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How many Surjective functions are the for $A = a,b,c,d,e,f$ to $B = 1,2,3,4$ [duplicate]

Number of surjections from $1,…,m$ to $1,…,n$Surjective, Injective, Bijective Functions from $mathbb Z$ to itselfWhat is the practical benefit of a function being injective? surjective?Calculating the total number of surjective functionsProving functions are injective and surjectiveProve that the following functions defined in $Rto R$ are neither injective nor surjective.Cardinality of the Domain vs Codomain in Surjective (non-injective) & Injective (non-surjective) functionsCould someone explain injectivity and surjectivity of functions?How many injective and surjective functions are there from $A$ to $B$?Number of injective, surjective, bijective functions.Why is the composition of a surjective and injective function neither surjective nor injective?

0

$begingroup$

This question already has an answer here:

• 
  Number of surjections from $1,…,m$ to $1,…,n$
  
  3 answers
  
  

I am having trouble understanding this question.

I know the total number of function is $4^6 = 4096$. There are no injective functions as the domain > co domain.

I know there is a part that I'm missing that has to do with the amount of functions that are neither surjective nor injective.

My train of thought was along the lines of (total - injective (=0) - not injective not surjective) = total number of surjective.

combinatorics functions elementary-set-theory

share|cite|improve this question

edited Mar 30 at 10:35

N. F. Taussig

45.1k103358

asked Mar 30 at 3:22

lowy95lowy95

1

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marked as duplicate by Mike Earnest, Eevee Trainer, N. F. Taussig combinatorics
Users with the  combinatorics badge can single-handedly close combinatorics questions as duplicates and reopen them as needed.

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This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Your thought is completely illogical. Coming to your question, there is a long formula to find this (uses inclusion-exclusion principle). You can find it up on internet.
  $endgroup$
  – Tojrah
  Mar 30 at 3:28
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  okay thanks.... thought it might be on the wrong track. im very new to this and am a little confused
  $endgroup$
  – lowy95
  Mar 30 at 3:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hint: you will need to consider two types of cases: 3 elements of $A$ map to 1 element of $B$ and the other 3 elements of $A$ map to the other 3 elements of $B$; and 2 elements of $A$ map to 1 element of $B$ other 2 elements of $A$ map to another 1 element of $B$ and the remaining 2 elements of $A$ map to 2 elements of $B$.
  $endgroup$
  – Ertxiem
  Mar 30 at 3:37
  

  
  
  
  

add a comment | 

0

$begingroup$

This question already has an answer here:

• 
  Number of surjections from $1,…,m$ to $1,…,n$
  
  3 answers
  
  

I am having trouble understanding this question.

I know the total number of function is $4^6 = 4096$. There are no injective functions as the domain > co domain.

I know there is a part that I'm missing that has to do with the amount of functions that are neither surjective nor injective.

My train of thought was along the lines of (total - injective (=0) - not injective not surjective) = total number of surjective.

combinatorics functions elementary-set-theory

share|cite|improve this question

edited Mar 30 at 10:35

N. F. Taussig

45.1k103358

asked Mar 30 at 3:22

lowy95lowy95

1

$endgroup$

marked as duplicate by Mike Earnest, Eevee Trainer, N. F. Taussig combinatorics
Users with the  combinatorics badge can single-handedly close combinatorics questions as duplicates and reopen them as needed.

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Mar 30 at 10:40

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Your thought is completely illogical. Coming to your question, there is a long formula to find this (uses inclusion-exclusion principle). You can find it up on internet.
  $endgroup$
  – Tojrah
  Mar 30 at 3:28
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  okay thanks.... thought it might be on the wrong track. im very new to this and am a little confused
  $endgroup$
  – lowy95
  Mar 30 at 3:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hint: you will need to consider two types of cases: 3 elements of $A$ map to 1 element of $B$ and the other 3 elements of $A$ map to the other 3 elements of $B$; and 2 elements of $A$ map to 1 element of $B$ other 2 elements of $A$ map to another 1 element of $B$ and the remaining 2 elements of $A$ map to 2 elements of $B$.
  $endgroup$
  – Ertxiem
  Mar 30 at 3:37
  

  
  
  
  

add a comment | 

$begingroup$

This question already has an answer here:

• 
  Number of surjections from $1,…,m$ to $1,…,n$
  
  3 answers
  
  

I am having trouble understanding this question.

I know the total number of function is $4^6 = 4096$. There are no injective functions as the domain > co domain.

I know there is a part that I'm missing that has to do with the amount of functions that are neither surjective nor injective.

