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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?
$begingroup$
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
asked 3 hours ago
ZakuZaku
1879
$endgroup$
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
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:
- 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
asked 3 hours ago
ZakuZaku
1879
$endgroup$
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
combinatorics
asked 3 hours ago
ZakuZaku
1879
asked 3 hours ago
ZakuZaku
1879
asked 3 hours ago
ZakuZaku
1879
asked 3 hours ago
ZakuZaku
1879
asked 3 hours ago
ZakuZaku
1879
1879
$begingroup$
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 |
$begingroup$
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
$begingroup$
There are $5$ objects, not $4$. One object is the double, but that doesn't change anything,
$endgroup$
– lulu
3 hours ago
$begingroup$
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
$begingroup$
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
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
$begingroup$
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
$begingroup$
" 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 |
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 3 hours ago
CopyPasteItCopyPasteIt
4,3271828
$endgroup$
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 3 hours ago
fleabloodfleablood
73.9k22891
$endgroup$
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 $binom 62binom 514!=(15)(5)(4!)=(15)(5!)=1800.$
answered 18 mins ago
DanielWainfleetDanielWainfleet
35.8k31648
$endgroup$
add a comment |
Your Answer
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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 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: $15 times 120 = 1800$
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: $15 times 120 = 1800$
answered 3 hours ago
CopyPasteItCopyPasteIt
4,3271828
$endgroup$
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 3 hours ago
CopyPasteItCopyPasteIt
4,3271828
answered 3 hours ago
CopyPasteItCopyPasteIt
4,3271828
answered 3 hours ago
CopyPasteItCopyPasteIt
4,3271828
answered 3 hours ago
CopyPasteItCopyPasteIt
4,3271828
4,3271828
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 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.
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.
answered 3 hours ago
fleabloodfleablood
73.9k22891
$endgroup$
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 3 hours ago
fleabloodfleablood
73.9k22891
answered 3 hours ago
fleabloodfleablood
73.9k22891
answered 3 hours ago
fleabloodfleablood
73.9k22891
answered 3 hours ago
fleabloodfleablood
73.9k22891
73.9k22891
add a comment |
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 $binom 62binom 514!=(15)(5)(4!)=(15)(5!)=1800.$
answered 18 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 $binom 62binom 514!=(15)(5)(4!)=(15)(5!)=1800.$
answered 18 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 $binom 62binom 514!=(15)(5)(4!)=(15)(5!)=1800.$
answered 18 mins ago
DanielWainfleetDanielWainfleet
35.8k31648
$endgroup$
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 $binom 62binom 514!=(15)(5)(4!)=(15)(5!)=1800.$
answered 18 mins ago
DanielWainfleetDanielWainfleet
35.8k31648
answered 18 mins ago
DanielWainfleetDanielWainfleet
35.8k31648
answered 18 mins ago
DanielWainfleetDanielWainfleet
35.8k31648
answered 18 mins ago
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
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