How many possible ways there are to assign $30$ students to $3$ instructors, where each instructor is...












7












$begingroup$


As the title states.



We have 30 people that need to be divided into 3 groups of same size. How many possibilities are there?



I found this problem when looking through old exams of my current subject.



Note: The title is exactly the translated formulation of the question.



It seemed simple enough but my answer doesn't match the solution.
The correct solution is 5,550,996,791,340. How does one solve this?






combinatorics







share|cite|improve this question











$endgroup$












  • $begingroup$
    For more information, look into multinomial coefficients.
    $endgroup$
    – Michael Seifert
    Feb 9 at 14:38
















7












$begingroup$


As the title states.



We have 30 people that need to be divided into 3 groups of same size. How many possibilities are there?



I found this problem when looking through old exams of my current subject.



Note: The title is exactly the translated formulation of the question.



It seemed simple enough but my answer doesn't match the solution.
The correct solution is 5,550,996,791,340. How does one solve this?






combinatorics







share|cite|improve this question











$endgroup$












  • $begingroup$
    For more information, look into multinomial coefficients.
    $endgroup$
    – Michael Seifert
    Feb 9 at 14:38














7












7








7


1



$begingroup$


As the title states.



We have 30 people that need to be divided into 3 groups of same size. How many possibilities are there?



I found this problem when looking through old exams of my current subject.



Note: The title is exactly the translated formulation of the question.



It seemed simple enough but my answer doesn't match the solution.
The correct solution is 5,550,996,791,340. How does one solve this?






combinatorics







share|cite|improve this question











$endgroup$




As the title states.



We have 30 people that need to be divided into 3 groups of same size. How many possibilities are there?



I found this problem when looking through old exams of my current subject.



Note: The title is exactly the translated formulation of the question.



It seemed simple enough but my answer doesn't match the solution.
The correct solution is 5,550,996,791,340. How does one solve this?






combinatorics




combinatorics






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Feb 9 at 23:59









miracle173

7,38022247




7,38022247










asked Feb 9 at 14:11









KroyerKroyer

456




456












  • $begingroup$
    For more information, look into multinomial coefficients.
    $endgroup$
    – Michael Seifert
    Feb 9 at 14:38


















  • $begingroup$
    For more information, look into multinomial coefficients.
    $endgroup$
    – Michael Seifert
    Feb 9 at 14:38
















$begingroup$
For more information, look into multinomial coefficients.
$endgroup$
– Michael Seifert
Feb 9 at 14:38




$begingroup$
For more information, look into multinomial coefficients.
$endgroup$
– Michael Seifert
Feb 9 at 14:38










3 Answers
3






active

oldest

votes


















12












$begingroup$

First, we need to form the first group of $10$ people. There is $C_{30}^{10}$ ways to do this. Then, from the remaining $20$ people you form the second group of $10$ people. You can do it in $C_{20}^{10}$ ways. The remaining $10$ people all go to the third group. So the total number of ways to do this is by fundamental theorem of combinatorics $C_{30}^{10} C_{20}^{10} = frac{30!}{10!20!}frac{20!}{10!10!} = frac{30!}{{(10!)}^3}$






share|cite|improve this answer











$endgroup$









  • 4




    $begingroup$
    For reference, it's $5 550 996 791 340$.
    $endgroup$
    – Eric Duminil
    Feb 9 at 15:52





















11












$begingroup$

It's $frac {30!} {(10!)^3}$ - for each of $30!$ permutations, you set the people in row according to that order, first ten go to the first group, second ten to the second and third to third. In each of these groups order doesn't matter, so you divide by the number of possible orderings.






share|cite|improve this answer









$endgroup$





















    7












    $begingroup$

    The answer on this is $$frac{30!}{10!10!10!}$$



    First place the students on a row ($30!$ possibilities) and assign the first $10$ to instructor $1$, the second $10$ to instructor $2$ and the third $10$ to instructor $3$.



    Now realize that every possible splitup will be counted $10!10!10!$ times.



