Permutations with same repeated objects












0












$begingroup$


Suppose I have these numbers:



$$1,1,1,1,0.$$



When I arrange them to find permutations I will get:



$01111$
$10111$
$11011$
$11101$
$11110$



Is this is the idea what is commonly known as "Permutations with repetition"?



If so, then I am not getting the excepted answer: e.g. n! = 5! = 120 permutations.



E.g:



$$n!/(n-r)! = 120/(5-5)! = 120$$



I am not getting an excepted answer. Is this case something different? Is there a formula?



Is what I am trying to do known as "combinations with repetition"?






combinatorics permutations







share|cite|improve this question









$endgroup$












  • $begingroup$
    This is a permutation of a multiset problem since each number appears a fixed number of times.
    $endgroup$
    – N. F. Taussig
    Jan 16 at 11:27
















0












$begingroup$


Suppose I have these numbers:



$$1,1,1,1,0.$$



When I arrange them to find permutations I will get:



$01111$
$10111$
$11011$
$11101$
$11110$



Is this is the idea what is commonly known as "Permutations with repetition"?



If so, then I am not getting the excepted answer: e.g. n! = 5! = 120 permutations.



E.g:



$$n!/(n-r)! = 120/(5-5)! = 120$$



I am not getting an excepted answer. Is this case something different? Is there a formula?



Is what I am trying to do known as "combinations with repetition"?






combinatorics permutations







share|cite|improve this question









$endgroup$












  • $begingroup$
    This is a permutation of a multiset problem since each number appears a fixed number of times.
    $endgroup$
    – N. F. Taussig
    Jan 16 at 11:27














0












0








0





$begingroup$


Suppose I have these numbers:



$$1,1,1,1,0.$$



When I arrange them to find permutations I will get:



$01111$
$10111$
$11011$
$11101$
$11110$



Is this is the idea what is commonly known as "Permutations with repetition"?



If so, then I am not getting the excepted answer: e.g. n! = 5! = 120 permutations.



E.g:



$$n!/(n-r)! = 120/(5-5)! = 120$$



I am not getting an excepted answer. Is this case something different? Is there a formula?



Is what I am trying to do known as "combinations with repetition"?






combinatorics permutations







share|cite|improve this question









$endgroup$




Suppose I have these numbers:



$$1,1,1,1,0.$$



When I arrange them to find permutations I will get:



$01111$
$10111$
$11011$
$11101$
$11110$



Is this is the idea what is commonly known as "Permutations with repetition"?



If so, then I am not getting the excepted answer: e.g. n! = 5! = 120 permutations.



E.g:



$$n!/(n-r)! = 120/(5-5)! = 120$$



I am not getting an excepted answer. Is this case something different? Is there a formula?



Is what I am trying to do known as "combinations with repetition"?






combinatorics permutations




combinatorics permutations






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 16 at 7:33









user963241user963241

17518




17518












  • $begingroup$
    This is a permutation of a multiset problem since each number appears a fixed number of times.
    $endgroup$
    – N. F. Taussig
    Jan 16 at 11:27


















  • $begingroup$
    This is a permutation of a multiset problem since each number appears a fixed number of times.
    $endgroup$
    – N. F. Taussig
    Jan 16 at 11:27
















$begingroup$
This is a permutation of a multiset problem since each number appears a fixed number of times.
$endgroup$
– N. F. Taussig
Jan 16 at 11:27




$begingroup$
This is a permutation of a multiset problem since each number appears a fixed number of times.
$endgroup$
– N. F. Taussig
Jan 16 at 11:27










3 Answers
3






active

oldest

votes


















2












$begingroup$

Going backward:



Take any one of your permutations, say



$11110.$



Assume you had $a,b,c,d$ distinct digits ($not =0$)



Look at $abcd0$. The number of distinct permutations with the $0$ fixed in the last slot is:



$4!$ .



