Find the number of surjections from A to B.












2












$begingroup$


Where A = {1,2,3,4,5,6} and B = {a,b,c,d,e}.



My book says it's:




  1. Select a two-element subset of A.

  2. Assign images without repetition to the two-element subset and the four
    remaining individual elements of A.


This shows that the total number of surjections from A to B is C(6, 2)5! = 1800.



I'm confused at why it's multiplied by 5! and not by 4!. Also in part 2, when we assign images, do they mean images in B?






combinatorics







share|cite|improve this question









$endgroup$












  • $begingroup$
    There are $5$ objects, not $4$. One object is the double, but that doesn't change anything,
    $endgroup$
    – lulu
    3 hours ago










  • $begingroup$
    I thought since we have a subset of 2, we multiply by 4! since there are 4 elements left in A.
    $endgroup$
    – Zaku
    3 hours ago






  • 1




    $begingroup$
    It's not a question of what's left in $A$. Having paired, say, $1,2$ we now need to count the surjections of ${P,3,4,5,6}$ onto ${a,b,c,d,e}$, where $P$ denotes the pair $(1,2)$. There are clearly $5!$ such surjections.
    $endgroup$
    – lulu
    3 hours ago










  • $begingroup$
    " I thought ..., we multiply by 4! since there are 4 elements left in A." But you haven't chosen which of the 5 elements that subset of 2 map to. Would it make more sense if we said the (number of ways to chose the two that aren't distinct)(choices for that pair)(choices for what is left) $={6choose 2}*5*4! $? That's actually the same thing as (number of ways to chose the two that aren't distinct)(number of choices for the four distinct and the pair)$={6choose 2}*5! $.
    $endgroup$
    – fleablood
    3 hours ago
















2












$begingroup$


Where A = {1,2,3,4,5,6} and B = {a,b,c,d,e}.



My book says it's:




  1. Select a two-element subset of A.

  2. Assign images without repetition to the two-element subset and the four
    remaining individual elements of A.


This shows that the total number of surjections from A to B is C(6, 2)5! = 1800.



I'm confused at why it's multiplied by 5! and not by 4!. Also in part 2, when we assign images, do they mean images in B?






combinatorics







share|cite|improve this question









$endgroup$












  • $begingroup$
    There are $5$ objects, not $4$. One object is the double, but that doesn't change anything,
    $endgroup$
    – lulu
    3 hours ago










  • $begingroup$
    I thought since we have a subset of 2, we multiply by 4! since there are 4 elements left in A.
    $endgroup$
    – Zaku
    3 hours ago






  • 1




    $begingroup$
    It's not a question of what's left in $A$. Having paired, say, $1,2$ we now need to count the surjections of ${P,3,4,5,6}$ onto ${a,b,c,d,e}$, where $P$ denotes the pair $(1,2)$. There are clearly $5!$ such surjections.
    $endgroup$
    – lulu
    3 hours ago










  • $begingroup$
    " I thought ..., we multiply by 4! since there are 4 elements left in A." But you haven't chosen which of the 5 elements that subset of 2 map to. Would it make more sense if we said the (number of ways to chose the two that aren't distinct)(choices for that pair)(choices for what is left) $={6choose 2}*5*4! $? That's actually the same thing as (number of ways to chose the two that aren't distinct)(number of choices for the four distinct and the pair)$={6choose 2}*5! $.
    $endgroup$
    – fleablood
    3 hours ago














2












2








2


2



$begingroup$


Where A = {1,2,3,4,5,6} and B = {a,b,c,d,e}.



My book says it's:




  1. Select a two-element subset of A.

  2. Assign images without repetition to the two-element subset and the four
    remaining individual elements of A.


This shows that the total number of surjections from A to B is C(6, 2)5! = 1800.



I'm confused at why it's multiplied by 5! and not by 4!. Also in part 2, when we assign images, do they mean images in B?






combinatorics







share|cite|improve this question









$endgroup$




Where A = {1,2,3,4,5,6} and B = {a,b,c,d,e}.



