How to calculate a subgroup of some permutation?












0














I think is easier to explain it with a specific problem (but would be nice a generic answer):



We have 2 options, $A$ and $B$, counting permutations of $N$ items is easy, but how to calculate the subgroup where we have two $A$'s?

Example with $N = 3$:
$AAA$
$AAB$
$ABA$
$BAA$
$ABB$
$BBA$
$BAB$
$BBB$



To get the $8$ options is just $2^3$, but what is the formula to get "$3$" (permutation with 2 $A$)?










share|cite|improve this question
























  • Welcome to MSE! For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
    – platty
    Dec 1 at 3:45
















0














I think is easier to explain it with a specific problem (but would be nice a generic answer):



We have 2 options, $A$ and $B$, counting permutations of $N$ items is easy, but how to calculate the subgroup where we have two $A$'s?

Example with $N = 3$:
$AAA$
$AAB$
$ABA$
$BAA$
$ABB$
$BBA$
$BAB$
$BBB$



To get the $8$ options is just $2^3$, but what is the formula to get "$3$" (permutation with 2 $A$)?










share|cite|improve this question
























  • Welcome to MSE! For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
    – platty
    Dec 1 at 3:45














0












0








0







I think is easier to explain it with a specific problem (but would be nice a generic answer):



We have 2 options, $A$ and $B$, counting permutations of $N$ items is easy, but how to calculate the subgroup where we have two $A$'s?

Example with $N = 3$:
$AAA$
$AAB$
$ABA$
$BAA$
$ABB$
$BBA$
$BAB$
$BBB$



To get the $8$ options is just $2^3$, but what is the formula to get "$3$" (permutation with 2 $A$)?










share|cite|improve this question















I think is easier to explain it with a specific problem (but would be nice a generic answer):



We have 2 options, $A$ and $B$, counting permutations of $N$ items is easy, but how to calculate the subgroup where we have two $A$'s?

Example with $N = 3$:
$AAA$
$AAB$
$ABA$
$BAA$
$ABB$
$BBA$
$BAB$
$BBB$



To get the $8$ options is just $2^3$, but what is the formula to get "$3$" (permutation with 2 $A$)?







combinatorics permutations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 1 at 3:45









platty

3,360320




3,360320










asked Dec 1 at 3:32









Enrique

1032




1032












  • Welcome to MSE! For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
    – platty
    Dec 1 at 3:45


















  • Welcome to MSE! For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
    – platty
    Dec 1 at 3:45
















Welcome to MSE! For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– platty
Dec 1 at 3:45




Welcome to MSE! For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– platty
Dec 1 at 3:45










1 Answer
1






active

oldest

votes


















4














I might be oversimplifying this, but if order does not matter, then it's simply a question of how many ways can you place two A's in a list of three, which is ${3 choose 2} = 3.$ In general, for N objects with k number of A's it'd be ${N choose k} = frac{N!}{k!(N-k)!}$.






share|cite|improve this answer





















    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%2f3020954%2fhow-to-calculate-a-subgroup-of-some-permutation%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    4














    I might be oversimplifying this, but if order does not matter, then it's simply a question of how many ways can you place two A's in a list of three, which is ${3 choose 2} = 3.$ In general, for N objects with k number of A's it'd be ${N choose k} = frac{N!}{k!(N-k)!}$.






    share|cite|improve this answer


























      4














      I might be oversimplifying this, but if order does not matter, then it's simply a question of how many ways can you place two A's in a list of three, which is ${3 choose 2} = 3.$ In general, for N objects with k number of A's it'd be ${N choose k} = frac{N!}{k!(N-k)!}$.






      share|cite|improve this answer
























        4












        4








        4






        I might be oversimplifying this, but if order does not matter, then it's simply a question of how many ways can you place two A's in a list of three, which is ${3 choose 2} = 3.$ In general, for N objects with k number of A's it'd be ${N choose k} = frac{N!}{k!(N-k)!}$.






        share|cite|improve this answer












        I might be oversimplifying this, but if order does not matter, then it's simply a question of how many ways can you place two A's in a list of three, which is ${3 choose 2} = 3.$ In general, for N objects with k number of A's it'd be ${N choose k} = frac{N!}{k!(N-k)!}$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 1 at 3:54









        eKoontz

        562




        562






























            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.





            Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


            Please pay close attention to the following guidance:


            • 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.


            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%2f3020954%2fhow-to-calculate-a-subgroup-of-some-permutation%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

            Wiesbaden

            Marschland

            Dieringhausen