Now you have $5$ permutations listed with four $1$'s and one $0$:



Look at



$abcd0, abc0d,ab0cd, a0bcd, 0abcd.$



where $a,b,c,d$ are distinct ($not =0$).



You get $5 × 4!=5!$ distinct permutations (Why?).



Finally :



If you have five objects to permute where $4$ are identical, you get:



$5!/4!=5$ distinct permutations .



Can you generalize for $n$ objects , $r$ of which are identical? How many distinct permutations do you get?






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks. Can you say whether these are permutations with repetition or no repetition?
    $endgroup$
    – user963241
    Jan 16 at 8:03










  • $begingroup$
    I'll try. 5 distinct! objects permuted: abcde, eabcd, ..etc. certainly no repetitions. a=b, then abcde is the same as bacde(ok?), since a=b, I.e you get aacde and aacde.The same.To get rid of repetitions (overcounting) divide by 2! since a=b .
    $endgroup$
    – Peter Szilas
    Jan 16 at 8:12










  • $begingroup$
    So, these are permutations with no repetition even though I have the same 4 objects (1's). Right @Peter?
    $endgroup$
    – user963241
    Jan 16 at 8:56










  • $begingroup$
    Yes!!!Sorry for the roundabout way.You look at your list: 5 permutations without rep.(OK?).With 5 distinct !! objects you get 5! perm, no rep.Now let 4 be identical, you divide 5!/4!= 5(your list) to have #of perm without !! rep.
    $endgroup$
    – Peter Szilas
    Jan 16 at 9:40






  • 1




    $begingroup$
    user963241Looks fine:)
    $endgroup$
    – Peter Szilas
    Jan 16 at 14:43



















1












$begingroup$

Yes, it is known as permutations with repetitions, and it has a formula. Say you take 5 different numbers: 1,2,3,4,5,then you can arrange them in 5!=120 days. But when you take 1,1,1,1,0, the 120 permutations will contain repeated formations. e.g. take the permutation 11110. It remains the same even when you interchange the first two 1's. In fact it remains the same even if you arrange the first four 1's in 4!=24 ways. So, 5!/4!=120/24=5 is the answer.
Now you can derive the formula yourself.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    This is a permutation with a multiset since each number in the sequence appears a fixed number of times.
    $endgroup$
    – N. F. Taussig
    Jan 16 at 11:29



















0












$begingroup$

The formula for Permutations with same repeated objects is :-
(total number of letters)!/(number of repeats)!



So in your case :- 1,1,1,1,0
total numer of letters is 5
number of repeats is 4
So the answer according to formula is 5!/4!



Also to clarify things further lets see one more example :-
if it was 1,1,1,1,0,0
Then
it would be :-
total number of letters is 6
number of repeats is 4 and 2 (1 is present 4 times and 0 is present 2 times)
So according to formula the answer is 6!/(4! x 2!)



Hope this clears the concept . It is becasue if you arrange two similar objects it is the same like arranging 4 identical white balls among themselves






share|cite|improve this answer









$endgroup$














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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    Going backward:



    Take any one of your permutations, say



    $11110.$



    Assume you had $a,b,c,d$ distinct digits ($not =0$)



    Look at $abcd0$. The number of distinct permutations with the $0$ fixed in the last slot is:



    $4!$ .



    Now you have $5$ permutations listed with four $1$'s and one $0$:



    Look at



    $abcd0, abc0d,ab0cd, a0bcd, 0abcd.$



    where $a,b,c,d$ are distinct ($not =0$).



    You get $5 × 4!=5!$ distinct permutations (Why?).



    Finally :



    If you have five objects to permute where $4$ are identical, you get:



    $5!/4!=5$ distinct permutations .



    Can you generalize for $n$ objects , $r$ of which are identical? How many distinct permutations do you get?