My book says it's:




  1. Select a two-element subset of A.

  2. Assign images without repetition to the two-element subset and the four
    remaining individual elements of A.


This shows that the total number of surjections from A to B is C(6, 2)5! = 1800.



I'm confused at why it's multiplied by 5! and not by 4!. Also in part 2, when we assign images, do they mean images in B?






combinatorics




combinatorics






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 3 hours ago









ZakuZaku

1879




1879












  • $begingroup$
    There are $5$ objects, not $4$. One object is the double, but that doesn't change anything,
    $endgroup$
    – lulu
    3 hours ago










  • $begingroup$
    I thought since we have a subset of 2, we multiply by 4! since there are 4 elements left in A.
    $endgroup$
    – Zaku
    3 hours ago






  • 1




    $begingroup$
    It's not a question of what's left in $A$. Having paired, say, $1,2$ we now need to count the surjections of ${P,3,4,5,6}$ onto ${a,b,c,d,e}$, where $P$ denotes the pair $(1,2)$. There are clearly $5!$ such surjections.
    $endgroup$
    – lulu
    3 hours ago










  • $begingroup$
    " I thought ..., we multiply by 4! since there are 4 elements left in A." But you haven't chosen which of the 5 elements that subset of 2 map to. Would it make more sense if we said the (number of ways to chose the two that aren't distinct)(choices for that pair)(choices for what is left) $={6choose 2}*5*4! $? That's actually the same thing as (number of ways to chose the two that aren't distinct)(number of choices for the four distinct and the pair)$={6choose 2}*5! $.
    $endgroup$
    – fleablood
    3 hours ago


















  • $begingroup$
    There are $5$ objects, not $4$. One object is the double, but that doesn't change anything,
    $endgroup$
    – lulu
    3 hours ago










  • $begingroup$
    I thought since we have a subset of 2, we multiply by 4! since there are 4 elements left in A.
    $endgroup$
    – Zaku
    3 hours ago






  • 1




    $begingroup$
    It's not a question of what's left in $A$. Having paired, say, $1,2$ we now need to count the surjections of ${P,3,4,5,6}$ onto ${a,b,c,d,e}$, where $P$ denotes the pair $(1,2)$. There are clearly $5!$ such surjections.
    $endgroup$
    – lulu
    3 hours ago










  • $begingroup$
    " I thought ..., we multiply by 4! since there are 4 elements left in A." But you haven't chosen which of the 5 elements that subset of 2 map to. Would it make more sense if we said the (number of ways to chose the two that aren't distinct)(choices for that pair)(choices for what is left) $={6choose 2}*5*4! $? That's actually the same thing as (number of ways to chose the two that aren't distinct)(number of choices for the four distinct and the pair)$={6choose 2}*5! $.
    $endgroup$
    – fleablood
    3 hours ago
















$begingroup$
There are $5$ objects, not $4$. One object is the double, but that doesn't change anything,
$endgroup$
– lulu
3 hours ago




$begingroup$
There are $5$ objects, not $4$. One object is the double, but that doesn't change anything,
$endgroup$
– lulu
3 hours ago












$begingroup$
I thought since we have a subset of 2, we multiply by 4! since there are 4 elements left in A.
$endgroup$
– Zaku
3 hours ago




$begingroup$
I thought since we have a subset of 2, we multiply by 4! since there are 4 elements left in A.
$endgroup$
– Zaku
3 hours ago




1




1




$begingroup$
It's not a question of what's left in $A$. Having paired, say, $1,2$ we now need to count the surjections of ${P,3,4,5,6}$ onto ${a,b,c,d,e}$, where $P$ denotes the pair $(1,2)$. There are clearly $5!$ such surjections.
$endgroup$
– lulu
3 hours ago




$begingroup$
It's not a question of what's left in $A$. Having paired, say, $1,2$ we now need to count the surjections of ${P,3,4,5,6}$ onto ${a,b,c,d,e}$, where $P$ denotes the pair $(1,2)$. There are clearly $5!$ such surjections.
$endgroup$
– lulu
3 hours ago












