understanding when “order” is implied by the multiplication principle











up vote
1
down vote

favorite












I feel really stupid posting this, but I keep turning in circles, so I'm really hoping someone can help elucidate the correct reasoning for the following problem:



Statement of the Question:




An agricultural researcher has produced multiple varieties of different strains of a particular plant: $color{blue}{text{six varieties of strain}}$ ${color{blue} A}$, $color{green}{text{three varieties of strain}}$ $color{green}{B}$ and $color{orange}{text{two varieties of strain}}$ $color{orange}{C}$. For a particular experiment she has to select four different varieties, (exactly) two of which must be varieties of strain $A$, and no variety is repeated. How many possible groups are there to choose from?






My Attempt & the Source of My Uncertainty:



What I can't settle for myself is whether (or how) order is accounted for. From my understanding of the multiplication principle, we effectively have 4 choices to make:



$$
underset{small{ A_1}}{underline{hspace{1cm}}} quad
%
underset{small{ A_2}}{underline{hspace{1cm}}} quad
%
underset{small{ A_1'}}{underline{hspace{1cm}}} quad
%
underset{small{ A_2'}}{underline{hspace{1cm}}} quad
$$



where there are 6 choices for the first variety of strain $A$; 5 choices for the second variety of strain $A$; then 5 choices for the third variety (as it can be either strain $B$ or strain $C$) and 4 choices for the fourth variety, i.e.



$$
underset{small{ A_1}}{underline{~6~}} ; cdot ;
%
underset{small{ A_2}}{underline{~5~}} ; cdot ;
%
underset{small{ A_1'}}{underline{~5~}} ; cdot ;
%
underset{small{ A_2'}}{underline{~4~}} ; ;
$$



This is where I get confused: does the above give me the number of permutations or the number of combinations?





My Attempt to Resolve My Uncertainty:



Combinations $_nC_r$ can be thought of as an adjustment to the value of a permutation $_nP_r = frac{n!}{(n-r)!}$, by dividing out the number of ways of $r$ distinct objects can be organized ( viz. $r!$):



$$; _nC_r = frac{_nP_r}{color{red}r!} = frac{n!}{(n-r)!color{red}{r!}},$$



so I reasoned that if the above expression did account for order, then I ought to divide it by $4!$, as there are 4 distinct varieties, so I'd end up with:



$$ (6 cdot 5 cdot 5 cdot 4 ) cdot tfrac{1}{4!}; $$



however, when I try reasoning through the problem using combinations, instead, I don't get the same result. In the case of considering combinations, there seem to be only two choices:




  1. choose 2 types of strain $A$: $; _6C_2 = frac{6!}{4!2!} = frac{6cdot 5}{2!}$;


  2. choose 2 types that are not of strain $A$: $; _5C_2 = frac{5!}{3!2!} = frac{5 cdot 4}{2!}$;



the total number of possible outcomes is then their product (again by the multiplication principle):



$$ _6C_2 cdot _5C_2 = frac{6 cdot 5}{2!} cdot frac{5 cdot 4}{2!} = (6cdot 5 cdot 5 cdot 4) cdot frac{1}{2!2!},$$



which is clearly not equal to the expression found, above.



What is really bothering me here is not only that I'm not certain about the final expression, but that I don't understand why I'm getting completely different values.



I can think of ways to resolve the differences (e.g. the permutations should each be divided by $2!$ because they are indistinguishable as $A$'s or $A'$'s, respectively), but doctoring the expression and reasoning backwards doesn't feel like I'm clarifying the actual source of my confusion.



I don't care about the final answer -- I just really want to clarify the reasoning here, as clearly there is something I'm not appreciating in the way these expressions work. Any help would be greatly appreciated





$color{red}{text{EDIT}}:$
I started to edit the post to clarify further, but think that the clarification constitutes a separate question, so will post it separately & link. Thank you kindly to all those who have responded so far.










share|cite|improve this question




























    up vote
    1
    down vote

    favorite












    I feel really stupid posting this, but I keep turning in circles, so I'm really hoping someone can help elucidate the correct reasoning for the following problem:



    Statement of the Question:




    An agricultural researcher has produced multiple varieties of different strains of a particular plant: $color{blue}{text{six varieties of strain}}$ ${color{blue} A}$, $color{green}{text{three varieties of strain}}$ $color{green}{B}$ and $color{orange}{text{two varieties of strain}}$ $color{orange}{C}$. For a particular experiment she has to select four different varieties, (exactly) two of which must be varieties of strain $A$, and no variety is repeated. How many possible groups are there to choose from?