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Thanks. Can you say whether these are permutations with repetition or no repetition?
      $endgroup$
      – user963241
      Jan 16 at 8:03










    • $begingroup$
      I'll try. 5 distinct! objects permuted: abcde, eabcd, ..etc. certainly no repetitions. a=b, then abcde is the same as bacde(ok?), since a=b, I.e you get aacde and aacde.The same.To get rid of repetitions (overcounting) divide by 2! since a=b .
      $endgroup$
      – Peter Szilas
      Jan 16 at 8:12










    • $begingroup$
      So, these are permutations with no repetition even though I have the same 4 objects (1's). Right @Peter?
      $endgroup$
      – user963241
      Jan 16 at 8:56










    • $begingroup$
      Yes!!!Sorry for the roundabout way.You look at your list: 5 permutations without rep.(OK?).With 5 distinct !! objects you get 5! perm, no rep.Now let 4 be identical, you divide 5!/4!= 5(your list) to have #of perm without !! rep.
      $endgroup$
      – Peter Szilas
      Jan 16 at 9:40






    • 1




      $begingroup$
      user963241Looks fine:)
      $endgroup$
      – Peter Szilas
      Jan 16 at 14:43
















    2












    $begingroup$

    Going backward:



    Take any one of your permutations, say



    $11110.$



    Assume you had $a,b,c,d$ distinct digits ($not =0$)



    Look at $abcd0$. The number of distinct permutations with the $0$ fixed in the last slot is:



    $4!$ .



    Now you have $5$ permutations listed with four $1$'s and one $0$:



    Look at



    $abcd0, abc0d,ab0cd, a0bcd, 0abcd.$



    where $a,b,c,d$ are distinct ($not =0$).



    You get $5 × 4!=5!$ distinct permutations (Why?).



    Finally :



    If you have five objects to permute where $4$ are identical, you get:



    $5!/4!=5$ distinct permutations .



    Can you generalize for $n$ objects , $r$ of which are identical? How many distinct permutations do you get?






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Thanks. Can you say whether these are permutations with repetition or no repetition?
      $endgroup$
      – user963241
      Jan 16 at 8:03










    • $begingroup$
      I'll try. 5 distinct! objects permuted: abcde, eabcd, ..etc. certainly no repetitions. a=b, then abcde is the same as bacde(ok?), since a=b, I.e you get aacde and aacde.The same.To get rid of repetitions (overcounting) divide by 2! since a=b .
      $endgroup$
      – Peter Szilas
      Jan 16 at 8:12










    • $begingroup$
      So, these are permutations with no repetition even though I have the same 4 objects (1's). Right @Peter?
      $endgroup$
      – user963241
      Jan 16 at 8:56










    • $begingroup$
      Yes!!!Sorry for the roundabout way.You look at your list: 5 permutations without rep.(OK?).With 5 distinct !! objects you get 5! perm, no rep.Now let 4 be identical, you divide 5!/4!= 5(your list) to have #of perm without !! rep.
      $endgroup$
      – Peter Szilas
      Jan 16 at 9:40






    • 1




      $begingroup$
      user963241Looks fine:)
      $endgroup$
      – Peter Szilas
      Jan 16 at 14:43














    2












    2








    2





    $begingroup$

    Going backward:



    Take any one of your permutations, say



    $11110.$



    Assume you had $a,b,c,d$ distinct digits ($not =0$)



    Look at $abcd0$. The number of distinct permutations with the $0$ fixed in the last slot is:



    $4!$ .



    Now you have $5$ permutations listed with four $1$'s and one $0$:



    Look at



    $abcd0, abc0d,ab0cd, a0bcd, 0abcd.$



    where $a,b,c,d$ are distinct ($not =0$).



    You get $5 × 4!=5!$ distinct permutations (Why?).



    Finally :



    If you have five objects to permute where $4$ are identical, you get:



    $5!/4!=5$ distinct permutations .



