Number of Permutations of All Elements/Proof 2
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Theorem
Let $S$ be a set of $n$ elements.
The number of permutations of $S$ is $n!$
Proof
We pick the elements of $S$ in an arbitrary order.
There are $n$ elements of $S$, so there are $n$ options for the first element.
Then there are $n - 1$ elements left in $S$ that we have not picked, so there are $n - 1$ options for the second element.
Then there are $n - 2$ elements left, so there are $n - 2$ options for the third element.
And so on, until there are $3$, then $2$, then $1$ remaining elements.
Each mapping is independent of the choices made for all the other mappings.
So, by the Product Rule for Counting, the total number of ordered selections from $S$:
\(\ds {}^n P_n\) | \(=\) | \(\ds n \paren {n - 1} \paren {n - 2} \cdots 3 \times 2 \times 1\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds n!\) | Definition of Factorial |
$\blacksquare$
Sources
- 1965: J.A. Green: Sets and Groups ... (previous) ... (next): $\S 3.6$. Products of bijective mappings. Permutations: Example $54$
- 1966: Richard A. Dean: Elements of Abstract Algebra ... (previous) ... (next): $\S 1.6$
- 1968: Ian D. Macdonald: The Theory of Groups ... (previous) ... (next): Appendix: Elementary set and number theory