Number of ways of distributing 2 distinct balls in 2 distinct bins is 4, but why not 6?

Question

No . ways of distributing 2 balls in 2 bins :

Balls $a,b$

Bins $1,2$

Taking ball configuration

$a$ $1 1 2 1 $

$b$ $1 2 1 2$ =4 correct

But taking bin configuration

$1$ $a b a,b b,a$

$2$ $b a a,b b,a $ =6 !

What is wrong in my understanding ?

Answer 1

Well. Suppose we have a red ball, denoted $R$, and a blue ball, denoted $B$. Assume we have two bins, Bin $1$ and Bin $2$.

Then we see that

Case 1. Bin $1$ has both $R$ and $B$; Bin $2$ has nothing.

Case 2. Bin $2$ has both $R$ and $B$; Bin $1$ has nothing.

Case 3. Bin $1$ has $R$; Bin $2$ has $B$.

Case 4. Bin $1$ has $B$; Bin $2$ has $R$.

There's no other case. So, $4$ cases in total.

Additional: This is also counting the number of functions mapping between two sets with $2$ elements, namely $2^2$.