    My Attempt & the Source of My Uncertainty:



    What I can't settle for myself is whether (or how) order is accounted for. From my understanding of the multiplication principle, we effectively have 4 choices to make:



    $$
    underset{small{ A_1}}{underline{hspace{1cm}}} quad
    %
    underset{small{ A_2}}{underline{hspace{1cm}}} quad
    %
    underset{small{ A_1'}}{underline{hspace{1cm}}} quad
    %
    underset{small{ A_2'}}{underline{hspace{1cm}}} quad
    $$



    where there are 6 choices for the first variety of strain $A$; 5 choices for the second variety of strain $A$; then 5 choices for the third variety (as it can be either strain $B$ or strain $C$) and 4 choices for the fourth variety, i.e.



    $$
    underset{small{ A_1}}{underline{~6~}} ; cdot ;
    %
    underset{small{ A_2}}{underline{~5~}} ; cdot ;
    %
    underset{small{ A_1'}}{underline{~5~}} ; cdot ;
    %
    underset{small{ A_2'}}{underline{~4~}} ; ;
    $$



    This is where I get confused: does the above give me the number of permutations or the number of combinations?





    My Attempt to Resolve My Uncertainty:



    Combinations $_nC_r$ can be thought of as an adjustment to the value of a permutation $_nP_r = frac{n!}{(n-r)!}$, by dividing out the number of ways of $r$ distinct objects can be organized ( viz. $r!$):



    $$; _nC_r = frac{_nP_r}{color{red}r!} = frac{n!}{(n-r)!color{red}{r!}},$$



    so I reasoned that if the above expression did account for order, then I ought to divide it by $4!$, as there are 4 distinct varieties, so I'd end up with:



    $$ (6 cdot 5 cdot 5 cdot 4 ) cdot tfrac{1}{4!}; $$



    however, when I try reasoning through the problem using combinations, instead, I don't get the same result. In the case of considering combinations, there seem to be only two choices:




    1. choose 2 types of strain $A$: $; _6C_2 = frac{6!}{4!2!} = frac{6cdot 5}{2!}$;


    2. choose 2 types that are not of strain $A$: $; _5C_2 = frac{5!}{3!2!} = frac{5 cdot 4}{2!}$;



    the total number of possible outcomes is then their product (again by the multiplication principle):



    $$ _6C_2 cdot _5C_2 = frac{6 cdot 5}{2!} cdot frac{5 cdot 4}{2!} = (6cdot 5 cdot 5 cdot 4) cdot frac{1}{2!2!},$$



    which is clearly not equal to the expression found, above.



    What is really bothering me here is not only that I'm not certain about the final expression, but that I don't understand why I'm getting completely different values.



    I can think of ways to resolve the differences (e.g. the permutations should each be divided by $2!$ because they are indistinguishable as $A$'s or $A'$'s, respectively), but doctoring the expression and reasoning backwards doesn't feel like I'm clarifying the actual source of my confusion.



    I don't care about the final answer -- I just really want to clarify the reasoning here, as clearly there is something I'm not appreciating in the way these expressions work. Any help would be greatly appreciated





    $color{red}{text{EDIT}}:$
    I started to edit the post to clarify further, but think that the clarification constitutes a separate question, so will post it separately & link. Thank you kindly to all those who have responded so far.










    share|cite|improve this question


























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      I feel really stupid posting this, but I keep turning in circles, so I'm really hoping someone can help elucidate the correct reasoning for the following problem:



      Statement of the Question:




      An agricultural researcher has produced multiple varieties of different strains of a particular plant: $color{blue}{text{six varieties of strain}}$ ${color{blue} A}$, $color{green}{text{three varieties of strain}}$ $color{green}{B}$ and $color{orange}{text{two varieties of strain}}$ $color{orange}{C}$. For a particular experiment she has to select four different varieties, (exactly) two of which must be varieties of strain $A$, and no variety is repeated. How many possible groups are there to choose from?






      My Attempt & the Source of My Uncertainty:



      What I can't settle for myself is whether (or how) order is accounted for. From my understanding of the multiplication principle, we effectively have 4 choices to make:



      $$
      underset{small{ A_1}}{underline{hspace{1cm}}} quad
      %
      underset{small{ A_2}}{underline{hspace{1cm}}} quad
      %
      underset{small{ A_1'}}{underline{hspace{1cm}}} quad
      %
      underset{small{ A_2'}}{underline{hspace{1cm}}} quad
      $$



      where there are 6 choices for the first variety of strain $A$; 5 choices for the second variety of strain $A$; then 5 choices for the third variety (as it can be either strain $B$ or strain $C$) and 4 choices for the fourth variety, i.e.



      $$
      underset{small{ A_1}}{underline{~6~}} ; cdot ;
      %
      underset{small{ A_2}}{underline{~5~}} ; cdot ;
      %
      underset{small{ A_1'}}{underline{~5~}} ; cdot ;
      %
      underset{small{ A_2'}}{underline{~4~}} ; ;
      $$



      This is where I get confused: does the above give me the number of permutations or the number of combinations?





