Mixing
Number of functions with certain property
$begingroup$
Let $f : {1,2,3}rightarrow{1,2,3}$ be a function. Then the number of functions $g : {1,2,3}rightarrow {1,2,3}$ such that $f(x) = g(x)$ for at least one $x$ belonging to ${1,2,3}$ is what?
My reasoning was that total number of functions $f$ will be $3^3$ and for each of these functions I can have $g$ take exactly the same values as $f$ does so the answer would be $27$ but this answer is not correct, can someone please provide the solution.
combinatorics permutations
edited Jan 25 at 10:58
N. F. Taussig
44.8k103358
asked Jan 25 at 3:55
user601297user601297
39719
$endgroup$
add a comment |
$begingroup$
Let $f : {1,2,3}rightarrow{1,2,3}$ be a function. Then the number of functions $g : {1,2,3}rightarrow {1,2,3}$ such that $f(x) = g(x)$ for at least one $x$ belonging to ${1,2,3}$ is what?
My reasoning was that total number of functions $f$ will be $3^3$ and for each of these functions I can have $g$ take exactly the same values as $f$ does so the answer would be $27$ but this answer is not correct, can someone please provide the solution.
combinatorics permutations
edited Jan 25 at 10:58
N. F. Taussig
44.8k103358
asked Jan 25 at 3:55
user601297user601297
39719
$endgroup$
add a comment |
$begingroup$
Let $f : {1,2,3}rightarrow{1,2,3}$ be a function. Then the number of functions $g : {1,2,3}rightarrow {1,2,3}$ such that $f(x) = g(x)$ for at least one $x$ belonging to ${1,2,3}$ is what?
My reasoning was that total number of functions $f$ will be $3^3$ and for each of these functions I can have $g$ take exactly the same values as $f$ does so the answer would be $27$ but this answer is not correct, can someone please provide the solution.
combinatorics permutations
edited Jan 25 at 10:58
N. F. Taussig
44.8k103358
asked Jan 25 at 3:55
user601297user601297
39719
$endgroup$
Let $f : {1,2,3}rightarrow{1,2,3}$ be a function. Then the number of functions $g : {1,2,3}rightarrow {1,2,3}$ such that $f(x) = g(x)$ for at least one $x$ belonging to ${1,2,3}$ is what?
My reasoning was that total number of functions $f$ will be $3^3$ and for each of these functions I can have $g$ take exactly the same values as $f$ does so the answer would be $27$ but this answer is not correct, can someone please provide the solution.
combinatorics permutations
combinatorics permutations
edited Jan 25 at 10:58
N. F. Taussig
44.8k103358
asked Jan 25 at 3:55
user601297user601297
39719
edited Jan 25 at 10:58
N. F. Taussig
44.8k103358
asked Jan 25 at 3:55
user601297user601297
39719
edited Jan 25 at 10:58
N. F. Taussig
44.8k103358
edited Jan 25 at 10:58
N. F. Taussig
44.8k103358
edited Jan 25 at 10:58
N. F. Taussig
44.8k103358
44.8k103358
asked Jan 25 at 3:55
user601297user601297
39719
asked Jan 25 at 3:55
user601297user601297
39719
asked Jan 25 at 3:55
user601297user601297
39719
39719
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
In general, the number of functions ${1,...,n} to {1,...,m}$ is $m^n$.
Hence the number of functions in this case is $3^3$.
Remember $f$ is one specific function.
It is easier to count the number of $g$s that do not equal $f$ anywhere.
In particular, the range of such a $g$ must have size $2$ not $3$, hence the number
of such functions is $2^3$.
Hence the number of the other $g$s, which is what you are looking for, is
$3^3-2^3$.
Elaboration: What I mean is that $g(k)$ takes the values ${1,2,3} setminus {f(k)}$, and for all $k$ we have $| {1,2,3} setminus {f(k)} | = 2$.
answered Jan 25 at 4:06
copper.hatcopper.hat
127k559160
$endgroup$
$begingroup$
By size do you mean the number of values $g$ takes ???