    Can you generalize for $n$ objects , $r$ of which are identical? How many distinct permutations do you get?






    share|cite|improve this answer











    $endgroup$



    Going backward:



    Take any one of your permutations, say



    $11110.$



    Assume you had $a,b,c,d$ distinct digits ($not =0$)



    Look at $abcd0$. The number of distinct permutations with the $0$ fixed in the last slot is:



    $4!$ .



    Now you have $5$ permutations listed with four $1$'s and one $0$:



    Look at



    $abcd0, abc0d,ab0cd, a0bcd, 0abcd.$



    where $a,b,c,d$ are distinct ($not =0$).



    You get $5 × 4!=5!$ distinct permutations (Why?).



    Finally :



    If you have five objects to permute where $4$ are identical, you get:



    $5!/4!=5$ distinct permutations .



    Can you generalize for $n$ objects , $r$ of which are identical? How many distinct permutations do you get?







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Jan 16 at 11:47

























    answered Jan 16 at 7:55









    Peter SzilasPeter Szilas

    11.7k2822




    11.7k2822












    • $begingroup$
      Thanks. Can you say whether these are permutations with repetition or no repetition?
      $endgroup$
      – user963241
      Jan 16 at 8:03










    • $begingroup$
      I'll try. 5 distinct! objects permuted: abcde, eabcd, ..etc. certainly no repetitions. a=b, then abcde is the same as bacde(ok?), since a=b, I.e you get aacde and aacde.The same.To get rid of repetitions (overcounting) divide by 2! since a=b .
      $endgroup$
      – Peter Szilas
      Jan 16 at 8:12










    • $begingroup$
      So, these are permutations with no repetition even though I have the same 4 objects (1's). Right @Peter?
      $endgroup$
      – user963241
      Jan 16 at 8:56










    • $begingroup$
      Yes!!!Sorry for the roundabout way.You look at your list: 5 permutations without rep.(OK?).With 5 distinct !! objects you get 5! perm, no rep.Now let 4 be identical, you divide 5!/4!= 5(your list) to have #of perm without !! rep.
      $endgroup$
      – Peter Szilas
      Jan 16 at 9:40






    • 1




      $begingroup$
      user963241Looks fine:)
      $endgroup$
      – Peter Szilas
      Jan 16 at 14:43


















    • $begingroup$
      Thanks. Can you say whether these are permutations with repetition or no repetition?
      $endgroup$
      – user963241
      Jan 16 at 8:03










    • $begingroup$
      I'll try. 5 distinct! objects permuted: abcde, eabcd, ..etc. certainly no repetitions. a=b, then abcde is the same as bacde(ok?), since a=b, I.e you get aacde and aacde.The same.To get rid of repetitions (overcounting) divide by 2! since a=b .
      $endgroup$
      – Peter Szilas
      Jan 16 at 8:12










    • $begingroup$
      So, these are permutations with no repetition even though I have the same 4 objects (1's). Right @Peter?
      $endgroup$
      – user963241
      Jan 16 at 8:56










    • $begingroup$
      Yes!!!Sorry for the roundabout way.You look at your list: 5 permutations without rep.(OK?).With 5 distinct !! objects you get 5! perm, no rep.Now let 4 be identical, you divide 5!/4!= 5(your list) to have #of perm without !! rep.
      $endgroup$
      – Peter Szilas
      Jan 16 at 9:40






    • 1




      $begingroup$
      user963241Looks fine:)
      $endgroup$
      – Peter Szilas
      Jan 16 at 14:43
















    $begingroup$
    Thanks. Can you say whether these are permutations with repetition or no repetition?
    $endgroup$
    – user963241
    Jan 16 at 8:03




    $begingroup$
    Thanks. Can you say whether these are permutations with repetition or no repetition?
    $endgroup$
    – user963241
    Jan 16 at 8:03












    $begingroup$
    I'll try. 5 distinct! objects permuted: abcde, eabcd, ..etc. certainly no repetitions. a=b, then abcde is the same as bacde(ok?), since a=b, I.e you get aacde and aacde.The same.To get rid of repetitions (overcounting) divide by 2! since a=b .
    $endgroup$
    – Peter Szilas
    Jan 16 at 8:12




