There is an unlimited supply of [congruent](https://artofproblemsolving.com/wiki/index.php/Congruent) [equilateral triangles](https://artofproblemsolving.com/wiki/index.php/Equilateral_triangle) made of colored paper. Each [triangle](https://artofproblemsolving.com/wiki/index.php/Triangle) is a solid color with the same color on both sides of the paper. A large equilateral triangle is constructed from four of these paper triangles. Two large triangles are considered distinguishable if it is not possible to place one on the other, using translations, rotations, and/or reflections, so that their corresponding small triangles are of the same color.
Given that there are six different colors of triangles from which to choose, how many distinguishable large equilateral triangles may be formed?
[asy] pair A,B; A=(0,0); B=(2,0); pair C=rotate(60,A)*B; pair D, E, F; D = (1,0); E=rotate(60,A)*D; F=rotate(60,C)*E; draw(C--A--B--cycle); draw(D--E--F--cycle); [/asy]
If two of our big equilateral triangles have the same color for their center triangle and the same [multiset](https://artofproblemsolving.com/wiki/index.php/Multiset) of colors for their outer three triangles, we can carry one onto the other by a combination of rotation and reflection.  Thus, to make two triangles distinct, they must differ either in their center triangle or in the collection of colors which make up their outer three triangles.
There are 6 possible colors for the center triangle.

There are ${6\choose3} = 20$ possible choices for the three outer triangles, if all three have different colors.
There are $6\cdot 5 = 30$ (or $2 {6\choose2}$) possible choices for the three outer triangles, if two are one color and the third is a different color.
There are ${6\choose1} = 6$ possible choices for the three outer triangles, if all three are the same color.
Thus, in total we have $6\cdot(20 + 30 + 6) = \boxed{336}$ total possibilities.