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












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 $B$?






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
    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 $B$?






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
    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 $B$?






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



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









    $endgroup$





















      0












      $begingroup$

      (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 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 yesterday









                    fleabloodfleablood

                    73.9k22891




                    73.9k22891























                        0












                        $begingroup$

                        (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











                        $endgroup$


















                          0












                          $begingroup$

                          (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











                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            (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 22 hours ago

























                            answered 23 hours ago









                            DanielWainfleetDanielWainfleet

                            35.8k31648




                            35.8k31648






























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                                How to label and detect the document text images

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                                1 $begingroup$ This is what I mean as document text image: I want to label the texts in image as separate blocks and my model should detect these labels as classes. NOTE: This is how the end result should be like: The labels like Block 1, Block 2, Block 3,.. should be Logo, Title, Date,.. Others, etc. Work done: First approach : I tried to implement this method via Object Detection, it didn't work. It didn't even detect any text. Second approach : Then I tried it using PixelLink. As this model is build for scene text detection, it detected each and every text in the image. But this method can detect multiple lines of text if the threshold values are increased. But I have no idea how do I add labels to the text blocks. PIXEL_CLS_WEIGHT_all_ones = 'PIXEL_CLS_WEIGHT_all_ones' PIXEL_C...
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                                Vallis Paradisi

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                                Coordinata: .mw-parser-output .geo-default,.mw-parser-output .geo-dms,.mw-parser-output .geo-dec{display:inline}.mw-parser-output .geo-nondefault,.mw-parser-output .geo-multi-punct{display:none}.mw-parser-output .longitude,.mw-parser-output .latitude{white-space:nowrap} 33°03′S 71°37′W  /  33.05°S 71.617°W  / -33.05; -71.617 Vallis Paradisi Res apud Vicidata repertae: Civitas : Chilia Numerus incolarum : 296 655 Zona horaria : UTC-4 Situs interretialis Nomen officiale : Valparaíso Gubernium Consilium : municipal council of Valparaíso Geographia Superficies : 401 600 000 Metrum quadratum Coniunctiones urbium Urbes gemellae : Ovetum , Badalona , Santa Fe , Ora longa , Sciamhaevum , Quantunia , Callao , Bat Yam , Melaka , Novorossiysk , Busan , Corduba , Odessa , Rosario , Malacca , Barcino , Soteropolis , Franciscopolis (California) , Guanajuato, Silao Tabula aut despectus Despectus in portum sinumque et urbem Vallis Paradisi Vallis Para...
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                                Tabula Rosettana

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                                Tabula multilinguis Rosettana in Museo Britannico ostenditur. Tabula Rosettana, [1] etiam titulo OGIS 90 agnita, est stela decreto de rebus sacris in Aegypto anno 196 a.C.n. lato inscripta. Tabula iuxta Rosettam Aegypti, urbem in delta Nili et ad oram maris Mediterranei iacentem, anno 1799 a milite Francico reperta est. Inventio stelae, linguis duabus scripturisque tribus inscriptae, eruditis Instituti Aegypti statim nuntiata est; ibi enim iussu imperatoris Napoleonis eruditi omnium scientiarum (sub aegide Commissionis Scientiarum et Artium) properaverant cum expeditione Francica. Qua a Britannis mox debellata, tabula Rosettana Londinium missa hodie apud Museum Britannicum iacet. Textus Graecus cito lectus interpretationi textuum Aegyptiorum (in formis hieroglyphica et demotica expressorum) gradatim adiuvit. Denique textum plene interpretatus est Ioannes Franciscus Champollion. Ab opere eruditorum cumulativo coepit hodiernus scripturae hieroglyphicae linguaeque Aegyptiae a...
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