Mixing
How many functions can be made with the following sets?
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Lets say we have a function $f:Xrightarrow Y$ where $X={a,b,c,d}$ and $Y={x,y}$. Can we assign each $xin{X}$ to $yin{Y}$ in $4^2$ different possible ways? that is, create $16$ possible functions? below is my answer. Is there an easier or quicker method to do this or is my answer sufficient for this type of question?
$$f:Xrightarrow Yquad f(a)=x , f(b)=x, f(c)=x ,f(d)=x;$$ $$f:Xrightarrow Yquad f(a)=x , f(b)=x, f(c)=x ,f(d)=y;$$ $$f:Xrightarrow Yquad f(a)=x , f(b)=x, f(c)=y ,f(d)=y;$$ $$f:Xrightarrow Yquad f(a)=x , f(b)=y, f(c)=y ,f(d)=y;$$ $$f:Xrightarrow Yquad f(a)=y , f(b)=y, f(c)=y ,f(d)=y;$$ $$f:Xrightarrow Yquad f(a)=y , f(b)=y, f(c)=y ,f(d)=x;$$ $$f:Xrightarrow Yquad f(a)=y , f(b)=y, f(c)=x ,f(d)=x;$$ $$f:Xrightarrow Yquad f(a)=y , f(b)=x, f(c)=x ,f(d)=x;$$ $$f:Xrightarrow Yquad f(a)=x , f(b)=x, f(c)=y ,f(d)=x;$$ $$f:Xrightarrow Yquad f(a)=x , f(b)=y, f(c)=y ,f(d)=x;$$ $$f:Xrightarrow Yquad f(a)=y , f(b)=x, f(c)=y ,f(d)=x;$$ $$f:Xrightarrow Yquad f(a)=x , f(b)=y, f(c)=x ,f(d)=y;$$ $$f:Xrightarrow Yquad f(a)=y , f(b)=x, f(c)=x ,f(d)=y;$$ $$f:Xrightarrow Yquad f(a)=y , f(b)=x, f(c)=y ,f(d)=y;$$ $$f:Xrightarrow Yquad f(a)=y , f(b)=y, f(c)=x ,f(d)=y;$$ $$f:Xrightarrow Yquad f(a)=x , f(b)=y, f(c)=x ,f(d)=x;$$
functions proof-verification
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$begingroup$
Lets say we have a function $f:Xrightarrow Y$ where $X={a,b,c,d}$ and $Y={x,y}$. Can we assign each $xin{X}$ to $yin{Y}$ in $4^2$ different possible ways? that is, create $16$ possible functions? below is my answer. Is there an easier or quicker method to do this or is my answer sufficient for this type of question?
$$f:Xrightarrow Yquad f(a)=x , f(b)=x, f(c)=x ,f(d)=x;$$ $$f:Xrightarrow Yquad f(a)=x , f(b)=x, f(c)=x ,f(d)=y;$$ $$f:Xrightarrow Yquad f(a)=x , f(b)=x, f(c)=y ,f(d)=y;$$ $$f:Xrightarrow Yquad f(a)=x , f(b)=y, f(c)=y ,f(d)=y;$$ $$f:Xrightarrow Yquad f(a)=y , f(b)=y, f(c)=y ,f(d)=y;$$ $$f:Xrightarrow Yquad f(a)=y , f(b)=y, f(c)=y ,f(d)=x;$$ $$f:Xrightarrow Yquad f(a)=y , f(b)=y, f(c)=x ,f(d)=x;$$ $$f:Xrightarrow Yquad f(a)=y , f(b)=x, f(c)=x ,f(d)=x;$$ $$f:Xrightarrow Yquad f(a)=x , f(b)=x, f(c)=y ,f(d)=x;$$ $$f:Xrightarrow Yquad f(a)=x , f(b)=y, f(c)=y ,f(d)=x;$$ $$f:Xrightarrow Yquad f(a)=y , f(b)=x, f(c)=y ,f(d)=x;$$ $$f:Xrightarrow Yquad f(a)=x , f(b)=y, f(c)=x ,f(d)=y;$$ $$f:Xrightarrow Yquad f(a)=y , f(b)=x, f(c)=x ,f(d)=y;$$ $$f:Xrightarrow Yquad f(a)=y , f(b)=x, f(c)=y ,f(d)=y;$$ $$f:Xrightarrow Yquad f(a)=y , f(b)=y, f(c)=x ,f(d)=y;$$ $$f:Xrightarrow Yquad f(a)=x , f(b)=y, f(c)=x ,f(d)=x;$$
functions proof-verification
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1
$begingroup$
See this
$endgroup$
– John Douma
Jan 17 at 21:32
add a comment |
$begingroup$
Lets say we have a function $f:Xrightarrow Y$ where $X={a,b,c,d}$ and $Y={x,y}$. Can we assign each $xin{X}$ to $yin{Y}$ in $4^2$ different possible ways? that is, create $16$ possible functions? below is my answer. Is there an easier or quicker method to do this or is my answer sufficient for this type of question?
