How many ways can you color this?

There are $7^2 = 49$ smaller triangles in a $7$ -check triangle, and there are $_{49}\text{C}_{3}$ ways to choose $3$ smaller triangles out of the total $49$ .

However, all of these will have $3$ duplicates through rotational symmetry except for $_{16}\text{C}_{1}$ of them, the number of ways to choose $1$ triangle out of the $16$ triangles that make up a third symmetrical section of the whole.

So the maximum number of elements in a corresponding "okay" set will be $\frac{_{49}\text{C}_{3} - _{16}\text{C}_{1}}{3} + _{16}\text{C}_{1} = \frac{18424 - 16}{3} + 16 = \boxed{652}$ .

It was kinda not that tough as it seems so Let x be those arrangements in which after one-third rotation it gives different image and y be those arrangements which gives the same image after one -third rotation So according to question 3x+y=49C3 Now for one block there correspond two other block except for the centre This implies 3y=49-1=48 y=16 Now solving we get x =6136 Required Sum=x+y This implies x+y=6152