Number of surjections from $1,2,3,4,5,6$ to $a,b,c,d,e$ The 2019 Stack Overflow Developer Survey Results Are InPascal's relation theorem from the book Combinatorics, R. Merris; need some help in clarificationGet the number of subset.Comparing probabilities of drawing balls of certain color, with and without replacementDifferent ways of picking sets producing different results?Number of possibilities of permutation with repetitions with additional equal elements addedStuck trying to understand N Choose K formulaUnderstanding difference between ordered sequences with repetition and unordered sequences with repetitionIs there a relation between the triangular numbers and the combinations with repetition?How many equivalence classes are over $4$-digit strings from $1,2,3,4,5,6$ if strings are in relation of they differ in order or are the same?A subset of three distinct positive integers, each less than 20, is selected. How many subsets will contain exactly one even number?

What do hard-Brexiteers want with respect to the Irish border?

Are USB sockets on wall outlets live all the time, even when the switch is off?

On the insanity of kings as an argument against monarchy

I looked up a future colleague on LinkedIn before I started a job. I told my colleague about it and he seemed surprised. Should I apologize?

How are circuits which use complex ICs normally simulated?

How to manage monthly salary

What is a mixture ratio of propellant?

How to answer pointed "are you quitting" questioning when I don't want them to suspect

What is the use of option -o in the useradd command?

Why is Grand Jury testimony secret?

Which Sci-Fi work first showed weapon of galactic-scale mass destruction?

The difference between dialogue marks

How can I create a character who can assume the widest possible range of creature sizes?

Does duplicating a spell with Wish count as casting that spell?

Why do UK politicians seemingly ignore opinion polls on Brexit?

Why isn't airport relocation done gradually?

Spanish for "widget"

Geography at the pixel level

What does Linus Torvalds mean when he says that Git "never ever" tracks a file?

Inflated grade on resume at previous job, might former employer tell new employer?

Is this food a bread or a loaf?

What is the meaning of Triage in Cybersec world?

Is there a name of the flying bionic bird?

How to create dashed lines/arrows in Illustrator



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



The 2019 Stack Overflow Developer Survey Results Are InPascal's relation theorem from the book Combinatorics, R. Merris; need some help in clarificationGet the number of subset.Comparing probabilities of drawing balls of certain color, with and without replacementDifferent ways of picking sets producing different results?Number of possibilities of permutation with repetitions with additional equal elements addedStuck trying to understand N Choose K formulaUnderstanding difference between ordered sequences with repetition and unordered sequences with repetitionIs there a relation between the triangular numbers and the combinations with repetition?How many equivalence classes are over $4$-digit strings from $1,2,3,4,5,6$ if strings are in relation of they differ in order or are the same?A subset of three distinct positive integers, each less than 20, is selected. How many subsets will contain exactly one even number?










1












$begingroup$


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 $B$?










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
    yesterday










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






  • 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
    yesterday










  • $begingroup$
    " 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
    yesterday















1












$begingroup$


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 $B$?










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
    yesterday










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






  • 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
    yesterday










  • $begingroup$
    " 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
    yesterday













1












1








1


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 $B$?










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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 16 hours ago









N. F. Taussig

45.1k103358




45.1k103358










asked yesterday









ZakuZaku

1829




1829











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










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






  • 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
    yesterday










  • $begingroup$
    " 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
    yesterday
















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










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






  • 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
    yesterday










  • $begingroup$
    " 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
    yesterday















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




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












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




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




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
yesterday




$begingroup$
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
yesterday












$begingroup$
" 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
yesterday




$begingroup$
" 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
yesterday










3 Answers
3






active

oldest

votes


















4












$begingroup$

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



Answer: $binom62 = 15$



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









    $endgroup$




















      0












      $begingroup$

      (i). Select a $2$-member $A_1subset A.$ There are $binom 62$ ways to do this. Select a $1$-member $B_1subset B.$ There are $binom 51$ ways to do this. There are $binom 62binom 51$ 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 $binom 62binom 514!=(15)(5)(4!)=(15)(5!)=1800$ 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 $binom 51$ of $B_1$'s and the number $4$! of bijections from a $4$-member set to another.