    So division by $10!10!10!$ will repair.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thank you too, It was so simple after all.
      $endgroup$
      – Kroyer
      Feb 9 at 14:22










    • $begingroup$
      Hey! After splitting them into 3 groups of 10 each won’t we multiply the multinomial by 3! For the number of ways in which these groups can be assigned to the instructors?
      $endgroup$
      – user601297
      Feb 9 at 15:10






    • 2




      $begingroup$
      No. That is already included in our first move where we place the students in a row. To get more grip on this divide $3$ persons instead of $30$. Then the answer is $frac{3!}{1!1!1!}=6$ which is clear. No multiplication needed anymore.
      $endgroup$
      – drhab
      Feb 9 at 16:39










    • $begingroup$
      @user601297 I answered the question in your comment but see now that I forgot to address. So decided to do that afterwards.
      $endgroup$
      – drhab
      Feb 10 at 7:40












    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%2f3106218%2fhow-many-possible-ways-there-are-to-assign-30-students-to-3-instructors-whe%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









    12












    $begingroup$

    First, we need to form the first group of $10$ people. There is $C_{30}^{10}$ ways to do this. Then, from the remaining $20$ people you form the second group of $10$ people. You can do it in $C_{20}^{10}$ ways. The remaining $10$ people all go to the third group. So the total number of ways to do this is by fundamental theorem of combinatorics $C_{30}^{10} C_{20}^{10} = frac{30!}{10!20!}frac{20!}{10!10!} = frac{30!}{{(10!)}^3}$






    share|cite|improve this answer











    $endgroup$









    • 4




      $begingroup$
      For reference, it's $5 550 996 791 340$.
      $endgroup$
      – Eric Duminil
      Feb 9 at 15:52


















    12












    $begingroup$

    First, we need to form the first group of $10$ people. There is $C_{30}^{10}$ ways to do this. Then, from the remaining $20$ people you form the second group of $10$ people. You can do it in $C_{20}^{10}$ ways. The remaining $10$ people all go to the third group. So the total number of ways to do this is by fundamental theorem of combinatorics $C_{30}^{10} C_{20}^{10} = frac{30!}{10!20!}frac{20!}{10!10!} = frac{30!}{{(10!)}^3}$






    share|cite|improve this answer











    $endgroup$









    • 4




      $begingroup$
      For reference, it's $5 550 996 791 340$.
      $endgroup$
      – Eric Duminil
      Feb 9 at 15:52
















    12












    12








    12





    $begingroup$

    First, we need to form the first group of $10$ people. There is $C_{30}^{10}$ ways to do this. Then, from the remaining $20$ people you form the second group of $10$ people. You can do it in $C_{20}^{10}$ ways. The remaining $10$ people all go to the third group. So the total number of ways to do this is by fundamental theorem of combinatorics $C_{30}^{10} C_{20}^{10} = frac{30!}{10!20!}frac{20!}{10!10!} = frac{30!}{{(10!)}^3}$






    share|cite|improve this answer











    $endgroup$



    First, we need to form the first group of $10$ people. There is $C_{30}^{10}$ ways to do this. Then, from the remaining $20$ people you form the second group of $10$ people. You can do it in $C_{20}^{10}$ ways. The remaining $10$ people all go to the third group. So the total number of ways to do this is by fundamental theorem of combinatorics $C_{30}^{10} C_{20}^{10} = frac{30!}{10!20!}frac{20!}{10!10!} = frac{30!}{{(10!)}^3}$







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Feb 9 at 22:21









    jvdhooft

    5,65961641




    5,65961641










    answered Feb 9 at 14:22









    Yanior WegYanior Weg

    2,72711446




    2,72711446








    • 4




      $begingroup$
      For reference, it's $5 550 996 791 340$.
      $endgroup$
      – Eric Duminil
      Feb 9 at 15:52
















    • 4




      $begingroup$
      For reference, it's $5 550 996 791 340$.
      $endgroup$
      – Eric Duminil
      Feb 9 at 15:52










    4




    4




    $begingroup$
    For reference, it's $5 550 996 791 340$.
    $endgroup$
    – Eric Duminil
    Feb 9 at 15:52






    $begingroup$
    For reference, it's $5 550 996 791 340$.
    $endgroup$
    – Eric Duminil
    Feb 9 at 15:52













    11












    $begingroup$

    It's $frac {30!} {(10!)^3}$ - for each of $30!$ permutations, you set the people in row according to that order, first ten go to the first group, second ten to the second and third to third. In each of these groups order doesn't matter, so you divide by the number of possible orderings.






    share|cite|improve this answer









    $endgroup$


















      11












      $begingroup$

      It's $frac {30!} {(10!)^3}$ - for each of $30!$ permutations, you set the people in row according to that order, first ten go to the first group, second ten to the second and third to third. In each of these groups order doesn't matter, so you divide by the number of possible orderings.






      share|cite|improve this answer









      $endgroup$
















        11












        11








        11





        $begingroup$

        It's $frac {30!} {(10!)^3}$ - for each of $30!$ permutations, you set the people in row according to that order, first ten go to the first group, second ten to the second and third to third. In each of these groups order doesn't matter, so you divide by the number of possible orderings.