      My Attempt to Resolve My Uncertainty:



      Combinations $_nC_r$ can be thought of as an adjustment to the value of a permutation $_nP_r = frac{n!}{(n-r)!}$, by dividing out the number of ways of $r$ distinct objects can be organized ( viz. $r!$):



      $$; _nC_r = frac{_nP_r}{color{red}r!} = frac{n!}{(n-r)!color{red}{r!}},$$



      so I reasoned that if the above expression did account for order, then I ought to divide it by $4!$, as there are 4 distinct varieties, so I'd end up with:



      $$ (6 cdot 5 cdot 5 cdot 4 ) cdot tfrac{1}{4!}; $$



      however, when I try reasoning through the problem using combinations, instead, I don't get the same result. In the case of considering combinations, there seem to be only two choices:




      1. choose 2 types of strain $A$: $; _6C_2 = frac{6!}{4!2!} = frac{6cdot 5}{2!}$;


      2. choose 2 types that are not of strain $A$: $; _5C_2 = frac{5!}{3!2!} = frac{5 cdot 4}{2!}$;



      the total number of possible outcomes is then their product (again by the multiplication principle):



      $$ _6C_2 cdot _5C_2 = frac{6 cdot 5}{2!} cdot frac{5 cdot 4}{2!} = (6cdot 5 cdot 5 cdot 4) cdot frac{1}{2!2!},$$



      which is clearly not equal to the expression found, above.



      What is really bothering me here is not only that I'm not certain about the final expression, but that I don't understand why I'm getting completely different values.



      I can think of ways to resolve the differences (e.g. the permutations should each be divided by $2!$ because they are indistinguishable as $A$'s or $A'$'s, respectively), but doctoring the expression and reasoning backwards doesn't feel like I'm clarifying the actual source of my confusion.



      I don't care about the final answer -- I just really want to clarify the reasoning here, as clearly there is something I'm not appreciating in the way these expressions work. Any help would be greatly appreciated





      $color{red}{text{EDIT}}:$
      I started to edit the post to clarify further, but think that the clarification constitutes a separate question, so will post it separately & link. Thank you kindly to all those who have responded so far.










      share|cite|improve this question















      I feel really stupid posting this, but I keep turning in circles, so I'm really hoping someone can help elucidate the correct reasoning for the following problem:



      Statement of the Question:




      An agricultural researcher has produced multiple varieties of different strains of a particular plant: $color{blue}{text{six varieties of strain}}$ ${color{blue} A}$, $color{green}{text{three varieties of strain}}$ $color{green}{B}$ and $color{orange}{text{two varieties of strain}}$ $color{orange}{C}$. For a particular experiment she has to select four different varieties, (exactly) two of which must be varieties of strain $A$, and no variety is repeated. How many possible groups are there to choose from?






      My Attempt & the Source of My Uncertainty:



      What I can't settle for myself is whether (or how) order is accounted for. From my understanding of the multiplication principle, we effectively have 4 choices to make:



      $$
      underset{small{ A_1}}{underline{hspace{1cm}}} quad
      %
      underset{small{ A_2}}{underline{hspace{1cm}}} quad
      %
      underset{small{ A_1'}}{underline{hspace{1cm}}} quad
      %
      underset{small{ A_2'}}{underline{hspace{1cm}}} quad
      $$



      where there are 6 choices for the first variety of strain $A$; 5 choices for the second variety of strain $A$; then 5 choices for the third variety (as it can be either strain $B$ or strain $C$) and 4 choices for the fourth variety, i.e.



      $$
      underset{small{ A_1}}{underline{~6~}} ; cdot ;
      %
      underset{small{ A_2}}{underline{~5~}} ; cdot ;
      %
      underset{small{ A_1'}}{underline{~5~}} ; cdot ;
      %
      underset{small{ A_2'}}{underline{~4~}} ; ;
      $$



      This is where I get confused: does the above give me the number of permutations or the number of combinations?





