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









      $endgroup$














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        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 19 mins ago









                            DanielWainfleetDanielWainfleet

                            35.8k31648




                            35.8k31648






























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