$begingroup$
" I thought ..., we multiply by 4! since there are 4 elements left in A." But you haven't chosen which of the 5 elements that subset of 2 map to. Would it make more sense if we said the (number of ways to chose the two that aren't distinct)(choices for that pair)(choices for what is left) $={6choose 2}*5*4! $? That's actually the same thing as (number of ways to chose the two that aren't distinct)(number of choices for the four distinct and the pair)$={6choose 2}*5! $.
$endgroup$
– fleablood
3 hours ago




$begingroup$
" I thought ..., we multiply by 4! since there are 4 elements left in A." But you haven't chosen which of the 5 elements that subset of 2 map to. Would it make more sense if we said the (number of ways to chose the two that aren't distinct)(choices for that pair)(choices for what is left) $={6choose 2}*5*4! $? That's actually the same thing as (number of ways to chose the two that aren't distinct)(number of choices for the four distinct and the pair)$={6choose 2}*5! $.
$endgroup$
– fleablood
3 hours ago










3 Answers
3






active

oldest

votes


















2












$begingroup$

How many ways can $A$ be partitioned into $5$ blocks?



Answer: $binom{6}{2} = 15$



Given any $5text{-block}$ partition of $A$, in how many ways can the blocks be bijectively
assigned to the $5$ element set $B$?



Answer: $5! =120$



How many surjective functions from $A$ onto $B$ are there?



Answer: $15 times 120 = 1800$






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    Think of it this way:



    There is a pair of terms that get mapped to the same element. Call that pair $alpha $. There are four terms remaining. Call them $beta,gamma,delta$ and $epsilon $.



    There are ${6choose 2} $ possible pairs that can be $alpha $.



    And we must map $alpha,beta,gamma,delta,epsilon $ to $a,b,c,d,e $. There is $5! $ ways to do that.






    share|cite|improve this answer









    $endgroup$





















      1












      $begingroup$

      Select a $2$-member $A_1subset A.$ There are $binom {6}{2}$ ways to do this. Select a $1$-member $B_1subset B.$ There are $binom {5}{1}$ ways to do this. For each pair $(A_1,B_1)$ there are $4!$ surjections $f:Ato B$ such that ${f(x):xin A_1}=B_1.$ So we get a total of $binom {6}{2}binom {5}{1}4!=(15)(5)(4!)=(15)(5!)=1800.$






      share|cite|improve this answer









      $endgroup$














        Your Answer





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

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

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

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


        }
        });














        draft saved

        draft discarded


















        StackExchange.ready(
        function () {
        StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3180474%2ffind-the-number-of-surjections-from-a-to-b%23new-answer', 'question_page');
        }
        );

        Post as a guest















        Required, but never shown

























        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        2












        $begingroup$

        How many ways can $A$ be partitioned into $5$ blocks?



        Answer: $binom{6}{2} = 15$



        Given any $5text{-block}$ partition of $A$, in how many ways can the blocks be bijectively
        assigned to the $5$ element set $B$?



        Answer: $5! =120$



        How many surjective functions from $A$ onto $B$ are there?



        Answer: $15 times 120 = 1800$






        share|cite|improve this answer









        $endgroup$


















          2












          $begingroup$

          How many ways can $A$ be partitioned into $5$ blocks?



          Answer: $binom{6}{2} = 15$



          Given any $5text{-block}$ partition of $A$, in how many ways can the blocks be bijectively
          assigned to the $5$ element set $B$?



          Answer: $5! =120$



          How many surjective functions from $A$ onto $B$ are there?



          Answer: $15 times 120 = 1800$






          share|cite|improve this answer









          $endgroup$
















            2












            2








            2





            $begingroup$

            How many ways can $A$ be partitioned into $5$ blocks?



            Answer: $binom{6}{2} = 15$



            Given any $5text{-block}$ partition of $A$, in how many ways can the blocks be bijectively
            assigned to the $5$ element set $B$?



            Answer: $5! =120$



            How many surjective functions from $A$ onto $B$ are there?



            Answer: $15 times 120 = 1800$






            share|cite|improve this answer









            $endgroup$



            How many ways can $A$ be partitioned into $5$ blocks?



