Number of surjections from ${1,2,3,4,5,6}$ to ${a,b,c,d,e}$












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 functions







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
    11 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
    11 hours ago






  • 2




    $begingroup$
    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
    11 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
    11 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 functions







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
    11 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
    11 hours ago






  • 2




    $begingroup$
    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
    11 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
    11 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 functions







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 functions




combinatorics functions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 3 hours ago









N. F. Taussig

45.1k103358




45.1k103358










asked 11 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
    11 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
    11 hours ago






  • 2




    $begingroup$
    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
    11 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
    11 hours ago


















  • $begingroup$
    There are $5$ objects, not $4$. One object is the double, but that doesn't change anything,
    $endgroup$
    – lulu
    11 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
    11 hours ago






  • 2




    $begingroup$
    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
    11 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
    11 hours ago
















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




$begingroup$
There are $5$ objects, not $4$. One object is the double, but that doesn't change anything,
$endgroup$
– lulu
11 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
11 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
11 hours ago




2




2




$begingroup$
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
11 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
11 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
11 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
11 hours ago










3 Answers
3






active

oldest

votes


















4












$begingroup$

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









$endgroup$





















    4












    $begingroup$

    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












      $begingroup$

      (i). 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. There are $binom {6}{2}binom {5}{1}$ such pairs $(A_1,B_1)$ and for each pair there is a set $F(A_1,B_1)$ of $4!$ surjections $f:Ato B$ such that ${f(x):xin A_1}=B_1.$



      And if $(A_1, B_1)ne (A'_1, B'_1)$ then the sets $F(A_1,B_1), F(A'_1,B'_1)$ are disjoint.



      So there are at least $binom {6}{2}binom {5}{1}4!=(15)(5)(4!)=(15)(5!)=1800$ surjections.



      (ii). Every surjection $f:Ato B$ belongs to some $F(A_1,B_1)$ so there are at most $1800$ surjections.



      In other words: (i) we didn't count any $f$ more than once, and (ii) we didn't fail to count any $f$.



      Remark. The "mysterious" $5!$ came from two sources: The product of the number $binom {5}{1}$ of $B_1$'s and the number $4$! of bijections from a $4$-member set to another.






      share|cite|improve this answer











      $endgroup$














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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        4












        $begingroup$

        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









        $endgroup$


















          4












          $begingroup$

          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









          $endgroup$
















            4












            4








            4





            $begingroup$

            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 11 hours ago









            CopyPasteItCopyPasteIt

            4,3471828




            4,3471828























                4












                $begingroup$

                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









                $endgroup$


















                  4












                  $begingroup$

                  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









                  $endgroup$
















                    4












                    4








                    4





                    $begingroup$

                    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 10 hours ago









                    fleabloodfleablood

                    73.9k22891




                    73.9k22891























                        1












                        $begingroup$

                        (i). 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. There are $binom {6}{2}binom {5}{1}$ such pairs $(A_1,B_1)$ and for each pair there is a set $F(A_1,B_1)$ of $4!$ surjections $f:Ato B$ such that ${f(x):xin A_1}=B_1.$



                        And if $(A_1, B_1)ne (A'_1, B'_1)$ then the sets $F(A_1,B_1), F(A'_1,B'_1)$ are disjoint.



                        So there are at least $binom {6}{2}binom {5}{1}4!=(15)(5)(4!)=(15)(5!)=1800$ surjections.



                        (ii). Every surjection $f:Ato B$ belongs to some $F(A_1,B_1)$ so there are at most $1800$ surjections.



                        In other words: (i) we didn't count any $f$ more than once, and (ii) we didn't fail to count any $f$.



                        Remark. The "mysterious" $5!$ came from two sources: The product of the number $binom {5}{1}$ of $B_1$'s and the number $4$! of bijections from a $4$-member set to another.






                        share|cite|improve this answer











                        $endgroup$


















                          1












                          $begingroup$

                          (i). 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. There are $binom {6}{2}binom {5}{1}$ such pairs $(A_1,B_1)$ and for each pair there is a set $F(A_1,B_1)$ of $4!$ surjections $f:Ato B$ such that ${f(x):xin A_1}=B_1.$



                          And if $(A_1, B_1)ne (A'_1, B'_1)$ then the sets $F(A_1,B_1), F(A'_1,B'_1)$ are disjoint.



                          So there are at least $binom {6}{2}binom {5}{1}4!=(15)(5)(4!)=(15)(5!)=1800$ surjections.



                          (ii). Every surjection $f:Ato B$ belongs to some $F(A_1,B_1)$ so there are at most $1800$ surjections.



                          In other words: (i) we didn't count any $f$ more than once, and (ii) we didn't fail to count any $f$.



