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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                                Contact image not getting when fetch all contact list from iPhone by CNContact

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                                up vote 0 down vote favorite I know this question already asked but not getting solution. From this code I will get all the information from the contact but image not found when open vcf files on mac os, also not getting when share this file. I use this stackoverflow link here but It's not help full. var contacts = [CNContact]() let keys = [CNContactVCardSerialization.descriptorForRequiredKeys() ] as [Any] let request = CNContactFetchRequest(keysToFetch: keys as! [CNKeyDescriptor]) do { try self.contactStore.enumerateContacts(with: request) { (contact, stop) in // Array containing all unified contacts from everywhere contacts.append(contact) } } catch { print("unable to fetch contacts") } do { let data = try
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                                Idjassú

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                                Idjassú Informação geral Prática Esporte de combate Foco Agarramentos Dureza Contato pleno Outras informações Esporte olímpico Não
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                                count number of partitions of a set with n elements into k subsets

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                                up vote 6 down vote favorite 1 This program is for count number of partitions of a set with n elements into k subsets I am confusing here return k*countP(n-1, k) + countP(n-1, k-1); can some one explain what is happening here? why we are multiplying with k? NOTE->I know this is not the best way to calculate number of partitions that would be DP // A C++ program to count number of partitions // of a set with n elements into k subsets #include<iostream> using namespace std; // Returns count of different partitions of n // elements in k subsets int countP(int n, int k) { // Base cases if (n == 0 || k == 0 || k > n) return 0; if (k == 1 || k == n) return 1; // S(n+1, k) = k*S(n, k) + S(n, k-1) return k*countP(n-1, k) + countP(n-1, k-1); } // Driver program int ma
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