            Answer: $binom{6}{2} = 15$



            Given any $5text{-block}$ partition of $A$, in how many ways can the blocks be bijectively
            assigned to the $5$ element set $B$?



            Answer: $5! =120$



            How many surjective functions from $A$ onto $B$ are there?



            Answer: $15 times 120 = 1800$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 3 hours ago









            CopyPasteItCopyPasteIt

            4,3271828




            4,3271828























                2












                $begingroup$

                Think of it this way:



                There is a pair of terms that get mapped to the same element. Call that pair $alpha $. There are four terms remaining. Call them $beta,gamma,delta$ and $epsilon $.



                There are ${6choose 2} $ possible pairs that can be $alpha $.



                And we must map $alpha,beta,gamma,delta,epsilon $ to $a,b,c,d,e $. There is $5! $ ways to do that.






                share|cite|improve this answer









                $endgroup$


















                  2












                  $begingroup$

                  Think of it this way:



                  There is a pair of terms that get mapped to the same element. Call that pair $alpha $. There are four terms remaining. Call them $beta,gamma,delta$ and $epsilon $.



                  There are ${6choose 2} $ possible pairs that can be $alpha $.



                  And we must map $alpha,beta,gamma,delta,epsilon $ to $a,b,c,d,e $. There is $5! $ ways to do that.






                  share|cite|improve this answer









                  $endgroup$
















                    2












                    2








                    2





                    $begingroup$

                    Think of it this way:



                    There is a pair of terms that get mapped to the same element. Call that pair $alpha $. There are four terms remaining. Call them $beta,gamma,delta$ and $epsilon $.



                    There are ${6choose 2} $ possible pairs that can be $alpha $.



                    And we must map $alpha,beta,gamma,delta,epsilon $ to $a,b,c,d,e $. There is $5! $ ways to do that.






                    share|cite|improve this answer









                    $endgroup$



                    Think of it this way:



                    There is a pair of terms that get mapped to the same element. Call that pair $alpha $. There are four terms remaining. Call them $beta,gamma,delta$ and $epsilon $.



                    There are ${6choose 2} $ possible pairs that can be $alpha $.



                    And we must map $alpha,beta,gamma,delta,epsilon $ to $a,b,c,d,e $. There is $5! $ ways to do that.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered 3 hours ago









                    fleabloodfleablood

                    73.9k22891




                    73.9k22891























                        1












                        $begingroup$

                        Select a $2$-member $A_1subset A.$ There are $binom {6}{2}$ ways to do this. Select a $1$-member $B_1subset B.$ There are $binom {5}{1}$ ways to do this. For each pair $(A_1,B_1)$ there are $4!$ surjections $f:Ato B$ such that ${f(x):xin A_1}=B_1.$ So we get a total of $binom {6}{2}binom {5}{1}4!=(15)(5)(4!)=(15)(5!)=1800.$






                        share|cite|improve this answer









                        $endgroup$


















                          1












                          $begingroup$

                          Select a $2$-member $A_1subset A.$ There are $binom {6}{2}$ ways to do this. Select a $1$-member $B_1subset B.$ There are $binom {5}{1}$ ways to do this. For each pair $(A_1,B_1)$ there are $4!$ surjections $f:Ato B$ such that ${f(x):xin A_1}=B_1.$ So we get a total of $binom {6}{2}binom {5}{1}4!=(15)(5)(4!)=(15)(5!)=1800.$






                          share|cite|improve this answer









                          $endgroup$
















                            1












                            1








                            1





                            $begingroup$

                            Select a $2$-member $A_1subset A.$ There are $binom {6}{2}$ ways to do this. Select a $1$-member $B_1subset B.$ There are $binom {5}{1}$ ways to do this. For each pair $(A_1,B_1)$ there are $4!$ surjections $f:Ato B$ such that ${f(x):xin A_1}=B_1.$ So we get a total of $binom {6}{2}binom {5}{1}4!=(15)(5)(4!)=(15)(5!)=1800.$






