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Find the number of surjections from A to B.

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Find the number of surjections from A to B.

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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

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asked 3 hours ago

ZakuZaku

1879

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  

  
  
  
  

add a comment | 

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

asked 3 hours ago

ZakuZaku

1879

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  

  
  
  
  

add a comment | 

$begingroup$

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

asked 3 hours ago

ZakuZaku

1879

$endgroup$

share|cite|improve this question

asked 3 hours ago

ZakuZaku

1879

share|cite|improve this question

asked 3 hours ago

ZakuZaku

1879

asked 3 hours ago

ZakuZaku

1879

asked 3 hours ago

ZakuZaku

1879

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

answered 3 hours ago

CopyPasteItCopyPasteIt

4,3271828

$endgroup$

add a comment | 

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

answered 3 hours ago

fleabloodfleablood

73.9k22891

$endgroup$

add a comment | 

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

answered 16 mins ago

DanielWainfleetDanielWainfleet

35.8k31648

$endgroup$

add a comment | 

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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

answered 3 hours ago

CopyPasteItCopyPasteIt

4,3271828

$endgroup$

add a comment | 

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

answered 3 hours ago

CopyPasteItCopyPasteIt

4,3271828

$endgroup$

add a comment | 

$begingroup$

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

answered 3 hours ago

CopyPasteItCopyPasteIt

4,3271828

$endgroup$

share|cite|improve this answer

answered 3 hours ago

CopyPasteItCopyPasteIt

4,3271828

answered 3 hours ago

CopyPasteItCopyPasteIt

4,3271828

answered 3 hours ago

CopyPasteItCopyPasteIt

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

answered 3 hours ago

fleabloodfleablood

73.9k22891

$endgroup$

add a comment | 

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

answered 3 hours ago

fleabloodfleablood

73.9k22891

$endgroup$

add a comment | 

$begingroup$

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

answered 3 hours ago

fleabloodfleablood

73.9k22891

$endgroup$

share|cite|improve this answer

answered 3 hours ago

fleabloodfleablood

73.9k22891

answered 3 hours ago

fleabloodfleablood

73.9k22891

answered 3 hours ago

fleabloodfleablood

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

answered 16 mins ago

DanielWainfleetDanielWainfleet

35.8k31648

$endgroup$

add a comment | 

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

answered 16 mins ago

DanielWainfleetDanielWainfleet

35.8k31648

$endgroup$

add a comment | 

$begingroup$

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

answered 16 mins ago

DanielWainfleetDanielWainfleet

35.8k31648

$endgroup$

share|cite|improve this answer

answered 16 mins ago

DanielWainfleetDanielWainfleet

35.8k31648

answered 16 mins ago

DanielWainfleetDanielWainfleet

35.8k31648

answered 16 mins ago

DanielWainfleetDanielWainfleet

35.8k31648