- Let $X$ be a non-empty set and let $\{a_1,a_2,\ldots,a_n\}$ be a finite set with $n\ge 1$ elements.
- A given map $f:\{a,b\}\to X$ determines uniquely an ordered n-tuple $(x_1,x_2,\ldots,x_n)$ of elements $x_i\in X$ by setting $x_i:=f(a_i)$ for $i=1,\ldots,n.$
- Vice versa, every ordered n-tuple $(x_1,\ldots,x_n)$ uniquely determines a map $f:\{a_1,\ldots,a_b\}\to X$ such that $f(a_i):=x_i$ for $i=1,\ldots,n.$
- Therefore, the number of different maps of the set $\{a_1,a_2,\ldots,a_n\}$ to the set $X$ is equipotent to the Cartesian product $$\underbrace{X\times X\times\cdots\times X}_{n\text{ times}},$$ since the latter consists exactly of the ordered n-tuples $(x_1,x_2,\ldots,x_n).$
- Now, let $X$ be finite with the cardinality $|X|=m.$
- According to the fundamental counting principle, the cardinality of the $n$-fold Cartesian product is $m^n.$

q.e.d

| | | | created: 2019-09-07 08:37:50 | modified: 2019-09-07 08:37:50 | by: *bookofproofs* | references: [8297]

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[8297] **Flachsmeyer, Jürgen**: “Kombinatorik”, VEB Deutscher Verlag der Wissenschaften, 1972, Dritte Auflage