The binomial coefficient answer given is correct as a general formula and I'd recommend you program that in. If you'd like intuition for why this is the case, think about the number of choices you are making. First, choose your first rune (call it Rune A). You have 13 choices for this rune. Now set rune A aside and choose rune B from the remaining 12 runes. You have 12 options. Finally, choose rune C from the remaining 11 runes. This gives you 13x12x11 = 1716 choices. That being said, you are not done, as some of these choices are redundant. For example, you could have chosen the 9th, 4th, and 3rd rune in that order, or chosen the 4th rune, 9th rune then the 3rd rune. This yields the same collection of 3 runes, but done in a different way.

To cut out these possibilities, you need to count how many redundancies there are. This corresponds to the number of orderings in which you can choose three runes at random, since the order you chose the runes is irrelevant. For runes A,B, and C, you can choose then in orders ABC, ACB, BAC, BCA, CAB, CBA. In particular, you can choose one of A,B,C first, then choose one of the two remaining ones second, and the last one has to be the last chosen. That's 3 x 2 x 1 = 6 possible orders. for any given choice you have 5 redundant choices. That means, you get the true count, you need to toss 5 choices for every 1 you make. This is done by dividing by 6. Thus you get 1716/6 = 286 as your final answer.

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This agrees with the formula because you just computed (13 x 12 x 11)/3!, which is 13!/(3! x 10!) = (13 choose 3).