The Knights Quest round table











up vote
0
down vote

favorite












pg. 20 of The Art of Enumerative Combinatorics, George E. Martin.



How many ways are there to select 4 knights from 15 knights sitting at a round table if no adjacent knights can be chosen?



On the next page he generalizes the problem to the following:



Suppose there are k knights and that q are to be chosen for the quest with no adjacent knights chosen. If we calculate the number of quests that have Lancelot as a member and multiply this number by k, then this product equals q times the number that we are seeking.



I don't understand why this last statement is true. I understand how to get the number of quests with lucky Lancelot as a member
${k-q-1} choose{q-1}$ but why should this number multiplied by k equal q times the total number of quests? I do not know how to visualize the second part of the problem.






combinatorics discrete-mathematics







share|cite|improve this question






















  • You should be dividing by $q$ for the number of quests.
    – Ross Millikan
    11 hours ago















up vote
0
down vote

favorite












pg. 20 of The Art of Enumerative Combinatorics, George E. Martin.



How many ways are there to select 4 knights from 15 knights sitting at a round table if no adjacent knights can be chosen?



On the next page he generalizes the problem to the following:



Suppose there are k knights and that q are to be chosen for the quest with no adjacent knights chosen. If we calculate the number of quests that have Lancelot as a member and multiply this number by k, then this product equals q times the number that we are seeking.



I don't understand why this last statement is true. I understand how to get the number of quests with lucky Lancelot as a member
${k-q-1} choose{q-1}$ but why should this number multiplied by k equal q times the total number of quests? I do not know how to visualize the second part of the problem.






combinatorics discrete-mathematics







share|cite|improve this question






















  • You should be dividing by $q$ for the number of quests.
    – Ross Millikan
    11 hours ago













up vote
0
down vote

favorite









up vote
0
down vote

favorite











pg. 20 of The Art of Enumerative Combinatorics, George E. Martin.



How many ways are there to select 4 knights from 15 knights sitting at a round table if no adjacent knights can be chosen?



On the next page he generalizes the problem to the following:



Suppose there are k knights and that q are to be chosen for the quest with no adjacent knights chosen. If we calculate the number of quests that have Lancelot as a member and multiply this number by k, then this product equals q times the number that we are seeking.



I don't understand why this last statement is true. I understand how to get the number of quests with lucky Lancelot as a member
${k-q-1} choose{q-1}$ but why should this number multiplied by k equal q times the total number of quests? I do not know how to visualize the second part of the problem.






combinatorics discrete-mathematics







share|cite|improve this question













pg. 20 of The Art of Enumerative Combinatorics, George E. Martin.



How many ways are there to select 4 knights from 15 knights sitting at a round table if no adjacent knights can be chosen?



On the next page he generalizes the problem to the following:



Suppose there are k knights and that q are to be chosen for the quest with no adjacent knights chosen. If we calculate the number of quests that have Lancelot as a member and multiply this number by k, then this product equals q times the number that we are seeking.



I don't understand why this last statement is true. I understand how to get the number of quests with lucky Lancelot as a member
${k-q-1} choose{q-1}$ but why should this number multiplied by k equal q times the total number of quests? I do not know how to visualize the second part of the problem.






combinatorics discrete-mathematics




combinatorics discrete-mathematics






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 12 hours ago









Jason Butler

62




62












  • You should be dividing by $q$ for the number of quests.
    – Ross Millikan
    11 hours ago


















  • You should be dividing by $q$ for the number of quests.
    – Ross Millikan
    11 hours ago
















You should be dividing by $q$ for the number of quests.
– Ross Millikan
11 hours ago




You should be dividing by $q$ for the number of quests.
– Ross Millikan
11 hours ago










1 Answer
1






active

oldest

votes

















up vote
0
down vote













Once you know the number of combinations that include Lancelot, imagine sending each combination (both those including Lancelot and those that do not) out once. By symmetry, each knight is sent out the same number of times, so the total number of trips is $k{k-q-1 choose q-1}$. On each quest there are $q$ knights, so there are $frac kq{k-q-1 choose q-1}$ quests.






