Probability for fair coin I apologize for asking simple question but iam bit confused
Let's say we toss a coin $5$ times well, we have five choices for the first slot, four for the second slot, and so on, resulting in $5⋅4⋅3⋅2⋅1=5!=120$ different re-arrangements,where letter are consider unique
If we toss a coin $2$ times and if consider each letter unique so we get 12 ways to arrange it  so the factorial of 2 doesn't work here to find the ways to arrange it why so?
$
\begin{array}{l|l|l} 
  \ H1 & H2  \\ \hline
  \ H2 & H1   \\ \hline
  \ T1 & T2  \\ \hline
  \ T2 & T1 \\ \hline
  \ H1 & T1 \\ \hline
  \ H2 & T2 \\ \hline
  \ T1 & H1 \\ \hline
  \ T2 & H2 \\ \hline
  \ H1 & T2 \\ \hline
  \ H2 & T1 \\ \hline
  \ T1 & H2 \\ \hline
  \ T2 & H1 \\ \hline
\end{array}
$
 A: *

*We toss a coin 5 times $\rm\to(H,T)_1\times(H,T)_2\times(H,T)_3\times(H,T)_4\times(H,T)_5\to2^5$

*We toss a coin 2 times $\to2^2$
A: 
Let's say we toss a coin 5 times well, we have five choices for the first slot, four for the second slot, and so on, ...

No.  That is counting $5$ selections from a set of $5$ choices with no repetition allowed.  You'd use this for cases like: picking five coloured balls from a bag, one by one.
What you need here is to count $5$ selections from a set of $2$ choices with repetition allowed. 
The first 'slot' has two choices: head or tail.  The second 'slot' has two choices: head or tail.  ... and so on.
There are $2^5$ distinct results from tossing a coin five times.

If we toss a coin 2 times and if consider each letter unique so we get 12 ways ...

There are $2^2$ distinct results of tossing a coin two times.
$$\sf \{ H_1H_2, H_1T_2, T_1H_2, T_1T_2\} $$
Note: the results would be indexed by the time of tossing, so a listing like $\sf H_2H_2$ in your table makes little sense, and $\rm T_1H_2$ would be the same as $\rm H_2T_1$ ( a tail on the first toss and a head on the second).
