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









$endgroup$














    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3075421%2fpermutations-with-same-repeated-objects%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    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




        264






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3075421%2fpermutations-with-same-repeated-objects%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            Human spaceflight

            Image
            Human spaceflight From Wikipedia, the free encyclopedia Jump to navigation Jump to search "Space traveler" redirects here. For other uses, see Space traveler (disambiguation). @media all and (max-width:720px){.mw-parser-output .tmulti>.thumbinner{width:100%!important;max-width:none!important}.mw-parser-output .tmulti .tsingle{float:none!important;max-width:none!important;width:100%!important;text-align:center}} Apollo 11 crewmember Buzz Aldrin on the Moon, 1969 International Space Station crewmember Tracy Caldwell Dyson views the Earth, 2010 Space Shuttle Discovery heads into space with a crew aboard, STS-121 in 2006 Inside a space suit on the Canadarm, 1993 Human spaceflight (also referred to as crewed spaceflight or manned spaceflight ) is space travel with a crew or passengers aboard the spacecraft. Spacecraft carrying people may be operated directly, by human crew, or it may be e
            Read more

            Can not write log (Is /dev/pts mounted?) - openpty in Ubuntu-on-Windows?

            Image
            0 1 What exactly does the error message beginning with E: mean? I assume it has to do with the file structure on Ubuntu-on-Windows, but what exactly? The following package was automatically installed and is no longer required: os-prober Use 'apt-get autoremove' to remove it. The following extra packages will be installed: libxslt1.1 The following NEW packages will be installed: libxslt1.1 xmlstarlet 0 upgraded, 2 newly installed, 0 to remove and 0 not upgraded. Need to get 435 kB of archives. After this operation, 1023 kB of additional disk space will be used. Get:1 http://archive.ubuntu.com/ubuntu/ trusty/main libxslt1.1 amd64 1.1.28-2build1 [145 kB] Get:2 http://archive.ubuntu.com/ubuntu/ trusty/universe xmlstarlet amd64 1.5.0-1 [290 kB]
            Read more

            張江高科駅

            Image
            張江高科駅 ちょうこうこうか Zhangjiang Hi-Tech Park 所在地 中国上海市浦東新区 所属事業者 上海地鉄運営有限公司（上海軌道交通・駅詳細） 上海浦東現代有軌交通有限公司（駅詳細） テンプレートを表示 張江高科駅 各種表記 繁体字： 張江高科站 簡体字： 张江高科站 拼音 ： Zhàngjiāng Gāokē Zhàn 発音： ヂャンジャン ガオクー ヂャン 英文： Zhangjiang Hi-Tech Park Station テンプレートを表示 張江高科駅 （ちょうこうこうか-えき）は中華人民共和国上海市浦東新区に位置する、上海軌道交通2号線の駅である。 なお、張江高科駅前にある張江軌道電車の 張江地鉄駅駅 についてもここで述べる。 目次 1 利用可能な鉄道路線 2 沿革 3 駅構造 3.1 上海軌道交通 3.2 張江軌道電車 4 駅周辺 5 隣の駅 6 注釈 7 関連項目 利用可能な鉄道路線 上海軌道交通 ■ 2号線 上海浦東現代有軌交通有限公司 ■ 張江軌道電車 沿革 高架駅時代の姿 2000年12月27日 - 2号線の駅が高架駅として開業。 2010年 1月1日 - 張江軌道電車の開業に伴い、張江地鉄駅駅（停留所）を設置。 2月13日 - 駅地下化に伴い高架駅の運用停止。 2月24日 - 10日間の閉鎖を経て地下新駅の開業。 2017年2月9日 - リニューアル工事を経て再開業。乗降客の増加に対応し、ホームへのエレベーター、エスカレーターを増設、券売機エリアを拡張した [1] 。 駅構造 上海軌道交通 上海軌道交通 張江高科駅 5号口 ちょうこうこうか Zhangjiang Hi-Tech Park ◄ 竜陽路 (2.60km) (1.61km) 金科路 ► 所在地 中国上海市浦東新区松涛路 祖冲之路 （中国語版） 北緯31度12分11秒 東経121度34分50秒  /  北緯31.2
            Read more