      My Attempt to Resolve My Uncertainty:



      Combinations $_nC_r$ can be thought of as an adjustment to the value of a permutation $_nP_r = frac{n!}{(n-r)!}$, by dividing out the number of ways of $r$ distinct objects can be organized ( viz. $r!$):



      $$; _nC_r = frac{_nP_r}{color{red}r!} = frac{n!}{(n-r)!color{red}{r!}},$$



      so I reasoned that if the above expression did account for order, then I ought to divide it by $4!$, as there are 4 distinct varieties, so I'd end up with:



      $$ (6 cdot 5 cdot 5 cdot 4 ) cdot tfrac{1}{4!}; $$



      however, when I try reasoning through the problem using combinations, instead, I don't get the same result. In the case of considering combinations, there seem to be only two choices:




      1. choose 2 types of strain $A$: $; _6C_2 = frac{6!}{4!2!} = frac{6cdot 5}{2!}$;


      2. choose 2 types that are not of strain $A$: $; _5C_2 = frac{5!}{3!2!} = frac{5 cdot 4}{2!}$;



      the total number of possible outcomes is then their product (again by the multiplication principle):



      $$ _6C_2 cdot _5C_2 = frac{6 cdot 5}{2!} cdot frac{5 cdot 4}{2!} = (6cdot 5 cdot 5 cdot 4) cdot frac{1}{2!2!},$$



      which is clearly not equal to the expression found, above.



      What is really bothering me here is not only that I'm not certain about the final expression, but that I don't understand why I'm getting completely different values.



      I can think of ways to resolve the differences (e.g. the permutations should each be divided by $2!$ because they are indistinguishable as $A$'s or $A'$'s, respectively), but doctoring the expression and reasoning backwards doesn't feel like I'm clarifying the actual source of my confusion.



      I don't care about the final answer -- I just really want to clarify the reasoning here, as clearly there is something I'm not appreciating in the way these expressions work. Any help would be greatly appreciated





      $color{red}{text{EDIT}}:$
      I started to edit the post to clarify further, but think that the clarification constitutes a separate question, so will post it separately & link. Thank you kindly to all those who have responded so far.







      combinatorics permutations combinations






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 21 at 22:28

























      asked Nov 21 at 21:51









      Rax Adaam

      311111




      311111






















          3 Answers
          3






          active

          oldest

          votes

















          up vote
          2
          down vote
















          An agricultural researcher has produced multiple varieties of different strains of a particular plant: $color{blue}{text{six varieties of strain}}$ ${color{blue} A}$, $color{green}{text{three varieties of strain}}$ $color{green}{B}$ and $color{orange}{text{two varieties of strain}}$ $color{orange}{C}$. For a particular experiment she has to select four different varieties, (exactly) two of which must be varieties of strain $A$, and no variety is repeated. How many possible groups are there to choose from?





          My Attempt to Resolve My Uncertainty:



          Combinations $_nC_r$ can be thought of as an adjustment to the value of a permutation $_nP_r = tfrac{n!}{(n-r)!}$, by dividing out the number of ways of $r$ distinct objects can be organized ( viz. $r!$):




          I find it more intuitive to think of it the other way around. Counting permutations involves counting ways to select the items, then to arrange them. Thus we have ${^nmathrm P_r}={^nmathrm C_r}cdot{r!}$



          All we need do here is select two from six varieties of A, and two from the five other varieties. We have no need to arrange the selections. The answer is thus simply ${^6mathrm C_2}{^5mathrm C_2}$.






          share|cite|improve this answer





















          • Yes: I realized that part of the issue was that I was conflating permutations with the multiplication principle. This was encouraged by they way the material was presented to me (first the multiplication principle, then permutations as a natural way to concisely express the results obtained -- which is why they became conflated in my mind --, then combinations as "permutations with order removed." The reverse perspective you've proposed makes it easier for me to reason about it. Thank you very much. I'll wait just a bit more before choosing an answer, in case there are any other helpful tips!
            – Rax Adaam
            Nov 22 at 19:39


















          up vote
          1
          down vote













          $6.5.5.4$ gives you the number of permutations of four varieties satisfying the given condition in the question, where first two varieties are of strain A and rest two are of strain B.
          Note: $(A_{1}, A_{2}, B_{1}, C_{1)}$ is a permutation considered in '6.5.5.4' and so is $(A_{2}, A_{1}, B_{1}, C_{1)}$ but $(A_{2}, B_{1}, A_{1}, C_{1)}$ isn't. Clearly, as the four chosen varieties cannot permute amongst themselves in 'any' order, dividing it by $4!$ will be incorrect. The method you used in your last approach is alright.






          share|cite|improve this answer








          New contributor




          user50393 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.

























            up vote
            1
            down vote













            The solution you posed using combinations is correct.