My train of thought was along the lines of (total - injective (=0) - not injective not surjective) = total number of surjective.

combinatorics functions elementary-set-theory

share|cite|improve this question

edited Mar 30 at 10:35

N. F. Taussig

45.1k103358

asked Mar 30 at 3:22

lowy95lowy95

1

$endgroup$

share|cite|improve this question

edited Mar 30 at 10:35

N. F. Taussig

45.1k103358

asked Mar 30 at 3:22

lowy95lowy95

1

share|cite|improve this question

edited Mar 30 at 10:35

N. F. Taussig

45.1k103358

asked Mar 30 at 3:22

lowy95lowy95

1

edited Mar 30 at 10:35

N. F. Taussig

45.1k103358

edited Mar 30 at 10:35

N. F. Taussig

45.1k103358

asked Mar 30 at 3:22

lowy95lowy95

1

asked Mar 30 at 3:22

lowy95lowy95

1

1 Answer
1

active

oldest

votes

2

$begingroup$

Since the function is surjective, the pre-image of any element is non-empty. Thus, the maximum number of elements in any pre-image of any element is 3. So we break it up into cases:

Case 1 Some element in B has 3 elements in its preimage: There are $binom41 = 4$ ways to pick that element. Given that element there are $binom63$ = 20 ways to pick the 3 elements in A as the pre-image. Then there are 3!=6 ways to pick the pre-images of the remaining elements. So this case has counted 4*20*6 such functions.

Case 2 Two elements in B each have 2 elements in its preimage: Arguing similarly in Case 1, we have $binom42$=6 ways to pick these two elements. Given the two elements we have $binom62binom42 = 15*6$ ways to pick the pre-images. Then there are 2!=2 ways to pick the pre-image of the remaining elements. So this case has counted 6*15*6*2 such functions.

Now we have exhausted all possibilities, so we just add the numbers together and we are done.

share|cite|improve this answer

answered Mar 30 at 4:46

Joel PereiraJoel Pereira

83719

$endgroup$

add a comment | 

2

$begingroup$

Since the function is surjective, the pre-image of any element is non-empty. Thus, the maximum number of elements in any pre-image of any element is 3. So we break it up into cases:

Case 1 Some element in B has 3 elements in its preimage: There are $binom41 = 4$ ways to pick that element. Given that element there are $binom63$ = 20 ways to pick the 3 elements in A as the pre-image. Then there are 3!=6 ways to pick the pre-images of the remaining elements. So this case has counted 4*20*6 such functions.

Case 2 Two elements in B each have 2 elements in its preimage: Arguing similarly in Case 1, we have $binom42$=6 ways to pick these two elements. Given the two elements we have $binom62binom42 = 15*6$ ways to pick the pre-images. Then there are 2!=2 ways to pick the pre-image of the remaining elements. So this case has counted 6*15*6*2 such functions.

Now we have exhausted all possibilities, so we just add the numbers together and we are done.

share|cite|improve this answer

answered Mar 30 at 4:46

Joel PereiraJoel Pereira

83719

$endgroup$

add a comment | 

2

$begingroup$

Since the function is surjective, the pre-image of any element is non-empty. Thus, the maximum number of elements in any pre-image of any element is 3. So we break it up into cases:

Case 1 Some element in B has 3 elements in its preimage: There are $binom41 = 4$ ways to pick that element. Given that element there are $binom63$ = 20 ways to pick the 3 elements in A as the pre-image. Then there are 3!=6 ways to pick the pre-images of the remaining elements. So this case has counted 4*20*6 such functions.

Case 2 Two elements in B each have 2 elements in its preimage: Arguing similarly in Case 1, we have $binom42$=6 ways to pick these two elements. Given the two elements we have $binom62binom42 = 15*6$ ways to pick the pre-images. Then there are 2!=2 ways to pick the pre-image of the remaining elements. So this case has counted 6*15*6*2 such functions.

Now we have exhausted all possibilities, so we just add the numbers together and we are done.

share|cite|improve this answer

answered Mar 30 at 4:46

Joel PereiraJoel Pereira

83719

$endgroup$

add a comment | 

$begingroup$

Since the function is surjective, the pre-image of any element is non-empty. Thus, the maximum number of elements in any pre-image of any element is 3. So we break it up into cases:

Case 1 Some element in B has 3 elements in its preimage: There are $binom41 = 4$ ways to pick that element. Given that element there are $binom63$ = 20 ways to pick the 3 elements in A as the pre-image. Then there are 3!=6 ways to pick the pre-images of the remaining elements. So this case has counted 4*20*6 such functions.

Case 2 Two elements in B each have 2 elements in its preimage: Arguing similarly in Case 1, we have $binom42$=6 ways to pick these two elements. Given the two elements we have $binom62binom42 = 15*6$ ways to pick the pre-images. Then there are 2!=2 ways to pick the pre-image of the remaining elements. So this case has counted 6*15*6*2 such functions.

Now we have exhausted all possibilities, so we just add the numbers together and we are done.

share|cite|improve this answer

answered Mar 30 at 4:46

Joel PereiraJoel Pereira

83719

$endgroup$

share|cite|improve this answer

answered Mar 30 at 4:46

Joel PereiraJoel Pereira

83719

answered Mar 30 at 4:46

Joel PereiraJoel Pereira

83719

answered Mar 30 at 4:46

Joel PereiraJoel Pereira

83719