A Set, by definition, has no order, it is simply a collection of elements. As such there are no duplicate elements in a set, and:
We define a combination here as a subset of elements from a particular set. Let us define a set $A$, and a generic subset of $A$ as $S_A$. We might define $S_A$ algebraically as:
$\displaystyle{\begin{array}{l} S_A=\{\begin{array}{c}x_1,x_2,\ldots,x_n\end{array}\} \\\\ x_1\in A \\ x_2\in (A/\{\begin{array}{c}x_1\end{array}\}) \\ \vdots \\ x_n\in (A/\{\begin{array}{c}x_1,x_2,\ldots,x_{n-1}\end{array}\}) \end{array}}$
$\displaystyle{ \{\begin{array}{c}a,b,c,d\end{array}\}=\{\begin{array}{c}d,c,b,a\end{array}\}\neq\{\begin{array}{c}a,b\end{array}\} }$
A Permutation is a particular ordering of a set. You might think these as an ordered sequence rather than a set.
If $n=\|A\|$ and $r=\|S_A\|$ then we have:
$\displaystyle{ {}^nP_r=\frac{n!}{(n-r)!} }$
Think about the number of possible choices of elements from $A$, for each element of $S_A$, the total number of variations will be the product of the number of possible choices for each element.
$\displaystyle{ n\times (n-1)\times (n-2)\times\ldots\times(n-r) }$
This can be found with the formula given above (${}^nP_r$).
If the number of Permutations is given by $\displaystyle{{}^nP_r=\frac{n!}{(n-r)!}}$, ($\|A\|=n$) then the number of Combinations is given by
$\displaystyle{ {}^nC_r=\frac{n!}{r!(n-r)!} }$
This is because dividing by $r!$, the number of permutations of the subset $S_A$ ($\|S_A\|=r$), effectively eliminates all the duplicate subsets.