$$f:Xrightarrow Yquad f(a)=x , f(b)=x, f(c)=x ,f(d)=x;$$ $$f:Xrightarrow Yquad f(a)=x , f(b)=x, f(c)=x ,f(d)=y;$$ $$f:Xrightarrow Yquad f(a)=x , f(b)=x, f(c)=y ,f(d)=y;$$ $$f:Xrightarrow Yquad f(a)=x , f(b)=y, f(c)=y ,f(d)=y;$$ $$f:Xrightarrow Yquad f(a)=y , f(b)=y, f(c)=y ,f(d)=y;$$ $$f:Xrightarrow Yquad f(a)=y , f(b)=y, f(c)=y ,f(d)=x;$$ $$f:Xrightarrow Yquad f(a)=y , f(b)=y, f(c)=x ,f(d)=x;$$ $$f:Xrightarrow Yquad f(a)=y , f(b)=x, f(c)=x ,f(d)=x;$$ $$f:Xrightarrow Yquad f(a)=x , f(b)=x, f(c)=y ,f(d)=x;$$ $$f:Xrightarrow Yquad f(a)=x , f(b)=y, f(c)=y ,f(d)=x;$$ $$f:Xrightarrow Yquad f(a)=y , f(b)=x, f(c)=y ,f(d)=x;$$ $$f:Xrightarrow Yquad f(a)=x , f(b)=y, f(c)=x ,f(d)=y;$$ $$f:Xrightarrow Yquad f(a)=y , f(b)=x, f(c)=x ,f(d)=y;$$ $$f:Xrightarrow Yquad f(a)=y , f(b)=x, f(c)=y ,f(d)=y;$$ $$f:Xrightarrow Yquad f(a)=y , f(b)=y, f(c)=x ,f(d)=y;$$ $$f:Xrightarrow Yquad f(a)=x , f(b)=y, f(c)=x ,f(d)=x;$$
functions proof-verification
$endgroup$
Lets say we have a function $f:Xrightarrow Y$ where $X={a,b,c,d}$ and $Y={x,y}$. Can we assign each $xin{X}$ to $yin{Y}$ in $4^2$ different possible ways? that is, create $16$ possible functions? below is my answer. Is there an easier or quicker method to do this or is my answer sufficient for this type of question?
$$f:Xrightarrow Yquad f(a)=x , f(b)=x, f(c)=x ,f(d)=x;$$ $$f:Xrightarrow Yquad f(a)=x , f(b)=x, f(c)=x ,f(d)=y;$$ $$f:Xrightarrow Yquad f(a)=x , f(b)=x, f(c)=y ,f(d)=y;$$ $$f:Xrightarrow Yquad f(a)=x , f(b)=y, f(c)=y ,f(d)=y;$$ $$f:Xrightarrow Yquad f(a)=y , f(b)=y, f(c)=y ,f(d)=y;$$ $$f:Xrightarrow Yquad f(a)=y , f(b)=y, f(c)=y ,f(d)=x;$$ $$f:Xrightarrow Yquad f(a)=y , f(b)=y, f(c)=x ,f(d)=x;$$ $$f:Xrightarrow Yquad f(a)=y , f(b)=x, f(c)=x ,f(d)=x;$$ $$f:Xrightarrow Yquad f(a)=x , f(b)=x, f(c)=y ,f(d)=x;$$ $$f:Xrightarrow Yquad f(a)=x , f(b)=y, f(c)=y ,f(d)=x;$$ $$f:Xrightarrow Yquad f(a)=y , f(b)=x, f(c)=y ,f(d)=x;$$ $$f:Xrightarrow Yquad f(a)=x , f(b)=y, f(c)=x ,f(d)=y;$$ $$f:Xrightarrow Yquad f(a)=y , f(b)=x, f(c)=x ,f(d)=y;$$ $$f:Xrightarrow Yquad f(a)=y , f(b)=x, f(c)=y ,f(d)=y;$$ $$f:Xrightarrow Yquad f(a)=y , f(b)=y, f(c)=x ,f(d)=y;$$ $$f:Xrightarrow Yquad f(a)=x , f(b)=y, f(c)=x ,f(d)=x;$$
functions proof-verification
functions proof-verification
edited Jan 17 at 21:28
asked Jan 17 at 21:25
user606466
1
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See this
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– John Douma
Jan 17 at 21:32
add a comment |
1
$begingroup$
See this
$endgroup$
– John Douma
Jan 17 at 21:32
1
1
$begingroup$
See this
$endgroup$
– John Douma
Jan 17 at 21:32
$begingroup$
See this
$endgroup$
– John Douma
Jan 17 at 21:32
add a comment |
2 Answers
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$begingroup$
For each element in X, there are two elements in $Y$ that they can be mapped to an element in $Y$: $a$ can be mapped to $x$ or $y$; same thing for $b, c, d$. In all we have: $2cdot 2 cdot 2cdot 2 = 2^4 = 16$ ways to do that.
But you've explicitly written the list of all possible function assignments from $Xto Y$, which helps you understand this.