                          Remark. The "mysterious" $5!$ came from two sources: The product of the number $binom {5}{1}$ of $B_1$'s and the number $4$! of bijections from a $4$-member set to another.






                          share|cite|improve this answer











                          $endgroup$
















                            1












                            1








                            1





                            $begingroup$

                            (i). 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. There are $binom {6}{2}binom {5}{1}$ such pairs $(A_1,B_1)$ and for each pair there is a set $F(A_1,B_1)$ of $4!$ surjections $f:Ato B$ such that ${f(x):xin A_1}=B_1.$



                            And if $(A_1, B_1)ne (A'_1, B'_1)$ then the sets $F(A_1,B_1), F(A'_1,B'_1)$ are disjoint.



                            So there are at least $binom {6}{2}binom {5}{1}4!=(15)(5)(4!)=(15)(5!)=1800$ surjections.



                            (ii). Every surjection $f:Ato B$ belongs to some $F(A_1,B_1)$ so there are at most $1800$ surjections.



                            In other words: (i) we didn't count any $f$ more than once, and (ii) we didn't fail to count any $f$.



                            Remark. The "mysterious" $5!$ came from two sources: The product of the number $binom {5}{1}$ of $B_1$'s and the number $4$! of bijections from a $4$-member set to another.






                            share|cite|improve this answer











                            $endgroup$



                            (i). 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. There are $binom {6}{2}binom {5}{1}$ such pairs $(A_1,B_1)$ and for each pair there is a set $F(A_1,B_1)$ of $4!$ surjections $f:Ato B$ such that ${f(x):xin A_1}=B_1.$



                            And if $(A_1, B_1)ne (A'_1, B'_1)$ then the sets $F(A_1,B_1), F(A'_1,B'_1)$ are disjoint.



                            So there are at least $binom {6}{2}binom {5}{1}4!=(15)(5)(4!)=(15)(5!)=1800$ surjections.



                            (ii). Every surjection $f:Ato B$ belongs to some $F(A_1,B_1)$ so there are at most $1800$ surjections.



                            In other words: (i) we didn't count any $f$ more than once, and (ii) we didn't fail to count any $f$.



                            Remark. The "mysterious" $5!$ came from two sources: The product of the number $binom {5}{1}$ of $B_1$'s and the number $4$! of bijections from a $4$-member set to another.







                            share|cite|improve this answer














                            share|cite|improve this answer



                            share|cite|improve this answer








                            edited 7 hours ago

























                            answered 7 hours ago









                            DanielWainfleetDanielWainfleet

                            35.8k31648




                            35.8k31648






























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                                A CLEAN and SIMPLE way to add appendices to Table of Contents and bookmarks

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                                0 I found a lot of questions abount appendices and ToC. Many users want appendices to be grouped in an Appendix part, however some problems arise with ToC, hyperref, PDF viewer bookmarks, and so on. There are different solutions which require extra packages, command patching and other extra code, however none of them satisfies me. I almost found an easy way to accomplish a good result, where appendices are added to bookmarks in the right way and hyperref links point to the right page. However, the number of the "Appendix" part page is wrong (it's the number of appendix A). Is there any EASY way to fix that? This is a MWE: documentclass{book} usepackage[nottoc,notlot,notlof]{tocbibind} usepackage{hyperref} begin{document} frontmatter tableofcontents mainmatter part{First} chapter{...
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                                Calculate evaluation metrics using cross_val_predict sklearn

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                                1 In the sklearn.model_selection.cross_val_predict page it is stated: Generate cross-validated estimates for each input data point. It is not appropriate to pass these predictions into an evaluation metric. Can someone explain what does it mean? If this gives estimate of Y (y prediction) for every Y (true Y), why can't I calculate metrics such as RMSE or coefficient of determination using these results? python scikit-learn cross-validation share | improve this question edited Nov 28 '18 at 17:52 desertnaut 20.3k 7 43 79 ...
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                                Insert data from modal to MySQL (multiple modal on website)

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                                0 I have problem with inserting data to MySQL from modal. My modal: <a href="#" class="badge badge-pill badge-success">6 komentarzy</a> <a href="#" class="badge badge-pill badge-danger">brak komentarzy</a> <a data-toggle="modal" href="#add_desk_comm_{$desk_['desk_id']}" data-target="#add_desk_comm_{$desk_['desk_id']}" class="ediiit">(dodaj)</a> <div class="modal fade" id="add_desk_comm_{$desk_['desk_id']}" tabindex="-1" role="dialog" aria-labelledby="edit_printer" aria-hidden="true"> <div class="modal-dialog" role="document"> ...
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