            As for your question, as you stated, you first selected a variety of strain $A$, then selected another variety of strain $A$, then selected a variety selected from strains $B$ or $C$, and finally selected another variety selected from strains $B$ or $C$. That is, you selected the strains in that particular order. Therefore, it does not make sense to divide by the $4!$ orders in which you could have picked the same strains had you picked two varieties of strain $A$ and two varieties from strains $B$ and $C$ in any order.



            What we do care about is which strains we pick. Picking varieties $A_1$, $A_2$, $B_1$, and $C_1$ in that order results in the same set of varieties as selecting varieties $A_2$, $A_1$, $C_1$, and $B_1$ in that order. Thus, selecting two varieties of strain $A$ and then selecting two varieties from strains $B$ or $C$ in that order counts each possible selection four times since there are $2!$ orders in which the same varieties of strain $A$ could be selected and $2!$ orders in which the same varieties of strains $B$ or $C$ could be selected. Notice that
            $$frac{6 cdot 5}{2!} cdot frac{5 cdot 4}{2!} = frac{6!}{2!4!} cdot frac{5!}{2!3!} = binom{6}{2}binom{5}{2}$$






            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%2f3008424%2funderstanding-when-order-is-implied-by-the-multiplication-principle%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








              up vote
              2
              down vote
















              An agricultural researcher has produced multiple varieties of different strains of a particular plant: $color{blue}{text{six varieties of strain}}$ ${color{blue} A}$, $color{green}{text{three varieties of strain}}$ $color{green}{B}$ and $color{orange}{text{two varieties of strain}}$ $color{orange}{C}$. For a particular experiment she has to select four different varieties, (exactly) two of which must be varieties of strain $A$, and no variety is repeated. How many possible groups are there to choose from?





              My Attempt to Resolve My Uncertainty:



              Combinations $_nC_r$ can be thought of as an adjustment to the value of a permutation $_nP_r = tfrac{n!}{(n-r)!}$, by dividing out the number of ways of $r$ distinct objects can be organized ( viz. $r!$):




              I find it more intuitive to think of it the other way around. Counting permutations involves counting ways to select the items, then to arrange them. Thus we have ${^nmathrm P_r}={^nmathrm C_r}cdot{r!}$



              All we need do here is select two from six varieties of A, and two from the five other varieties. We have no need to arrange the selections. The answer is thus simply ${^6mathrm C_2}{^5mathrm C_2}$.






              share|cite|improve this answer





















              • Yes: I realized that part of the issue was that I was conflating permutations with the multiplication principle. This was encouraged by they way the material was presented to me (first the multiplication principle, then permutations as a natural way to concisely express the results obtained -- which is why they became conflated in my mind --, then combinations as "permutations with order removed." The reverse perspective you've proposed makes it easier for me to reason about it. Thank you very much. I'll wait just a bit more before choosing an answer, in case there are any other helpful tips!
                – Rax Adaam
                Nov 22 at 19:39















              up vote
              2
              down vote
















              An agricultural researcher has produced multiple varieties of different strains of a particular plant: $color{blue}{text{six varieties of strain}}$ ${color{blue} A}$, $color{green}{text{three varieties of strain}}$ $color{green}{B}$ and $color{orange}{text{two varieties of strain}}$ $color{orange}{C}$. For a particular experiment she has to select four different varieties, (exactly) two of which must be varieties of strain $A$, and no variety is repeated. How many possible groups are there to choose from?





              My Attempt to Resolve My Uncertainty:



              Combinations $_nC_r$ can be thought of as an adjustment to the value of a permutation $_nP_r = tfrac{n!}{(n-r)!}$, by dividing out the number of ways of $r$ distinct objects can be organized ( viz. $r!$):




              I find it more intuitive to think of it the other way around. Counting permutations involves counting ways to select the items, then to arrange them. Thus we have ${^nmathrm P_r}={^nmathrm C_r}cdot{r!}$



              All we need do here is select two from six varieties of A, and two from the five other varieties. We have no need to arrange the selections. The answer is thus simply ${^6mathrm C_2}{^5mathrm C_2}$.






