There is a formula for counting how many combinations you can make when repetition is allowed; if you are choosing k things from n, where order is irrelevant, and you are allowed repetition, you can make (n+k-1)!/[ (k!) (n-1)!] selections. So if you have five colors of stones, and you want to know how many ways we can choose three stones for a ring, if we don't care about order and we can use a color more than once, we have n = 5, k = 3, and thus (5+3-1)!/[3! * 4!] = 7!/(3!)(4!) = 35 possible selections.

You might want to know why this works. This is far beyond the scope of the GMAT, so for interest only! If we have, say, the five colors (I'll just use letters) R, G, B, Y, W, then if we make the selection of two Rs and one W, we could write this as a word:

RxxGBYWx

and if we select one G, one B and one Y, we could write the word:

RGxBxYxW

and so on. That is, we insert one 'x' after each letter for each time we wish to select that letter. Since the word will never begin with an x, there will be 5 locations where we can place the first x, 6 where we can place the second (since it can be immediately to the left or right of the first x), and 7 where we can place the third. Since the order of the 3 x's is irrelevant, we then must divide by 3!, so the x's can be placed in 7*6*5/3! = 35 ways.

There's a problem which, on the surface, looks quite different from the above, but actually boils down to the same fundamental principle - I posted a solution (scroll down) using a similar method:

https://www.beatthegmat.com/combination-t41362.html

That said, I've *never* seen a real GMAT question where you might be able to use this formula, and you can always break any similar problem into cases - for the ring problem, we can have three stones of the same color, two of the same color and one of a different color, or three of different colors. Adding the number of ways to make each type of selection will give the answer, as you found above.