does not matter. Since the number in numerator counts all permutations as if all the elements were distinct, it should be divided by numbers of all possible permutations in every group of identical elements (counted as if the group elements were distinct).

For $(1,2,2,3,3,3)$ it would be
$$
frac 6!1!2!3!$$
and so on.

share|cite|improve this answer

edited Mar 29 at 21:49

answered Mar 29 at 20:56

useruser

6,32411031

$endgroup$

share|cite|improve this answer

edited Mar 29 at 21:49

answered Mar 29 at 20:56

useruser

6,32411031

answered Mar 29 at 20:56

useruser

6,32411031

answered Mar 29 at 20:56

useruser

6,32411031