              share|cite|improve this answer





















              • Yes: I realized that part of the issue was that I was conflating permutations with the multiplication principle. This was encouraged by they way the material was presented to me (first the multiplication principle, then permutations as a natural way to concisely express the results obtained -- which is why they became conflated in my mind --, then combinations as "permutations with order removed." The reverse perspective you've proposed makes it easier for me to reason about it. Thank you very much. I'll wait just a bit more before choosing an answer, in case there are any other helpful tips!
                – Rax Adaam
                Nov 22 at 19:39













              up vote
              2
              down vote










              up vote
              2
              down vote












              An agricultural researcher has produced multiple varieties of different strains of a particular plant: $color{blue}{text{six varieties of strain}}$ ${color{blue} A}$, $color{green}{text{three varieties of strain}}$ $color{green}{B}$ and $color{orange}{text{two varieties of strain}}$ $color{orange}{C}$. For a particular experiment she has to select four different varieties, (exactly) two of which must be varieties of strain $A$, and no variety is repeated. How many possible groups are there to choose from?





              My Attempt to Resolve My Uncertainty:



              Combinations $_nC_r$ can be thought of as an adjustment to the value of a permutation $_nP_r = tfrac{n!}{(n-r)!}$, by dividing out the number of ways of $r$ distinct objects can be organized ( viz. $r!$):




              I find it more intuitive to think of it the other way around. Counting permutations involves counting ways to select the items, then to arrange them. Thus we have ${^nmathrm P_r}={^nmathrm C_r}cdot{r!}$



              All we need do here is select two from six varieties of A, and two from the five other varieties. We have no need to arrange the selections. The answer is thus simply ${^6mathrm C_2}{^5mathrm C_2}$.






              share|cite|improve this answer















              An agricultural researcher has produced multiple varieties of different strains of a particular plant: $color{blue}{text{six varieties of strain}}$ ${color{blue} A}$, $color{green}{text{three varieties of strain}}$ $color{green}{B}$ and $color{orange}{text{two varieties of strain}}$ $color{orange}{C}$. For a particular experiment she has to select four different varieties, (exactly) two of which must be varieties of strain $A$, and no variety is repeated. How many possible groups are there to choose from?





              My Attempt to Resolve My Uncertainty:



              Combinations $_nC_r$ can be thought of as an adjustment to the value of a permutation $_nP_r = tfrac{n!}{(n-r)!}$, by dividing out the number of ways of $r$ distinct objects can be organized ( viz. $r!$):




              I find it more intuitive to think of it the other way around. Counting permutations involves counting ways to select the items, then to arrange them. Thus we have ${^nmathrm P_r}={^nmathrm C_r}cdot{r!}$



              All we need do here is select two from six varieties of A, and two from the five other varieties. We have no need to arrange the selections. The answer is thus simply ${^6mathrm C_2}{^5mathrm C_2}$.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Nov 22 at 9:18









              Graham Kemp

              84.2k43378




              84.2k43378












              • Yes: I realized that part of the issue was that I was conflating permutations with the multiplication principle. This was encouraged by they way the material was presented to me (first the multiplication principle, then permutations as a natural way to concisely express the results obtained -- which is why they became conflated in my mind --, then combinations as "permutations with order removed." The reverse perspective you've proposed makes it easier for me to reason about it. Thank you very much. I'll wait just a bit more before choosing an answer, in case there are any other helpful tips!
                – Rax Adaam
                Nov 22 at 19:39


















              • Yes: I realized that part of the issue was that I was conflating permutations with the multiplication principle. This was encouraged by they way the material was presented to me (first the multiplication principle, then permutations as a natural way to concisely express the results obtained -- which is why they became conflated in my mind --, then combinations as "permutations with order removed." The reverse perspective you've proposed makes it easier for me to reason about it. Thank you very much. I'll wait just a bit more before choosing an answer, in case there are any other helpful tips!
                – Rax Adaam
                Nov 22 at 19:39
















              Yes: I realized that part of the issue was that I was conflating permutations with the multiplication principle. This was encouraged by they way the material was presented to me (first the multiplication principle, then permutations as a natural way to concisely express the results obtained -- which is why they became conflated in my mind --, then combinations as "permutations with order removed." The reverse perspective you've proposed makes it easier for me to reason about it. Thank you very much. I'll wait just a bit more before choosing an answer, in case there are any other helpful tips!
              – Rax Adaam
              Nov 22 at 19:39




