A unique factorization domain is typically defined as:

a domain $D$ such that each non-zero and non-invertible element $d$ can be factored as a product of irreducible elements.

And, for any factorizations $d=u_1 \dots u_m$ and $d = v_1 \dots v_n$ with irreducibles $u_i,v_i$ one has
that $m=n$ and there exists a permutation $\sigma$ of $\{1,\dots, n\}$ such that for all $i$ one has that $u_i$ and $v_{\sigma(i)}$ are associated.

Yet, this can be replaced by: 

And, for any factorizations $d=u_1 \dots u_n$ and $d = v_1 \dots v_n$ with irreducibles $u_i,v_i$ one has that there exists a permutation $\sigma$ of $\{1,\dots, n\}$ such that for all $i$ one has that $u_i$ and $v_{\sigma(i)}$ are associated.


In other words, it is sufficient that all factorizations of an element with the *same number of factors* are essentially equal. The fact that there, then, cannot be any factorization with a different number of factors can be proved and thus not have to be included in the definition.

Side note: If one replaces 'domain' by 'commutative cancellative semigroup with identity' the weakened definition actually is different.