In general, given a set $X$ with $n$ elements, and a set $Y$ with $m$ elements, then the number of functions $f: Xto Y$ is equal to $|Y|^{|X|} = m^n$.
answered Jan 17 at 21:33
jordan_glenjordan_glen
1
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You actually got the right answer for the wrong reason: you want $mid Y^Xmid=mid Ymid^{mid Xmid}=2^4=16$.
Though it looks like you listed them correctly.
answered Jan 17 at 21:39
Chris CusterChris Custer
13.8k3827
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2 Answers
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2 Answers
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$begingroup$
For each element in X, there are two elements in $Y$ that they can be mapped to an element in $Y$: $a$ can be mapped to $x$ or $y$; same thing for $b, c, d$. In all we have: $2cdot 2 cdot 2cdot 2 = 2^4 = 16$ ways to do that.
But you've explicitly written the list of all possible function assignments from $Xto Y$, which helps you understand this.
In general, given a set $X$ with $n$ elements, and a set $Y$ with $m$ elements, then the number of functions $f: Xto Y$ is equal to $|Y|^{|X|} = m^n$.
answered Jan 17 at 21:33
jordan_glenjordan_glen
1
$endgroup$
add a comment |
$begingroup$
For each element in X, there are two elements in $Y$ that they can be mapped to an element in $Y$: $a$ can be mapped to $x$ or $y$; same thing for $b, c, d$. In all we have: $2cdot 2 cdot 2cdot 2 = 2^4 = 16$ ways to do that.
But you've explicitly written the list of all possible function assignments from $Xto Y$, which helps you understand this.
In general, given a set $X$ with $n$ elements, and a set $Y$ with $m$ elements, then the number of functions $f: Xto Y$ is equal to $|Y|^{|X|} = m^n$.
answered Jan 17 at 21:33
jordan_glenjordan_glen
1
$endgroup$
add a comment |
$begingroup$
For each element in X, there are two elements in $Y$ that they can be mapped to an element in $Y$: $a$ can be mapped to $x$ or $y$; same thing for $b, c, d$. In all we have: $2cdot 2 cdot 2cdot 2 = 2^4 = 16$ ways to do that.
But you've explicitly written the list of all possible function assignments from $Xto Y$, which helps you understand this.
In general, given a set $X$ with $n$ elements, and a set $Y$ with $m$ elements, then the number of functions $f: Xto Y$ is equal to $|Y|^{|X|} = m^n$.
answered Jan 17 at 21:33
jordan_glenjordan_glen
1
$endgroup$
For each element in X, there are two elements in $Y$ that they can be mapped to an element in $Y$: $a$ can be mapped to $x$ or $y$; same thing for $b, c, d$. In all we have: $2cdot 2 cdot 2cdot 2 = 2^4 = 16$ ways to do that.
But you've explicitly written the list of all possible function assignments from $Xto Y$, which helps you understand this.
In general, given a set $X$ with $n$ elements, and a set $Y$ with $m$ elements, then the number of functions $f: Xto Y$ is equal to $|Y|^{|X|} = m^n$.
answered Jan 17 at 21:33
jordan_glenjordan_glen
1
edited Jan 17 at 21:40
answered Jan 17 at 21:33
jordan_glenjordan_glen
1
answered Jan 17 at 21:33
jordan_glenjordan_glen
1
answered Jan 17 at 21:33
jordan_glenjordan_glen
1
1
add a comment |
add a comment |
$begingroup$
You actually got the right answer for the wrong reason: you want $mid Y^Xmid=mid Ymid^{mid Xmid}=2^4=16$.
Though it looks like you listed them correctly.
answered Jan 17 at 21:39
Chris CusterChris Custer
13.8k3827
$endgroup$
add a comment |
$begingroup$
You actually got the right answer for the wrong reason: you want $mid Y^Xmid=mid Ymid^{mid Xmid}=2^4=16$.
Though it looks like you listed them correctly.
answered Jan 17 at 21:39
Chris CusterChris Custer
13.8k3827
$endgroup$
add a comment |
$begingroup$
You actually got the right answer for the wrong reason: you want $mid Y^Xmid=mid Ymid^{mid Xmid}=2^4=16$.
Though it looks like you listed them correctly.
answered Jan 17 at 21:39
Chris CusterChris Custer
13.8k3827
$endgroup$
You actually got the right answer for the wrong reason: you want $mid Y^Xmid=mid Ymid^{mid Xmid}=2^4=16$.
Though it looks like you listed them correctly.
answered Jan 17 at 21:39
Chris CusterChris Custer
13.8k3827
edited Jan 17 at 21:47
answered Jan 17 at 21:39
Chris CusterChris Custer
13.8k3827
answered Jan 17 at 21:39
Chris CusterChris Custer
13.8k3827
answered Jan 17 at 21:39
Chris CusterChris Custer
13.8k3827
13.8k3827
add a comment |
add a comment |
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$begingroup$
See this
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– John Douma
Jan 17 at 21:32