$endgroup$
– user601297
Jan 25 at 4:11
$begingroup$
@user601297: Yes, excuse my tired wording. The point is at any $x$, we remove the value $f(x)$ from $g$s range. We don't care about the particular value here, just how many remain, which is $2$ for each value of $x$.
$endgroup$
– copper.hat
Jan 25 at 4:12
$begingroup$
But the example that @Bram28 gave has $g$ take all 3 values.
$endgroup$
– user601297
Jan 25 at 4:20
$begingroup$
In the example $g(1)$ must be one of $2,3$, $f(2)$ must be one of $1,3$ and $g(3)$ must be one of $1,2$. In all cases, the size of the range is $3-1=2$.
$endgroup$
– copper.hat
Jan 25 at 4:23
$begingroup$
Your answer is definitely correct, I’m just having a hard time understanding this, thanks anyways.
$endgroup$
– user601297
Jan 25 at 4:34
|
show 2 more comments
$begingroup$
It is more important to understand why your answer does not work.
Let's say that $f(1)=1, f(2)=2$ and $f(3)=3$
Then if you have $g(1)=2, g(2)=3$ and $f(3)=1$ (this is one of the $27$ possble functions for $g$), you have no $x$ such that $f(x)=g(x)$. So, not all $27$ possible functions for $g$ will have at least one 'match' with $f$.
Do you see how you can correct for this?
answered Jan 25 at 4:00
Bram28Bram28
63.8k44793
$endgroup$
$begingroup$
I’m sorry but I still don’t get it, why can’t I just change one of the $g$ values to match with $f$?
$endgroup$
– user601297
Jan 25 at 4:03
$begingroup$
@user601297 Yes, you can, but that would be a different $g$: one of the other $26$. So, for some of the $27$ possible $g$'s you get a match, but for others you do not.
$endgroup$
– Bram28
Jan 25 at 4:04
$begingroup$
I would really be grateful if you could provide me with the process to come to the right number, I will probably understand it better that way.
$endgroup$
– user601297
Jan 25 at 4:06
add a comment |
$begingroup$
Probably it is easier to count the number $a$ of functions $g$ such that $g(x)not=f(x)$ for all $xin{1,2,3}$ first. Then the set of functions you are looking has cardinality $b-a$ where $b$ denotes the number of all functions mapping ${1,2,3}$ Info itself.
answered Jan 25 at 4:10
Jens SchwaigerJens Schwaiger
1,629138
$endgroup$
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
$begingroup$
In general, the number of functions ${1,...,n} to {1,...,m}$ is $m^n$.
Hence the number of functions in this case is $3^3$.
Remember $f$ is one specific function.
It is easier to count the number of $g$s that do not equal $f$ anywhere.
In particular, the range of such a $g$ must have size $2$ not $3$, hence the number
of such functions is $2^3$.
Hence the number of the other $g$s, which is what you are looking for, is
$3^3-2^3$.
Elaboration: What I mean is that $g(k)$ takes the values ${1,2,3} setminus {f(k)}$, and for all $k$ we have $| {1,2,3} setminus {f(k)} | = 2$.
answered Jan 25 at 4:06
copper.hatcopper.hat
127k559160
$endgroup$
$begingroup$
By size do you mean the number of values $g$ takes ???
$endgroup$
– user601297
Jan 25 at 4:11
$begingroup$
@user601297: Yes, excuse my tired wording. The point is at any $x$, we remove the value $f(x)$ from $g$s range. We don't care about the particular value here, just how many remain, which is $2$ for each value of $x$.
$endgroup$
– copper.hat
Jan 25 at 4:12
$begingroup$
But the example that @Bram28 gave has $g$ take all 3 values.