              Yes: I realized that part of the issue was that I was conflating permutations with the multiplication principle. This was encouraged by they way the material was presented to me (first the multiplication principle, then permutations as a natural way to concisely express the results obtained -- which is why they became conflated in my mind --, then combinations as "permutations with order removed." The reverse perspective you've proposed makes it easier for me to reason about it. Thank you very much. I'll wait just a bit more before choosing an answer, in case there are any other helpful tips!
              – Rax Adaam
              Nov 22 at 19:39










              up vote
              1
              down vote













              $6.5.5.4$ gives you the number of permutations of four varieties satisfying the given condition in the question, where first two varieties are of strain A and rest two are of strain B.
              Note: $(A_{1}, A_{2}, B_{1}, C_{1)}$ is a permutation considered in '6.5.5.4' and so is $(A_{2}, A_{1}, B_{1}, C_{1)}$ but $(A_{2}, B_{1}, A_{1}, C_{1)}$ isn't. Clearly, as the four chosen varieties cannot permute amongst themselves in 'any' order, dividing it by $4!$ will be incorrect. The method you used in your last approach is alright.






              share|cite|improve this answer








              New contributor




              user50393 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
              Check out our Code of Conduct.






















                up vote
                1
                down vote













                $6.5.5.4$ gives you the number of permutations of four varieties satisfying the given condition in the question, where first two varieties are of strain A and rest two are of strain B.
                Note: $(A_{1}, A_{2}, B_{1}, C_{1)}$ is a permutation considered in '6.5.5.4' and so is $(A_{2}, A_{1}, B_{1}, C_{1)}$ but $(A_{2}, B_{1}, A_{1}, C_{1)}$ isn't. Clearly, as the four chosen varieties cannot permute amongst themselves in 'any' order, dividing it by $4!$ will be incorrect. The method you used in your last approach is alright.






                share|cite|improve this answer








                New contributor




                user50393 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                Check out our Code of Conduct.




















                  up vote
                  1
                  down vote










                  up vote
                  1
                  down vote









                  $6.5.5.4$ gives you the number of permutations of four varieties satisfying the given condition in the question, where first two varieties are of strain A and rest two are of strain B.
                  Note: $(A_{1}, A_{2}, B_{1}, C_{1)}$ is a permutation considered in '6.5.5.4' and so is $(A_{2}, A_{1}, B_{1}, C_{1)}$ but $(A_{2}, B_{1}, A_{1}, C_{1)}$ isn't. Clearly, as the four chosen varieties cannot permute amongst themselves in 'any' order, dividing it by $4!$ will be incorrect. The method you used in your last approach is alright.






                  share|cite|improve this answer








                  New contributor




                  user50393 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                  Check out our Code of Conduct.









                  $6.5.5.4$ gives you the number of permutations of four varieties satisfying the given condition in the question, where first two varieties are of strain A and rest two are of strain B.
                  Note: $(A_{1}, A_{2}, B_{1}, C_{1)}$ is a permutation considered in '6.5.5.4' and so is $(A_{2}, A_{1}, B_{1}, C_{1)}$ but $(A_{2}, B_{1}, A_{1}, C_{1)}$ isn't. Clearly, as the four chosen varieties cannot permute amongst themselves in 'any' order, dividing it by $4!$ will be incorrect. The method you used in your last approach is alright.







                  share|cite|improve this answer








                  New contributor




                  user50393 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                  Check out our Code of Conduct.









                  share|cite|improve this answer



                  share|cite|improve this answer






                  New contributor




                  user50393 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                  Check out our Code of Conduct.









                  answered Nov 21 at 22:12









                  user50393

                  134




                  134




                  New contributor




                  user50393 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                  Check out our Code of Conduct.





                  New contributor





                  user50393 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                  Check out our Code of Conduct.






                  user50393 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                  Check out our Code of Conduct.






















                      up vote
                      1
                      down vote













                      The solution you posed using combinations is correct.



                      As for your question, as you stated, you first selected a variety of strain $A$, then selected another variety of strain $A$, then selected a variety selected from strains $B$ or $C$, and finally selected another variety selected from strains $B$ or $C$. That is, you selected the strains in that particular order. Therefore, it does not make sense to divide by the $4!$ orders in which you could have picked the same strains had you picked two varieties of strain $A$ and two varieties from strains $B$ and $C$ in any order.



                      What we do care about is which strains we pick. Picking varieties $A_1$, $A_2$, $B_1$, and $C_1$ in that order results in the same set of varieties as selecting varieties $A_2$, $A_1$, $C_1$, and $B_1$ in that order. Thus, selecting two varieties of strain $A$ and then selecting two varieties from strains $B$ or $C$ in that order counts each possible selection four times since there are $2!$ orders in which the same varieties of strain $A$ could be selected and $2!$ orders in which the same varieties of strains $B$ or $C$ could be selected. Notice that
                      $$frac{6 cdot 5}{2!} cdot frac{5 cdot 4}{2!} = frac{6!}{2!4!} cdot frac{5!}{2!3!} = binom{6}{2}binom{5}{2}$$






                      share|cite|improve this answer

























                        up vote
                        1
                        down vote













                        The solution you posed using combinations is correct.



