No FSM/Regex exists for this language right?
Expanding on Daniel Martin's answer.
None of the following interpretations yields a regular language:


$L = \{ w \in \{a,b\}^*  : \exists n \ge 0,  \; w = a^n b^n \}$
$L$ contains all the words $w \in \{a,b\}^*$ such that the number of $a$s in $w$ is not larger than the number of $b$s in $w$.
$L$ contains all the words $w \in \{a,b\}^*$ such that the number of $a$s in $w$ is equal than the number of $b$s in $w$.
$L$ contains all the words $w \in \{a,b\}^*$ such that there is a injective/bijective mapping that maps each a $a$ that appears in a generic position $i$ in $w$ with a $b$ in $w$ that appears in a position $j&gt;i$. 


The proof is the same for all cases: if $L$ was regular then,
for a sufficiently large $n$, the word $w = a^n b^n \in L$ could be written as $w = a^{n-k} a^k b^n$ (with $1 \le k \le n)$ in such a way that $a^{n-k}a^{k \cdot h} b^n \in L$ for all choices of $h \in \mathbb{N}$ (see pumping lemma for regular languages).
This is a contradiction since $a^{n-k}a^{2k} b^n = a^{n+k}b^{n} \not\in L$.