$endgroup$
– user601297
Jan 25 at 4:20
$begingroup$
In the example $g(1)$ must be one of $2,3$, $f(2)$ must be one of $1,3$ and $g(3)$ must be one of $1,2$. In all cases, the size of the range is $3-1=2$.
$endgroup$
– copper.hat
Jan 25 at 4:23
$begingroup$
Your answer is definitely correct, I’m just having a hard time understanding this, thanks anyways.
$endgroup$
– user601297
Jan 25 at 4:34
|
show 2 more comments
$begingroup$
In general, the number of functions ${1,...,n} to {1,...,m}$ is $m^n$.
Hence the number of functions in this case is $3^3$.
Remember $f$ is one specific function.
It is easier to count the number of $g$s that do not equal $f$ anywhere.
In particular, the range of such a $g$ must have size $2$ not $3$, hence the number
of such functions is $2^3$.
Hence the number of the other $g$s, which is what you are looking for, is
$3^3-2^3$.
Elaboration: What I mean is that $g(k)$ takes the values ${1,2,3} setminus {f(k)}$, and for all $k$ we have $| {1,2,3} setminus {f(k)} | = 2$.
answered Jan 25 at 4:06
copper.hatcopper.hat
127k559160
$endgroup$
$begingroup$
By size do you mean the number of values $g$ takes ???
$endgroup$
– user601297
Jan 25 at 4:11
$begingroup$
@user601297: Yes, excuse my tired wording. The point is at any $x$, we remove the value $f(x)$ from $g$s range. We don't care about the particular value here, just how many remain, which is $2$ for each value of $x$.
$endgroup$
– copper.hat
Jan 25 at 4:12
$begingroup$
But the example that @Bram28 gave has $g$ take all 3 values.
$endgroup$
– user601297
Jan 25 at 4:20
$begingroup$
In the example $g(1)$ must be one of $2,3$, $f(2)$ must be one of $1,3$ and $g(3)$ must be one of $1,2$. In all cases, the size of the range is $3-1=2$.
$endgroup$
– copper.hat
Jan 25 at 4:23
$begingroup$
Your answer is definitely correct, I’m just having a hard time understanding this, thanks anyways.
$endgroup$
– user601297
Jan 25 at 4:34
|
show 2 more comments
$begingroup$
In general, the number of functions ${1,...,n} to {1,...,m}$ is $m^n$.
Hence the number of functions in this case is $3^3$.
Remember $f$ is one specific function.
It is easier to count the number of $g$s that do not equal $f$ anywhere.
In particular, the range of such a $g$ must have size $2$ not $3$, hence the number
of such functions is $2^3$.
Hence the number of the other $g$s, which is what you are looking for, is
$3^3-2^3$.
Elaboration: What I mean is that $g(k)$ takes the values ${1,2,3} setminus {f(k)}$, and for all $k$ we have $| {1,2,3} setminus {f(k)} | = 2$.
answered Jan 25 at 4:06
copper.hatcopper.hat
127k559160
$endgroup$
In general, the number of functions ${1,...,n} to {1,...,m}$ is $m^n$.
Hence the number of functions in this case is $3^3$.
Remember $f$ is one specific function.
It is easier to count the number of $g$s that do not equal $f$ anywhere.
In particular, the range of such a $g$ must have size $2$ not $3$, hence the number
of such functions is $2^3$.
Hence the number of the other $g$s, which is what you are looking for, is
$3^3-2^3$.
Elaboration: What I mean is that $g(k)$ takes the values ${1,2,3} setminus {f(k)}$, and for all $k$ we have $| {1,2,3} setminus {f(k)} | = 2$.
answered Jan 25 at 4:06
copper.hatcopper.hat
127k559160
edited Jan 25 at 4:25
answered Jan 25 at 4:06
copper.hatcopper.hat
127k559160
answered Jan 25 at 4:06
copper.hatcopper.hat
127k559160
answered Jan 25 at 4:06
copper.hatcopper.hat
127k559160
127k559160
$begingroup$
By size do you mean the number of values $g$ takes ???