    $begingroup$
    I'll try. 5 distinct! objects permuted: abcde, eabcd, ..etc. certainly no repetitions. a=b, then abcde is the same as bacde(ok?), since a=b, I.e you get aacde and aacde.The same.To get rid of repetitions (overcounting) divide by 2! since a=b .
    $endgroup$
    – Peter Szilas
    Jan 16 at 8:12












    $begingroup$
    So, these are permutations with no repetition even though I have the same 4 objects (1's). Right @Peter?
    $endgroup$
    – user963241
    Jan 16 at 8:56




    $begingroup$
    So, these are permutations with no repetition even though I have the same 4 objects (1's). Right @Peter?
    $endgroup$
    – user963241
    Jan 16 at 8:56












    $begingroup$
    Yes!!!Sorry for the roundabout way.You look at your list: 5 permutations without rep.(OK?).With 5 distinct !! objects you get 5! perm, no rep.Now let 4 be identical, you divide 5!/4!= 5(your list) to have #of perm without !! rep.
    $endgroup$
    – Peter Szilas
    Jan 16 at 9:40




    $begingroup$
    Yes!!!Sorry for the roundabout way.You look at your list: 5 permutations without rep.(OK?).With 5 distinct !! objects you get 5! perm, no rep.Now let 4 be identical, you divide 5!/4!= 5(your list) to have #of perm without !! rep.
    $endgroup$
    – Peter Szilas
    Jan 16 at 9:40




    1




    1




    $begingroup$
    user963241Looks fine:)
    $endgroup$
    – Peter Szilas
    Jan 16 at 14:43




    $begingroup$
    user963241Looks fine:)
    $endgroup$
    – Peter Szilas
    Jan 16 at 14:43











    1












    $begingroup$

    Yes, it is known as permutations with repetitions, and it has a formula. Say you take 5 different numbers: 1,2,3,4,5,then you can arrange them in 5!=120 days. But when you take 1,1,1,1,0, the 120 permutations will contain repeated formations. e.g. take the permutation 11110. It remains the same even when you interchange the first two 1's. In fact it remains the same even if you arrange the first four 1's in 4!=24 ways. So, 5!/4!=120/24=5 is the answer.
    Now you can derive the formula yourself.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      This is a permutation with a multiset since each number in the sequence appears a fixed number of times.
      $endgroup$
      – N. F. Taussig
      Jan 16 at 11:29
















    1












    $begingroup$

    Yes, it is known as permutations with repetitions, and it has a formula. Say you take 5 different numbers: 1,2,3,4,5,then you can arrange them in 5!=120 days. But when you take 1,1,1,1,0, the 120 permutations will contain repeated formations. e.g. take the permutation 11110. It remains the same even when you interchange the first two 1's. In fact it remains the same even if you arrange the first four 1's in 4!=24 ways. So, 5!/4!=120/24=5 is the answer.
    Now you can derive the formula yourself.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      This is a permutation with a multiset since each number in the sequence appears a fixed number of times.
      $endgroup$
      – N. F. Taussig
      Jan 16 at 11:29














    1












    1








    1





    $begingroup$

    Yes, it is known as permutations with repetitions, and it has a formula. Say you take 5 different numbers: 1,2,3,4,5,then you can arrange them in 5!=120 days. But when you take 1,1,1,1,0, the 120 permutations will contain repeated formations. e.g. take the permutation 11110. It remains the same even when you interchange the first two 1's. In fact it remains the same even if you arrange the first four 1's in 4!=24 ways. So, 5!/4!=120/24=5 is the answer.
    Now you can derive the formula yourself.