                            share|cite|improve this answer









                            $endgroup$



                            Select a $2$-member $A_1subset A.$ There are $binom {6}{2}$ ways to do this. Select a $1$-member $B_1subset B.$ There are $binom {5}{1}$ ways to do this. For each pair $(A_1,B_1)$ there are $4!$ surjections $f:Ato B$ such that ${f(x):xin A_1}=B_1.$ So we get a total of $binom {6}{2}binom {5}{1}4!=(15)(5)(4!)=(15)(5!)=1800.$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered 17 mins ago









                            DanielWainfleetDanielWainfleet

                            35.8k31648




                            35.8k31648






























                                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%2f3180474%2ffind-the-number-of-surjections-from-a-to-b%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

                                Hivernacle

                                Image
                                Hivernacle Hivernacle a les cases del Salt a Alcoi Un hivernacle és una construcció més o menys permanent que permet modificar el microclima del seu interior i permetre el conreu de vegetals en millors condicions o fora d'època. [1] Són el nivell superior de l'anomenat conreu protegit que va des d'un mínim que correspon a l'ús de plàstics que cobreixen el terra (mulching) fins al màxim que serien els hivernacles de vidre i amb un sistema complet de calefacció. Etimològicament la paraula hivernacle es refereix al lloc on passen l'hivern les plantes. En la pràctica actual s'hi mantenen tota la temporada de conreu o més. Contingut 1 Funcionament 2 Evolució en l'ús dels hivernacles 3 Característiques constructives dels hivernacles 4 Hivernacles als Països Baixos 5 Referències Funcionament Els que no tenen calefacció aprofiten precisament l'efecte hivernacle que produeix un escalfame
                                Read more

                                Fluorita

                                Image
                                Fluorita Fluorita (blau) amb pirita (daurat). Fórmula química CaF 2 Epònim fluor Localitat tipus Jáchymov Classificació Categoria Halurs Nickel-Strunz 10a ed. 3.AB.25 Nickel-Strunz 9a ed. 3.AB.25 Nickel-Strunz 8a ed. III/A.08 Dana 9.2.1.1 Heys 8.4.7 Propietats Sistema cristal·lí Cúbic Hàbit cristal·lí Sol presentar cubs; menys freqüentment dodecaedres. De vegades hexaoctaedres i tetrahexaedres. Per la combinació d'aquestes formes, les arestes dels cubs són sovint modificades. De vegades els cristalls poden presentar diferències de creixement entre les cares. Sovint s'observen cristalls compostos per sobrecreixement. Pot ser massiva, compacta, terrosa, columnar (poc freqüent), globular en agregats o botroïdal. Estructura cristal·lina a = 5.4626Å Simetria Classe (H-M): m 3 m (4/ m 3 2/ m ) - Hexoctaèdric; Grup espacial: Fm 3 m Color Blanc, groc, verd, violeta, vermell, rosa, blau o negre Mac
                                Read more

                                Hulsita

                                Image
                                Hulsita Hulsita de la localitat tipus Fórmula química Fe 2+ 2 Fe 3+ O 2 BO 3 Epònim Alfred Hulse Brooks Localitat tipus mont Brooks, península de Seward, Nome Borough, Alaska, Estats Units Classificació Categoria borats Nickel-Strunz 10a ed. 6.AB.45 Nickel-Strunz 9a ed. 6.AB.45 Nickel-Strunz 8a ed. V/G.03 Dana 24.2.3.1 Heys 9.6.4 Propietats Sistema cristal·lí monoclínic Estructura cristal·lina a = 10,68Å; b = 3,09Å; c = 5,43Å; β = 94,15° Color negre Macles en [001]; un membre de la macla es troba girat 120° en relació a l'altre Exfoliació bona en {110} Duresa 3 Lluïssor vítria, submetàl·lica Densitat 4,28 g/cm 3 (mesurada); Més informació Estatus IMA mineral heretat (G) i mineral heretat (G) Any d'aprovació 1908 Referències [1] La hulsita és un mineral de la classe dels borats, que pertany al grup de la pinakiolita. Rep el seu nom en honor del nord-americà
                                Read more