$endgroup$
– user601297
Jan 25 at 4:11
$begingroup$
@user601297: Yes, excuse my tired wording. The point is at any $x$, we remove the value $f(x)$ from $g$s range. We don't care about the particular value here, just how many remain, which is $2$ for each value of $x$.
$endgroup$
– copper.hat
Jan 25 at 4:12
$begingroup$
But the example that @Bram28 gave has $g$ take all 3 values.
$endgroup$
– user601297
Jan 25 at 4:20
$begingroup$
In the example $g(1)$ must be one of $2,3$, $f(2)$ must be one of $1,3$ and $g(3)$ must be one of $1,2$. In all cases, the size of the range is $3-1=2$.
$endgroup$
– copper.hat
Jan 25 at 4:23
$begingroup$
Your answer is definitely correct, I’m just having a hard time understanding this, thanks anyways.
$endgroup$
– user601297
Jan 25 at 4:34
|
show 2 more comments
$begingroup$
By size do you mean the number of values $g$ takes ???
$endgroup$
– user601297
Jan 25 at 4:11
$begingroup$
@user601297: Yes, excuse my tired wording. The point is at any $x$, we remove the value $f(x)$ from $g$s range. We don't care about the particular value here, just how many remain, which is $2$ for each value of $x$.
$endgroup$
– copper.hat
Jan 25 at 4:12
$begingroup$
But the example that @Bram28 gave has $g$ take all 3 values.
$endgroup$
– user601297
Jan 25 at 4:20
$begingroup$
In the example $g(1)$ must be one of $2,3$, $f(2)$ must be one of $1,3$ and $g(3)$ must be one of $1,2$. In all cases, the size of the range is $3-1=2$.
$endgroup$
– copper.hat
Jan 25 at 4:23
$begingroup$
Your answer is definitely correct, I’m just having a hard time understanding this, thanks anyways.
$endgroup$
– user601297
Jan 25 at 4:34
$begingroup$
By size do you mean the number of values $g$ takes ???
$endgroup$
– user601297
Jan 25 at 4:11
$begingroup$
By size do you mean the number of values $g$ takes ???
$endgroup$
– user601297
Jan 25 at 4:11
$begingroup$
@user601297: Yes, excuse my tired wording. The point is at any $x$, we remove the value $f(x)$ from $g$s range. We don't care about the particular value here, just how many remain, which is $2$ for each value of $x$.
$endgroup$
– copper.hat
Jan 25 at 4:12
$begingroup$
@user601297: Yes, excuse my tired wording. The point is at any $x$, we remove the value $f(x)$ from $g$s range. We don't care about the particular value here, just how many remain, which is $2$ for each value of $x$.
$endgroup$
– copper.hat
Jan 25 at 4:12
$begingroup$
But the example that @Bram28 gave has $g$ take all 3 values.
$endgroup$
– user601297
Jan 25 at 4:20
$begingroup$
But the example that @Bram28 gave has $g$ take all 3 values.
$endgroup$
– user601297
Jan 25 at 4:20
$begingroup$
In the example $g(1)$ must be one of $2,3$, $f(2)$ must be one of $1,3$ and $g(3)$ must be one of $1,2$. In all cases, the size of the range is $3-1=2$.
$endgroup$
– copper.hat
Jan 25 at 4:23
$begingroup$
In the example $g(1)$ must be one of $2,3$, $f(2)$ must be one of $1,3$ and $g(3)$ must be one of $1,2$. In all cases, the size of the range is $3-1=2$.
$endgroup$
– copper.hat
Jan 25 at 4:23
$begingroup$
Your answer is definitely correct, I’m just having a hard time understanding this, thanks anyways.
$endgroup$
– user601297
Jan 25 at 4:34
$begingroup$
Your answer is definitely correct, I’m just having a hard time understanding this, thanks anyways.
