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In my discrete mathematics class our notes say that between set A (having 6 elements) and set b (having 8 elements), there are $8^6$ distinct functions that can be formed, in other words: $|b|^{|a|}$ distinct functions. But no explanation is offered and I can't seem to figure out why this is true. Can anyone elaborate?

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4 Answers 4

up vote 9 down vote accepted

Let set $A$ have $a$ elements and set $B$ have $b$ elements. Each element in $A$ has $b$ choices to be mapped to. Each such choice gives you a unique function. Since each element has $b$ choices, the total number of functions from $A$ to $B$ is $$\underbrace{b \times b \times b \times \cdots b}_{a \text{ times}} = b^a$$

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A function on a set involves running the function on every element of the set A, each one producing some result in the set B. So, for the first run, every element of A gets mapped to an element in B. The question becomes, how many different mappings, all using every element of the set A, can we come up with? Take this example, mapping a 2 element set A, to a 3 element set B. There are 9 different ways, all beginning with both 1 and 2, that result in some different combination of mappings over to B.

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The number of functions from A to B is |B|^|A|, or $3^2$ = 9.

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Very thorough. Sadly I doubt the original poster will see it though. –  Simon S Nov 8 '14 at 23:54

Let's say for concreteness that $A$ is the set $\{p,q,r,s,t,u\}$. Let's try to define a function $f:A\to B$.

What is $f(p)$? It could be any element of $B$, so we have 8 choices.

What is $f(q)$? It could be any element of $B$, so we have 8 choices.


What is $f(u)$? It could be any element of $B$, so we have 8 choices.

So there are $8\cdot8\cdot8\cdot8\cdot8\cdot8 = 8^6$ ways to choose values for $f$, and each possible set of choices defines a different function $f$. So that's how many functions there are.

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You know that a function gives a unique value for each entry, if the fuction $f\colon A\to B$ where $|A|=n, ~|B|=m$, then for $a\in A$, you have $m$ values to assign. then for every $a\in A$, you can take |B| values, since $|A|$ have $n$ elements, then you have $|B|^{|A|}$ choices.

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