# Permutations of groups with exclusive members

Suppose we have a set $$A = \{ a_1, a_2, ... , a_m \}$$ and we want to make sets of $$3$$ elements. We will have $$m \choose 3$$ different sets. Let's call this new combinations set as $$B=\{\{a_1, a_2, a_3\}, \{a_1, a_2, a_4\} , ... , \{a_{m-2},a_{m-1}, a_m\}\}$$. The elements of this set will be referred as $$b_i$$. Now what I want is to know the number of permutations of $$k$$ elements of $$B$$ such that for all the pair of sets $$b^p_i, b^p_j$$ belonging to the permutation $$p$$, all the pair of elements of the sets $$a^{b^p_i}_k, a^{b^p_j}_l$$ are distinct to each other.

For example, for $$k=2$$, the permutation $$\left(\{a_1, a_2, a_3\}, \{a_1, a_2, a_4\}\right)$$ would be invalid, because $$a_1$$ and $$a_2$$ are in both sets of the permutation. However, the permutation $$\left(\{a_1, a_2, a_3\}, \{a_4, a_5, a_6\}\right)$$ is valid since none of the inner elements repeats within the other set.

• In your question you ask for the number of permutations, and in your example you describe combinations. Can you clarify? Commented Jun 23 at 20:51
• @UriGeorgePeterzil I meant permutation, I have already edited the question. Commented Jun 24 at 7:39
• So, you have $m\choose3$ $3$-sets. Given any one of them, there are ${m-3\choose3}$ that are disjoint from it. So, for $k=2$, the answer is ${m\choose3}{m-3\choose3}$. Commented Jun 24 at 7:43
• Reasoning the same way (if I interpret the question the way you intend), for $k=3$ it would be ${m\choose3}{m-3\choose3}{m-6\choose3}$, and so on for any value of $k$. Commented Jun 24 at 7:45
• But perhaps I misunderstand. Is $(\{a,b,c),(a,d,e)\}$ OK (where $a,b,c,d,e$ are all different)? Commented Jun 24 at 7:47