        share|cite|improve this answer









        $endgroup$



        It's $frac {30!} {(10!)^3}$ - for each of $30!$ permutations, you set the people in row according to that order, first ten go to the first group, second ten to the second and third to third. In each of these groups order doesn't matter, so you divide by the number of possible orderings.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Feb 9 at 14:17









        enedilenedil

        1,479720




        1,479720























            7












            $begingroup$

            The answer on this is $$frac{30!}{10!10!10!}$$



            First place the students on a row ($30!$ possibilities) and assign the first $10$ to instructor $1$, the second $10$ to instructor $2$ and the third $10$ to instructor $3$.



            Now realize that every possible splitup will be counted $10!10!10!$ times.



            So division by $10!10!10!$ will repair.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Thank you too, It was so simple after all.
              $endgroup$
              – Kroyer
              Feb 9 at 14:22










            • $begingroup$
              Hey! After splitting them into 3 groups of 10 each won’t we multiply the multinomial by 3! For the number of ways in which these groups can be assigned to the instructors?
              $endgroup$
              – user601297
              Feb 9 at 15:10






            • 2




              $begingroup$
              No. That is already included in our first move where we place the students in a row. To get more grip on this divide $3$ persons instead of $30$. Then the answer is $frac{3!}{1!1!1!}=6$ which is clear. No multiplication needed anymore.
              $endgroup$
              – drhab
              Feb 9 at 16:39










            • $begingroup$
              @user601297 I answered the question in your comment but see now that I forgot to address. So decided to do that afterwards.
              $endgroup$
              – drhab
              Feb 10 at 7:40
















            7












            $begingroup$

            The answer on this is $$frac{30!}{10!10!10!}$$



            First place the students on a row ($30!$ possibilities) and assign the first $10$ to instructor $1$, the second $10$ to instructor $2$ and the third $10$ to instructor $3$.



            Now realize that every possible splitup will be counted $10!10!10!$ times.



            So division by $10!10!10!$ will repair.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Thank you too, It was so simple after all.
              $endgroup$
              – Kroyer
              Feb 9 at 14:22










            • $begingroup$
              Hey! After splitting them into 3 groups of 10 each won’t we multiply the multinomial by 3! For the number of ways in which these groups can be assigned to the instructors?
              $endgroup$
              – user601297
              Feb 9 at 15:10






            • 2




              $begingroup$
              No. That is already included in our first move where we place the students in a row. To get more grip on this divide $3$ persons instead of $30$. Then the answer is $frac{3!}{1!1!1!}=6$ which is clear. No multiplication needed anymore.
              $endgroup$
              – drhab
              Feb 9 at 16:39










            • $begingroup$
              @user601297 I answered the question in your comment but see now that I forgot to address. So decided to do that afterwards.
              $endgroup$
              – drhab
              Feb 10 at 7:40














            7












            7








            7





            $begingroup$

            The answer on this is $$frac{30!}{10!10!10!}$$



            First place the students on a row ($30!$ possibilities) and assign the first $10$ to instructor $1$, the second $10$ to instructor $2$ and the third $10$ to instructor $3$.



            Now realize that every possible splitup will be counted $10!10!10!$ times.



            So division by $10!10!10!$ will repair.






            share|cite|improve this answer









            $endgroup$



            The answer on this is $$frac{30!}{10!10!10!}$$



            First place the students on a row ($30!$ possibilities) and assign the first $10$ to instructor $1$, the second $10$ to instructor $2$ and the third $10$ to instructor $3$.



            Now realize that every possible splitup will be counted $10!10!10!$ times.



            So division by $10!10!10!$ will repair.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Feb 9 at 14:19









            drhabdrhab

            104k545136




            104k545136












            • $begingroup$
              Thank you too, It was so simple after all.
              $endgroup$
              – Kroyer
              Feb 9 at 14:22










            • $begingroup$
              Hey! After splitting them into 3 groups of 10 each won’t we multiply the multinomial by 3! For the number of ways in which these groups can be assigned to the instructors?
              $endgroup$
              – user601297
              Feb 9 at 15:10






            • 2




              $begingroup$
              No. That is already included in our first move where we place the students in a row. To get more grip on this divide $3$ persons instead of $30$. Then the answer is $frac{3!}{1!1!1!}=6$ which is clear. No multiplication needed anymore.
              $endgroup$
              – drhab
              Feb 9 at 16:39










            • $begingroup$
              @user601297 I answered the question in your comment but see now that I forgot to address. So decided to do that afterwards.
              $endgroup$
              – drhab
              Feb 10 at 7:40


















