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Find the number of surjections from A to B.

Pascal's relation theorem from the book Combinatorics, R. Merris; need some help in clarificationCombinatorics: How many solns to equation? (principle of inclusion / exclusion)Get 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?A subset of three distinct positive integers, each less than 20, is selected. How many subsets will contain exactly one even number?

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 4 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 4 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 4 hours ago

ZakuZaku

1879

$endgroup$

share|cite|improve this question

asked 4 hours ago

ZakuZaku

1879

share|cite|improve this question

asked 4 hours ago

ZakuZaku

1879

asked 4 hours ago

ZakuZaku

1879

asked 4 hours ago

ZakuZaku

1879

3 Answers
3

active

oldest

votes

2

$begingroup$

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

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 $binom 62$ ways to do this. Select a $1$-member $B_1subset B.$ There are $binom 51$ 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 $binom 62binom 514!=(15)(5)(4!)=(15)(5!)=1800.$

share|cite|improve this answer

answered 20 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: $binom62 = 15$

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

answered 3 hours ago

CopyPasteItCopyPasteIt

4,3271828

$endgroup$

add a comment | 

2

$begingroup$

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

answered 3 hours ago

CopyPasteItCopyPasteIt

4,3271828

$endgroup$

add a comment | 

$begingroup$

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

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 $binom 62$ ways to do this. Select a $1$-member $B_1subset B.$ There are $binom 51$ 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 $binom 62binom 514!=(15)(5)(4!)=(15)(5!)=1800.$

share|cite|improve this answer

answered 20 mins ago

DanielWainfleetDanielWainfleet

35.8k31648

$endgroup$

add a comment | 

1

$begingroup$

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. 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 $binom 62binom 514!=(15)(5)(4!)=(15)(5!)=1800.$

share|cite|improve this answer

answered 20 mins ago

DanielWainfleetDanielWainfleet

35.8k31648

$endgroup$

add a comment | 

$begingroup$

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. 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 $binom 62binom 514!=(15)(5)(4!)=(15)(5!)=1800.$

share|cite|improve this answer

answered 20 mins ago

DanielWainfleetDanielWainfleet

35.8k31648

$endgroup$

share|cite|improve this answer

answered 20 mins ago

DanielWainfleetDanielWainfleet

35.8k31648

answered 20 mins ago

DanielWainfleetDanielWainfleet

35.8k31648

answered 20 mins ago

DanielWainfleetDanielWainfleet

35.8k31648