                        As for your question, as you stated, you first selected a variety of strain $A$, then selected another variety of strain $A$, then selected a variety selected from strains $B$ or $C$, and finally selected another variety selected from strains $B$ or $C$. That is, you selected the strains in that particular order. Therefore, it does not make sense to divide by the $4!$ orders in which you could have picked the same strains had you picked two varieties of strain $A$ and two varieties from strains $B$ and $C$ in any order.



                        What we do care about is which strains we pick. Picking varieties $A_1$, $A_2$, $B_1$, and $C_1$ in that order results in the same set of varieties as selecting varieties $A_2$, $A_1$, $C_1$, and $B_1$ in that order. Thus, selecting two varieties of strain $A$ and then selecting two varieties from strains $B$ or $C$ in that order counts each possible selection four times since there are $2!$ orders in which the same varieties of strain $A$ could be selected and $2!$ orders in which the same varieties of strains $B$ or $C$ could be selected. Notice that
                        $$frac{6 cdot 5}{2!} cdot frac{5 cdot 4}{2!} = frac{6!}{2!4!} cdot frac{5!}{2!3!} = binom{6}{2}binom{5}{2}$$






                        share|cite|improve this answer























                          up vote
                          1
                          down vote










                          up vote
                          1
                          down vote









                          The solution you posed using combinations is correct.



                          As for your question, as you stated, you first selected a variety of strain $A$, then selected another variety of strain $A$, then selected a variety selected from strains $B$ or $C$, and finally selected another variety selected from strains $B$ or $C$. That is, you selected the strains in that particular order. Therefore, it does not make sense to divide by the $4!$ orders in which you could have picked the same strains had you picked two varieties of strain $A$ and two varieties from strains $B$ and $C$ in any order.



                          What we do care about is which strains we pick. Picking varieties $A_1$, $A_2$, $B_1$, and $C_1$ in that order results in the same set of varieties as selecting varieties $A_2$, $A_1$, $C_1$, and $B_1$ in that order. Thus, selecting two varieties of strain $A$ and then selecting two varieties from strains $B$ or $C$ in that order counts each possible selection four times since there are $2!$ orders in which the same varieties of strain $A$ could be selected and $2!$ orders in which the same varieties of strains $B$ or $C$ could be selected. Notice that
                          $$frac{6 cdot 5}{2!} cdot frac{5 cdot 4}{2!} = frac{6!}{2!4!} cdot frac{5!}{2!3!} = binom{6}{2}binom{5}{2}$$






                          share|cite|improve this answer












                          The solution you posed using combinations is correct.



                          As for your question, as you stated, you first selected a variety of strain $A$, then selected another variety of strain $A$, then selected a variety selected from strains $B$ or $C$, and finally selected another variety selected from strains $B$ or $C$. That is, you selected the strains in that particular order. Therefore, it does not make sense to divide by the $4!$ orders in which you could have picked the same strains had you picked two varieties of strain $A$ and two varieties from strains $B$ and $C$ in any order.



                          What we do care about is which strains we pick. Picking varieties $A_1$, $A_2$, $B_1$, and $C_1$ in that order results in the same set of varieties as selecting varieties $A_2$, $A_1$, $C_1$, and $B_1$ in that order. Thus, selecting two varieties of strain $A$ and then selecting two varieties from strains $B$ or $C$ in that order counts each possible selection four times since there are $2!$ orders in which the same varieties of strain $A$ could be selected and $2!$ orders in which the same varieties of strains $B$ or $C$ could be selected. Notice that
                          $$frac{6 cdot 5}{2!} cdot frac{5 cdot 4}{2!} = frac{6!}{2!4!} cdot frac{5!}{2!3!} = binom{6}{2}binom{5}{2}$$







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Nov 21 at 22:14









                          N. F. Taussig

                          42.7k93254




                          42.7k93254






























                               

                              draft saved


                              draft discarded



















































                               


                              draft saved


                              draft discarded














                              StackExchange.ready(
                              function () {
                              StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3008424%2funderstanding-when-order-is-implied-by-the-multiplication-principle%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

                              Berounka

                              Sphinx de Gizeh

                              Different font size/position of beamer's navigation symbols template's content depending on regular/plain...