            • $begingroup$
              Thank you too, It was so simple after all.
              $endgroup$
              – Kroyer
              Feb 9 at 14:22










            • $begingroup$
              Hey! After splitting them into 3 groups of 10 each won’t we multiply the multinomial by 3! For the number of ways in which these groups can be assigned to the instructors?
              $endgroup$
              – user601297
              Feb 9 at 15:10






            • 2




              $begingroup$
              No. That is already included in our first move where we place the students in a row. To get more grip on this divide $3$ persons instead of $30$. Then the answer is $frac{3!}{1!1!1!}=6$ which is clear. No multiplication needed anymore.
              $endgroup$
              – drhab
              Feb 9 at 16:39










            • $begingroup$
              @user601297 I answered the question in your comment but see now that I forgot to address. So decided to do that afterwards.
              $endgroup$
              – drhab
              Feb 10 at 7:40
















            $begingroup$
            Thank you too, It was so simple after all.
            $endgroup$
            – Kroyer
            Feb 9 at 14:22




            $begingroup$
            Thank you too, It was so simple after all.
            $endgroup$
            – Kroyer
            Feb 9 at 14:22












            $begingroup$
            Hey! After splitting them into 3 groups of 10 each won’t we multiply the multinomial by 3! For the number of ways in which these groups can be assigned to the instructors?
            $endgroup$
            – user601297
            Feb 9 at 15:10




            $begingroup$
            Hey! After splitting them into 3 groups of 10 each won’t we multiply the multinomial by 3! For the number of ways in which these groups can be assigned to the instructors?
            $endgroup$
            – user601297
            Feb 9 at 15:10




            2




            2




            $begingroup$
            No. That is already included in our first move where we place the students in a row. To get more grip on this divide $3$ persons instead of $30$. Then the answer is $frac{3!}{1!1!1!}=6$ which is clear. No multiplication needed anymore.
            $endgroup$
            – drhab
            Feb 9 at 16:39




            $begingroup$
            No. That is already included in our first move where we place the students in a row. To get more grip on this divide $3$ persons instead of $30$. Then the answer is $frac{3!}{1!1!1!}=6$ which is clear. No multiplication needed anymore.
            $endgroup$
            – drhab
            Feb 9 at 16:39












            $begingroup$
            @user601297 I answered the question in your comment but see now that I forgot to address. So decided to do that afterwards.
            $endgroup$
            – drhab
            Feb 10 at 7:40




            $begingroup$
            @user601297 I answered the question in your comment but see now that I forgot to address. So decided to do that afterwards.
            $endgroup$
            – drhab
            Feb 10 at 7:40


















            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%2f3106218%2fhow-many-possible-ways-there-are-to-assign-30-students-to-3-instructors-whe%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

            Questions related to Moebius Transform of Characteristic Function of the Primes

            Image
            2 $begingroup$ Consider the function defined in (1) below related to the fundamental prime counting function $pi(x)$ . Note that A143519(n) is not multiplicative. (1) $quad f(x)=sumlimits_{n=1}^{x}A143519(n)$ https://oeis.org/A143519 The following plot illustrates $f(x)$ defined in formula (1) above. Figure (1) : Illustration of $f(x)$ defined in formula (1) The integer zeros of $f(x)$ for $xle 10,000$ are listed in (2) below. (2) $quad$ {1,6,9,12,19,30,79,80,81,116,193,201,287,288,291,668,673,679,680,685,686,1109} The zero crossings of $f(x)$ for $xle 10,000$ where $f(x)$ doesn't settle at zero are listed in (3) below. (3) $quad$ {14,21,33,114,115,118,195,286,290,295,442,445,665,667,670,671,678,682} Question (1) : Does $f(x)$ have a finite number of integer zeros, and if so ...
            Read more

            List of scandals in India

            Image
            List of scandals in India From Wikipedia, the free encyclopedia Jump to navigation Jump to search This is a dynamic list and may never be able to satisfy particular standards for completeness. You can help by expanding it with reliably sourced entries. This article has multiple issues. Please help improve it or discuss these issues on the talk page . (Learn how and when to remove these template messages) This article has an unclear citation style . The references used may be made clearer with a different or consistent style of citation and footnoting. ( January 2014 ) (Learn how and when to remove this template message) This list is in reverse order. As a consequence it appears to be slanted towards recent events. Please try to keep recent events in historical perspective and add more content related to non-recent events. ( May 2013 ) (Learn how and when to remove this template message) ...
            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