So we can say: the number of different triplet combinations from the set A is (Each time we have one less choice until there’s only 1 left) If there were more numbers (let’s call this number r), then the best way to calculate the number of different orderings would be There is one issue I didn’t raise: how did I know how many different ways there were of ordering x y and z? Well the simple answer is that it was fairly trivial to think up the different orderings, since there were only three numbers per ordering. So there are 165 unique triplets in B, which answers our first question. There will be 6p of these in B (ignoring their order). For each permutation, there are five others which contain exactly the same numbers.

We know there are 990 permutations in total. So our goal, is to remove all extra five elements from set B for all triplets x y z. Let’s call the numbers x y z.īecause set B is a set of permutations, it’ll also contain values: Let’s define a set B, which contains all of the different orderings of triplets from A.Ĭonsider ANY element in set B.

We start off with 11 options, followed by 10 left and then we are left with 9. Let’s start off by considering the number of different ways of ordering three numbers (permutations) from set A. 1 2 3 or 10 11 6), where triplets with exactly the same numbers count as one (e.g. Let’s take the numbers 1 2 3 4 5 6 7 8 9 10 11 and call it set AĪsk yourself: how many ways can I make triplets from A (e.g.

I copied my answer from Quora which I wrote on the same day as I saw this.

Permutation equation