      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%2fnumber-of-surjections-from-1-2-3-4-5-6-to-a-b-c-d-e%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









        4












        $begingroup$

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



        Answer: $binom62 = 15$



        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: $binom62 = 15$



          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: $binom62 = 15$



            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: $binom62 = 15$



            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 yesterday









            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 23 hours ago









                    fleabloodfleablood

                    73.9k22891




                    73.9k22891





















                        0












                        $begingroup$

                        (i). Select a $2$-member $A_1subset A.$ There are $binom 62$ ways to do this. Select a $1$-member $B_1subset B.$ There are $binom 51$ ways to do this. There are $binom 62binom 51$ 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 $binom 62binom 514!=(15)(5)(4!)=(15)(5!)=1800$ 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 $binom 51$ of $B_1$'s and the number $4$! of bijections from a $4$-member set to another.






                        share|cite|improve this answer











                        $endgroup$

















                          0












                          $begingroup$

                          (i). Select a $2$-member $A_1subset A.$ There are $binom 62$ ways to do this. Select a $1$-member $B_1subset B.$ There are $binom 51$ ways to do this. There are $binom 62binom 51$ 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 $binom 62binom 514!=(15)(5)(4!)=(15)(5!)=1800$ 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 $binom 51$ of $B_1$'s and the number $4$! of bijections from a $4$-member set to another.






                          share|cite|improve this answer











                          $endgroup$















                            0












                            0








                            0





                            $begingroup$

                            (i). Select a $2$-member $A_1subset A.$ There are $binom 62$ ways to do this. Select a $1$-member $B_1subset B.$ There are $binom 51$ ways to do this. There are $binom 62binom 51$ 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 $binom 62binom 514!=(15)(5)(4!)=(15)(5!)=1800$ 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 $binom 51$ 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 $binom 62$ ways to do this. Select a $1$-member $B_1subset B.$ There are $binom 51$ ways to do this. There are $binom 62binom 51$ 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 $binom 62binom 514!=(15)(5)(4!)=(15)(5!)=1800$ 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 $binom 51$ 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 20 hours ago

























                            answered 20 hours 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%2fnumber-of-surjections-from-1-2-3-4-5-6-to-a-b-c-d-e%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

                                Reverse int within the 32-bit signed integer range: [−2^31, 2^31 − 1]Combining two 32-bit integers into one 64-bit integerDetermine if an int is within rangeLossy packing 32 bit integer to 16 bitComputing the square root of a 64-bit integerKeeping integer addition within boundsSafe multiplication of two 64-bit signed integersLeetcode 10: Regular Expression MatchingSigned integer-to-ascii x86_64 assembler macroReverse the digits of an Integer“Add two numbers given in reverse order from a linked list”

                                Category:Fedor von Bock Media in category "Fedor von Bock"Navigation menuUpload mediaISNI: 0000 0000 5511 3417VIAF ID: 24712551GND ID: 119294796Library of Congress authority ID: n96068363BnF ID: 12534305fSUDOC authorities ID: 034604189Open Library ID: OL338253ANKCR AUT ID: jn19990000869National Library of Israel ID: 000514068National Thesaurus for Author Names ID: 341574317ReasonatorScholiaStatistics

                                Kiel Indholdsfortegnelse Historie | Transport og færgeforbindelser | Sejlsport og anden sport | Kultur | Kendte personer fra Kiel | Noter | Litteratur | Eksterne henvisninger | Navigationsmenuwww.kiel.de54°19′31″N 10°8′26″Ø / 54.32528°N 10.14056°Ø / 54.32528; 10.14056Oberbürgermeister Dr. Ulf Kämpferwww.statistik-nord.deDen danske Stats StatistikKiels hjemmesiderrrWorldCat312794080n790547494030481-4