Number of total orders on set with $n$ elements

Question

Show that there exist $n!$ total orders on a set with $n$ elements.

Could someone help me with this problem?

I get why this is when seeing an example, just not sure how to write a formal proof.

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Example:

For set $\{a,b,c\}$ with $3$ elements we get $3! = 6$ total orders:

$\{a<b<c, a<c<b, b<a<c, b<c<a, c<a<b, c<b<a\}$

Answer 1

The smallest element can be any of $n$. Once this has been selected, the next smallest can be any of $n-1$.

Continuing in this way we obtain

$n$ x $(n-1)$ x ... x $1$ possibilities

i.e. $n!$.

Answer 2

Hint: Any permutation of $n$ elements defines an order.