Number of surjections from $1,2,3,4,5,6$ to $a,b,c,d,e$ The 2019 Stack Overflow Developer Survey Results Are In Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Pascal'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?
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Number of surjections from $1,2,3,4,5,6$ to $a,b,c,d,e$
The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Pascal'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?
$begingroup$
Where $A = 1,2,3,4,5,6$ and $B = a,b,c,d,e$.
My book says it's:
- Select a two-element subset of $A$.
- 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
edited Apr 9 at 10:42
N. F. Taussig
45.2k103358
asked Apr 9 at 2:26
ZakuZaku
3089
$endgroup$
add a comment |
$begingroup$
Where $A = 1,2,3,4,5,6$ and $B = a,b,c,d,e$.
My book says it's:
- Select a two-element subset of $A$.
- 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
edited Apr 9 at 10:42
N. F. Taussig
45.2k103358
asked Apr 9 at 2:26
ZakuZaku
3089
$endgroup$
$begingroup$
There are $5$ objects, not $4$. One object is the double, but that doesn't change anything,
$endgroup$
– lulu
Apr 9 at 2:29
$begingroup$
I thought since we have a subset of 2, we multiply by 4! since there are 4 elements left in A.
$endgroup$
– Zaku
Apr 9 at 2:31
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
Apr 9 at 2:34
$begingroup$
" 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
Apr 9 at 2:50
add a comment |
$begingroup$
Where $A = 1,2,3,4,5,6$ and $B = a,b,c,d,e$.
My book says it's:
- Select a two-element subset of $A$.
- 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
edited Apr 9 at 10:42
N. F. Taussig
45.2k103358
asked Apr 9 at 2:26
ZakuZaku
3089
$endgroup$
Where $A = 1,2,3,4,5,6$ and $B = a,b,c,d,e$.
My book says it's:
- Select a two-element subset of $A$.
- 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
edited Apr 9 at 10:42
N. F. Taussig
45.2k103358
asked Apr 9 at 2:26
ZakuZaku
3089
edited Apr 9 at 10:42
N. F. Taussig
45.2k103358
asked Apr 9 at 2:26
ZakuZaku
3089
edited Apr 9 at 10:42
N. F. Taussig
45.2k103358
edited Apr 9 at 10:42
N. F. Taussig
45.2k103358
edited Apr 9 at 10:42
N. F. Taussig
45.2k103358
45.2k103358
asked Apr 9 at 2:26
ZakuZaku
3089
asked Apr 9 at 2:26
ZakuZaku
3089
asked Apr 9 at 2:26
ZakuZaku
3089
3089
$begingroup$
There are $5$ objects, not $4$. One object is the double, but that doesn't change anything,
$endgroup$
– lulu
Apr 9 at 2:29
$begingroup$
I thought since we have a subset of 2, we multiply by 4! since there are 4 elements left in A.
$endgroup$
– Zaku
Apr 9 at 2:31
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
Apr 9 at 2:34
$begingroup$
" 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
Apr 9 at 2:50
add a comment |
$begingroup$
There are $5$ objects, not $4$. One object is the double, but that doesn't change anything,
$endgroup$
– lulu
Apr 9 at 2:29
$begingroup$
I thought since we have a subset of 2, we multiply by 4! since there are 4 elements left in A.
$endgroup$
– Zaku
Apr 9 at 2:31
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
Apr 9 at 2:34
$begingroup$
" 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
Apr 9 at 2:50
$begingroup$
There are $5$ objects, not $4$. One object is the double, but that doesn't change anything,
$endgroup$
– lulu
Apr 9 at 2:29
$begingroup$
There are $5$ objects, not $4$. One object is the double, but that doesn't change anything,
$endgroup$
– lulu
Apr 9 at 2:29
$begingroup$
I thought since we have a subset of 2, we multiply by 4! since there are 4 elements left in A.
$endgroup$
– Zaku
Apr 9 at 2:31
$begingroup$
I thought since we have a subset of 2, we multiply by 4! since there are 4 elements left in A.
$endgroup$
– Zaku
Apr 9 at 2:31
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
Apr 9 at 2:34
$begingroup$
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
Apr 9 at 2:34
$begingroup$
" 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
Apr 9 at 2:50
$begingroup$
" 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
Apr 9 at 2:50
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
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: $15 times 120 = 1800$
answered Apr 9 at 2:48
CopyPasteItCopyPasteIt
4,3571828
$endgroup$
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.
answered Apr 9 at 2:59
fleabloodfleablood
1
$endgroup$
add a comment |
$begingroup$
(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.
answered Apr 9 at 6:07
DanielWainfleetDanielWainfleet
35.8k31648
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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: $15 times 120 = 1800$
answered Apr 9 at 2:48
CopyPasteItCopyPasteIt
4,3571828
$endgroup$
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: $15 times 120 = 1800$
answered Apr 9 at 2:48
CopyPasteItCopyPasteIt
4,3571828
$endgroup$
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: $15 times 120 = 1800$
answered Apr 9 at 2:48
CopyPasteItCopyPasteIt
4,3571828
$endgroup$
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: $15 times 120 = 1800$
answered Apr 9 at 2:48
CopyPasteItCopyPasteIt
4,3571828
answered Apr 9 at 2:48
CopyPasteItCopyPasteIt
4,3571828
answered Apr 9 at 2:48
CopyPasteItCopyPasteIt
4,3571828
answered Apr 9 at 2:48
CopyPasteItCopyPasteIt
4,3571828
4,3571828
add a comment |
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.
answered Apr 9 at 2:59
fleabloodfleablood
1
$endgroup$
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.
answered Apr 9 at 2:59
fleabloodfleablood
1
$endgroup$
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.
answered Apr 9 at 2:59
fleabloodfleablood
1
$endgroup$
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.
answered Apr 9 at 2:59
fleabloodfleablood
1
answered Apr 9 at 2:59
fleabloodfleablood
1
answered Apr 9 at 2:59
fleabloodfleablood
1
answered Apr 9 at 2:59
fleabloodfleablood
1
1
add a comment |
add a comment |
$begingroup$
(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.
answered Apr 9 at 6:07
DanielWainfleetDanielWainfleet
35.8k31648
$endgroup$
add a comment |
$begingroup$
(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.
answered Apr 9 at 6:07
DanielWainfleetDanielWainfleet
35.8k31648
$endgroup$
add a comment |
$begingroup$
(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.
answered Apr 9 at 6:07
DanielWainfleetDanielWainfleet
35.8k31648
$endgroup$
(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.
answered Apr 9 at 6:07
DanielWainfleetDanielWainfleet
35.8k31648
edited Apr 9 at 6:32
answered Apr 9 at 6:07
DanielWainfleetDanielWainfleet
35.8k31648
answered Apr 9 at 6:07
DanielWainfleetDanielWainfleet
35.8k31648
answered Apr 9 at 6:07
DanielWainfleetDanielWainfleet
35.8k31648
35.8k31648
add a comment |
add a comment |
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$begingroup$
There are $5$ objects, not $4$. One object is the double, but that doesn't change anything,
$endgroup$
– lulu
Apr 9 at 2:29
$begingroup$
I thought since we have a subset of 2, we multiply by 4! since there are 4 elements left in A.
$endgroup$
– Zaku
Apr 9 at 2:31
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
Apr 9 at 2:34
$begingroup$
" 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
Apr 9 at 2:50