in how many ways a $5\times5$ grid can be painted so that painted cells form a T-tetromino? My first question here..sorry if I'm not very specific but I try to be.
A T-tetromino has three connected blocks in a line and another one above the middle block. How many ways can one be painted on the grid if orientation matters? What about if it doesn't?
 A: Got an answer from somewhere, it says that the whole tetris can be painted in 12 ways by that pattern and as this can be done in 4 directions, hence 4 multiplied by 12 is 48 ways Any ideas?
A: Assume the point is facing up. The center can be eitherin the middle 9 blocks or the three middle ones on the bottom row. That means a specific orientation can have 12 ways to place it.

There are four orientations, so the total is $12 \times 4$, or 48.
A: Well, my study encompasses just the case where the orientation is dependent to each pattern, other case it is likely that no mathematical approach is valid and the solution is brute-forced.

From the picture above, see that each line contains minimally two tetrominos, where the red one fixed gives two eventual places for the other, 
which means there is $2^5$ in general.
When the blue tetris is fixed, just one position left, that sums it up to $3^5$ combinations with the fact that the grid is symmetrical (right to left), so there should be many cases to remove.
What is needing deep analysis is symmetricity, because amongst the $3^5$ ways of coloring the grid, these cases which are interleaving, or duplicated. so the only case where the symmetricity dosnt apply is the (red,green) couple, where the (blue,blue) and (red,red) couples are subject of being divided into 2, because one is vertically symmetrical to other.


*

*1st case : whene there is no (red,green) coloring in the grid


$U_0=\frac{2^5}{2}$


*

*2nd case : whene there is one (red,green) coloring in the grid


$U_1=5*\frac{2^4}{2}$


*

*3rd case : whene there is two (red,green) couples in the grid


$U_2=\binom25*\frac{2^3}{2}$


*

*4th case : whene there is three (red,green) couples in the grid


$U_3=\binom35*\frac{2^2}{2}$


*

*5th case : whene there is four (red,green) couples in the grid


$U_4=\binom45*\frac{2}{2}$


*

*6th case : whene there is five (red,green) couples in the grid


$U_5=\binom55$



*

*Summing all cases up gives:


$U_0+U_1+U_2+U_3+U_4+U_5=\frac{2^5}{2}+5*\frac{2^4}{2}+\binom25*\frac{2^3}{2}+\binom35*\frac{2^2}{2}+\binom45+\binom55$

(I dont know how did you guys find the number 48)