    share|cite|improve this answer









    $endgroup$



    Yes, it is known as permutations with repetitions, and it has a formula. Say you take 5 different numbers: 1,2,3,4,5,then you can arrange them in 5!=120 days. But when you take 1,1,1,1,0, the 120 permutations will contain repeated formations. e.g. take the permutation 11110. It remains the same even when you interchange the first two 1's. In fact it remains the same even if you arrange the first four 1's in 4!=24 ways. So, 5!/4!=120/24=5 is the answer.
    Now you can derive the formula yourself.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 16 at 7:54









    SaeeSaee

    488




    488












    • $begingroup$
      This is a permutation with a multiset since each number in the sequence appears a fixed number of times.
      $endgroup$
      – N. F. Taussig
      Jan 16 at 11:29


















    • $begingroup$
      This is a permutation with a multiset since each number in the sequence appears a fixed number of times.
      $endgroup$
      – N. F. Taussig
      Jan 16 at 11:29
















    $begingroup$
    This is a permutation with a multiset since each number in the sequence appears a fixed number of times.
    $endgroup$
    – N. F. Taussig
    Jan 16 at 11:29




    $begingroup$
    This is a permutation with a multiset since each number in the sequence appears a fixed number of times.
    $endgroup$
    – N. F. Taussig
    Jan 16 at 11:29











    0












    $begingroup$

    The formula for Permutations with same repeated objects is :-
    (total number of letters)!/(number of repeats)!



    So in your case :- 1,1,1,1,0
    total numer of letters is 5
    number of repeats is 4
    So the answer according to formula is 5!/4!



    Also to clarify things further lets see one more example :-
    if it was 1,1,1,1,0,0
    Then
    it would be :-
    total number of letters is 6
    number of repeats is 4 and 2 (1 is present 4 times and 0 is present 2 times)
    So according to formula the answer is 6!/(4! x 2!)



    Hope this clears the concept . It is becasue if you arrange two similar objects it is the same like arranging 4 identical white balls among themselves






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      The formula for Permutations with same repeated objects is :-
      (total number of letters)!/(number of repeats)!



      So in your case :- 1,1,1,1,0
      total numer of letters is 5
      number of repeats is 4
      So the answer according to formula is 5!/4!



      Also to clarify things further lets see one more example :-
      if it was 1,1,1,1,0,0
      Then
      it would be :-
      total number of letters is 6
      number of repeats is 4 and 2 (1 is present 4 times and 0 is present 2 times)
      So according to formula the answer is 6!/(4! x 2!)



      Hope this clears the concept . It is becasue if you arrange two similar objects it is the same like arranging 4 identical white balls among themselves






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        The formula for Permutations with same repeated objects is :-
        (total number of letters)!/(number of repeats)!



        So in your case :- 1,1,1,1,0
        total numer of letters is 5
        number of repeats is 4
        So the answer according to formula is 5!/4!



        Also to clarify things further lets see one more example :-
        if it was 1,1,1,1,0,0
        Then
        it would be :-
        total number of letters is 6
        number of repeats is 4 and 2 (1 is present 4 times and 0 is present 2 times)
        So according to formula the answer is 6!/(4! x 2!)



        Hope this clears the concept . It is becasue if you arrange two similar objects it is the same like arranging 4 identical white balls among themselves






        share|cite|improve this answer









        $endgroup$



        The formula for Permutations with same repeated objects is :-
        (total number of letters)!/(number of repeats)!



        So in your case :- 1,1,1,1,0
        total numer of letters is 5
        number of repeats is 4
        So the answer according to formula is 5!/4!



        Also to clarify things further lets see one more example :-
        if it was 1,1,1,1,0,0
        Then
        it would be :-
        total number of letters is 6
        number of repeats is 4 and 2 (1 is present 4 times and 0 is present 2 times)
        So according to formula the answer is 6!/(4! x 2!)



        Hope this clears the concept . It is becasue if you arrange two similar objects it is the same like arranging 4 identical white balls among themselves







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 16 at 8:09









        cognitivecognitive

        264




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