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Lets say we have a function $f:X\rightarrow Y$ where $X=\{a,b,c,d\}$ and $Y=\{x,y\}$. Can we assign each $x\in{X}$ to $y\in{Y}$ 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?

$$f:X\rightarrow Y\quad f(a)=x , f(b)=x, f(c)=x ,f(d)=x;$$ $$f:X\rightarrow Y\quad f(a)=x , f(b)=x, f(c)=x ,f(d)=y;$$ $$f:X\rightarrow Y\quad f(a)=x , f(b)=x, f(c)=y ,f(d)=y;$$ $$f:X\rightarrow Y\quad f(a)=x , f(b)=y, f(c)=y ,f(d)=y;$$ $$f:X\rightarrow Y\quad f(a)=y , f(b)=y, f(c)=y ,f(d)=y;$$ $$f:X\rightarrow Y\quad f(a)=y , f(b)=y, f(c)=y ,f(d)=x;$$ $$f:X\rightarrow Y\quad f(a)=y , f(b)=y, f(c)=x ,f(d)=x;$$ $$f:X\rightarrow Y\quad f(a)=y , f(b)=x, f(c)=x ,f(d)=x;$$ $$f:X\rightarrow Y\quad f(a)=x , f(b)=x, f(c)=y ,f(d)=x;$$ $$f:X\rightarrow Y\quad f(a)=x , f(b)=y, f(c)=y ,f(d)=x;$$ $$f:X\rightarrow Y\quad f(a)=y , f(b)=x, f(c)=y ,f(d)=x;$$ $$f:X\rightarrow Y\quad f(a)=x , f(b)=y, f(c)=x ,f(d)=y;$$ $$f:X\rightarrow Y\quad f(a)=y , f(b)=x, f(c)=x ,f(d)=y;$$ $$f:X\rightarrow Y\quad f(a)=y , f(b)=x, f(c)=y ,f(d)=y;$$ $$f:X\rightarrow Y\quad f(a)=y , f(b)=y, f(c)=x ,f(d)=y;$$ $$f:X\rightarrow Y\quad f(a)=x , f(b)=y, f(c)=x ,f(d)=x;$$

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    $\begingroup$ See this $\endgroup$ – John Douma Jan 17 at 21:32
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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: $2\cdot 2 \cdot 2\cdot 2 = 2^4 = 16$ ways to do that.

But you've explicitly written the list of all possible function assignments from $X\to Y$, 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 $f: X\to Y$ is equal to $|Y|^{|X|} = m^n$.

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You actually got the right answer for the wrong reason: you want $\mid Y^X\mid=\mid Y\mid^{\mid X\mid}=2^4=16$.

Though it looks like you listed them correctly.

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