$endgroup$
– user601297
Jan 25 at 4:34
|
show 2 more comments
$begingroup$
It is more important to understand why your answer does not work.
Let's say that $f(1)=1, f(2)=2$ and $f(3)=3$
Then if you have $g(1)=2, g(2)=3$ and $f(3)=1$ (this is one of the $27$ possble functions for $g$), you have no $x$ such that $f(x)=g(x)$. So, not all $27$ possible functions for $g$ will have at least one 'match' with $f$.
Do you see how you can correct for this?
answered Jan 25 at 4:00
Bram28Bram28
63.8k44793
$endgroup$
$begingroup$
I’m sorry but I still don’t get it, why can’t I just change one of the $g$ values to match with $f$?
$endgroup$
– user601297
Jan 25 at 4:03
$begingroup$
@user601297 Yes, you can, but that would be a different $g$: one of the other $26$. So, for some of the $27$ possible $g$'s you get a match, but for others you do not.
$endgroup$
– Bram28
Jan 25 at 4:04
$begingroup$
I would really be grateful if you could provide me with the process to come to the right number, I will probably understand it better that way.
$endgroup$
– user601297
Jan 25 at 4:06
add a comment |
$begingroup$
It is more important to understand why your answer does not work.
Let's say that $f(1)=1, f(2)=2$ and $f(3)=3$
Then if you have $g(1)=2, g(2)=3$ and $f(3)=1$ (this is one of the $27$ possble functions for $g$), you have no $x$ such that $f(x)=g(x)$. So, not all $27$ possible functions for $g$ will have at least one 'match' with $f$.
Do you see how you can correct for this?
answered Jan 25 at 4:00
Bram28Bram28
63.8k44793
$endgroup$
$begingroup$
I’m sorry but I still don’t get it, why can’t I just change one of the $g$ values to match with $f$?
$endgroup$
– user601297
Jan 25 at 4:03
$begingroup$
@user601297 Yes, you can, but that would be a different $g$: one of the other $26$. So, for some of the $27$ possible $g$'s you get a match, but for others you do not.
$endgroup$
– Bram28
Jan 25 at 4:04
$begingroup$
I would really be grateful if you could provide me with the process to come to the right number, I will probably understand it better that way.
$endgroup$
– user601297
Jan 25 at 4:06
add a comment |
$begingroup$
It is more important to understand why your answer does not work.
Let's say that $f(1)=1, f(2)=2$ and $f(3)=3$
Then if you have $g(1)=2, g(2)=3$ and $f(3)=1$ (this is one of the $27$ possble functions for $g$), you have no $x$ such that $f(x)=g(x)$. So, not all $27$ possible functions for $g$ will have at least one 'match' with $f$.
Do you see how you can correct for this?
answered Jan 25 at 4:00
Bram28Bram28
63.8k44793
$endgroup$
It is more important to understand why your answer does not work.
Let's say that $f(1)=1, f(2)=2$ and $f(3)=3$
Then if you have $g(1)=2, g(2)=3$ and $f(3)=1$ (this is one of the $27$ possble functions for $g$), you have no $x$ such that $f(x)=g(x)$. So, not all $27$ possible functions for $g$ will have at least one 'match' with $f$.
Do you see how you can correct for this?
answered Jan 25 at 4:00
Bram28Bram28
63.8k44793
answered Jan 25 at 4:00
Bram28Bram28
63.8k44793
answered Jan 25 at 4:00
Bram28Bram28
63.8k44793
answered Jan 25 at 4:00
Bram28Bram28
63.8k44793
63.8k44793
$begingroup$
I’m sorry but I still don’t get it, why can’t I just change one of the $g$ values to match with $f$?
$endgroup$
– user601297
Jan 25 at 4:03
$begingroup$
@user601297 Yes, you can, but that would be a different $g$: one of the other $26$. So, for some of the $27$ possible $g$'s you get a match, but for others you do not.