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',
    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%2f2995402%2fthe-knights-quest-round-table%23new-answer', 'question_page');
    }
    );

    Post as a guest


















    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    0
    down vote













    Once you know the number of combinations that include Lancelot, imagine sending each combination (both those including Lancelot and those that do not) out once. By symmetry, each knight is sent out the same number of times, so the total number of trips is $k{k-q-1 choose q-1}$. On each quest there are $q$ knights, so there are $frac kq{k-q-1 choose q-1}$ quests.






    share|cite|improve this answer

























      up vote
      0
      down vote













      Once you know the number of combinations that include Lancelot, imagine sending each combination (both those including Lancelot and those that do not) out once. By symmetry, each knight is sent out the same number of times, so the total number of trips is $k{k-q-1 choose q-1}$. On each quest there are $q$ knights, so there are $frac kq{k-q-1 choose q-1}$ quests.






      share|cite|improve this answer























        up vote
        0
        down vote










        up vote
        0
        down vote









        Once you know the number of combinations that include Lancelot, imagine sending each combination (both those including Lancelot and those that do not) out once. By symmetry, each knight is sent out the same number of times, so the total number of trips is $k{k-q-1 choose q-1}$. On each quest there are $q$ knights, so there are $frac kq{k-q-1 choose q-1}$ quests.






        share|cite|improve this answer












        Once you know the number of combinations that include Lancelot, imagine sending each combination (both those including Lancelot and those that do not) out once. By symmetry, each knight is sent out the same number of times, so the total number of trips is $k{k-q-1 choose q-1}$. On each quest there are $q$ knights, so there are $frac kq{k-q-1 choose q-1}$ quests.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 11 hours ago









        Ross Millikan

        286k23195363




        286k23195363






























             

            draft saved


            draft discarded



















































             


            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2995402%2fthe-knights-quest-round-table%23new-answer', 'question_page');
            }
            );

            Post as a guest






































































            -

            Popular posts from this blog

            How to change which sound is reproduced for terminal bell?

            Image
            3 Recently upgrade to 18.10 from 18.04, and using the terminal is becoming annoying: the new alert sound is a drip ( /usr/share/sounds/gnome/default/alerts/drip.ogg ), and the previous one is still located at ( /usr/share/sounds/ubuntu/stereo/bell.ogg ). My issue is that at the gnome-terminal, when autocompleting or pressing left BOTH sounds are reproduced, at the same time. I want to just have ubuntu/stereo/bell.ogg active, not both . At the moment I have been only able to reduce the volume of the alerts (but still both are played), or turn the alerts down (but I dislike both "solutions"). gnome-terminal 18.10 teminal-bell share | improve this question
            Read more

            Title Spacing in Bjornstrup Chapter, Removing Chapter Number From Contents

            Image
            up vote 5 down vote favorite 2 I would like to make it so that the vertical space below the title of each chapter is the same when using the Bjornstrup package (i.e. so that the descenders do not affect the spacing - as shown by the red arrow in the picture below). I think I'm fairly happy with the spacing around "A chapter", so I'd really like titles without descenders (e.g. "Contents") to have the same vertical spacing as "A chapter" below the title. I would also like to remove the large "0" in the contents title. Any help appreciated, as always. documentclass[11pt,a4paper]{book} % Packages[![enter image description here][1]][1] usepackage[left=3.5cm, right=3.5cm, top=4.3cm, bottom=4.25cm]{geometry} usepackage[utf8]{inputenc} usepackage{libertine} usepackage[libertine]{newtx
            Read more

            Can I use Tabulator js library in my java Spring + Thymeleaf project?

            Image
            0 I want to add Tabulator js to my java project. I added tabulator.min.js, tabulator.min.css to resources folder. After that added to my html page these dependencies as recommending in readme.md. <link rel="stylesheet" th:href="@{/css/tabulator.min.css}"> <script type="text/javascript" th:src="@{/js/tabulator.min.js}"></script> Now my tables view all elements, that I have in the html page, but not correct. In the documentation of Tabulator is example how to create header and table, var table = new Tabulator("#example-table", { columns:[ {title:"Name", field:"name", sorter:"string", width:200, editor:true}, {title:"Age", field:"age", sorter:"
            Read more