The $\triangle$ prefers to live in colors!

First, assign a colour to the middle triangle (which can be done in 4 ways).

Then, for each of the 3 triangles in the corners, they can be any one of the 3 remaining colours (independently of each other).

Therefore, there are $4 \times 3 \times 3 \times 3 = \boxed{108}$ ways.

We note that the outer three triangles can be filled with any of the four colors. Therefore, there are $4^3=64$ ways of filling them. If the outer triangles are filled with $n$ different colors. where $n=1, 2, 3$ , then the number of ways to filled the three outer triangles is $N_o (n) = N_c (n) N_p (n)$ , where $N_c(n) = \dbinom 4n$ is the number of ways to choose $n$ colors and $N_p (n)$ is the number of ways to permute the $n$ colors among the three triangles. Then, we have:

$\begin{array} {ccccccc} n & : & N_c(n) & \times & N_p(n) & = & N_o(n) \\ \hline 1 & : & 4 & \times & 1 & = & 4 \\ 2 & : & 6 & \times & 6 & = & 36 \\ 3 & : & 4 & \times & 6 & = & 24 \\ \hline & & & & & & 64 \end{array}$

The middle triangle however must be filled with a color different from that of any of the three outer triangles. Therefore, the number of ways to fill the middle triangle is $N_m (n) = 4-n$ . And the total number of ways of filling the four triangles is given by:

$\begin{array} {ccccccc} n & : & N_o(n) & \times & N_m(n) & = & N(n) \\ \hline 1 & : & 4 & \times & 3 & = & 12 \\ 2 & : & 36 & \times & 2 & = & 72 \\ 3 & : & 24 & \times & 1 & = & 24 \\ \hline & & & & & & \boxed{108} \end{array}$