$endgroup$
– Bram28
Jan 25 at 4:04
$begingroup$
I would really be grateful if you could provide me with the process to come to the right number, I will probably understand it better that way.
$endgroup$
– user601297
Jan 25 at 4:06
add a comment |
$begingroup$
I’m sorry but I still don’t get it, why can’t I just change one of the $g$ values to match with $f$?
$endgroup$
– user601297
Jan 25 at 4:03
$begingroup$
@user601297 Yes, you can, but that would be a different $g$: one of the other $26$. So, for some of the $27$ possible $g$'s you get a match, but for others you do not.
$endgroup$
– Bram28
Jan 25 at 4:04
$begingroup$
I would really be grateful if you could provide me with the process to come to the right number, I will probably understand it better that way.
$endgroup$
– user601297
Jan 25 at 4:06
$begingroup$
I’m sorry but I still don’t get it, why can’t I just change one of the $g$ values to match with $f$?
$endgroup$
– user601297
Jan 25 at 4:03
$begingroup$
I’m sorry but I still don’t get it, why can’t I just change one of the $g$ values to match with $f$?
$endgroup$
– user601297
Jan 25 at 4:03
$begingroup$
@user601297 Yes, you can, but that would be a different $g$: one of the other $26$. So, for some of the $27$ possible $g$'s you get a match, but for others you do not.
$endgroup$
– Bram28
Jan 25 at 4:04
$begingroup$
@user601297 Yes, you can, but that would be a different $g$: one of the other $26$. So, for some of the $27$ possible $g$'s you get a match, but for others you do not.
$endgroup$
– Bram28
Jan 25 at 4:04
$begingroup$
I would really be grateful if you could provide me with the process to come to the right number, I will probably understand it better that way.
$endgroup$
– user601297
Jan 25 at 4:06
$begingroup$
I would really be grateful if you could provide me with the process to come to the right number, I will probably understand it better that way.
$endgroup$
– user601297
Jan 25 at 4:06
add a comment |
$begingroup$
Probably it is easier to count the number $a$ of functions $g$ such that $g(x)not=f(x)$ for all $xin{1,2,3}$ first. Then the set of functions you are looking has cardinality $b-a$ where $b$ denotes the number of all functions mapping ${1,2,3}$ Info itself.
answered Jan 25 at 4:10
Jens SchwaigerJens Schwaiger
1,629138
$endgroup$
add a comment |
$begingroup$
Probably it is easier to count the number $a$ of functions $g$ such that $g(x)not=f(x)$ for all $xin{1,2,3}$ first. Then the set of functions you are looking has cardinality $b-a$ where $b$ denotes the number of all functions mapping ${1,2,3}$ Info itself.
answered Jan 25 at 4:10
Jens SchwaigerJens Schwaiger
1,629138
$endgroup$
add a comment |
$begingroup$
Probably it is easier to count the number $a$ of functions $g$ such that $g(x)not=f(x)$ for all $xin{1,2,3}$ first. Then the set of functions you are looking has cardinality $b-a$ where $b$ denotes the number of all functions mapping ${1,2,3}$ Info itself.
answered Jan 25 at 4:10
Jens SchwaigerJens Schwaiger
1,629138
$endgroup$
Probably it is easier to count the number $a$ of functions $g$ such that $g(x)not=f(x)$ for all $xin{1,2,3}$ first. Then the set of functions you are looking has cardinality $b-a$ where $b$ denotes the number of all functions mapping ${1,2,3}$ Info itself.
answered Jan 25 at 4:10
Jens SchwaigerJens Schwaiger
1,629138
answered Jan 25 at 4:10
Jens SchwaigerJens Schwaiger
1,629138
answered Jan 25 at 4:10
Jens SchwaigerJens Schwaiger
1,629138
answered Jan 25 at 4:10
Jens SchwaigerJens Schwaiger
1,629138
1,629138
add a comment |
add a comment |
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