Group of people? Combinatorics is simpler

This is about selecting two triples out of n people ,without regard to order within each group, but the order of the triplets must be taken into consideration, when there may be people who are common to both groups,but it can not be that the two groups are the same,

First division method :-

The number of triplets out of n people,without regard to order within each group is $\binom{n}{3}$ , we want to choose two of these triplets taking the order of triplets in consideration ,so we get $\binom{\binom{n}{3}}{2}2=2x$ ,

Second division method :-

The group are not the same , so the number of common people to the both groups can be ${\boxed{0,1,2}}$ ,

First case :- zero common members to both groups :- we want to choose $6$ out of $n$ people , and divide them into two groups , we get $\binom{n}{6}\binom{6}{3}$ .

Second case :-one common member to both groups :- we want to choose $6$ out of $n$ people , and divide them into two groups, where we have to choose one member to be common to both groups and then we have to choose to other members to the first group, and the other two members will be in the second group, we get $\binom{n}{5}\binom{5}{1}\binom{4}{2}$ .

Third case :- two common members to both groups :- As I solved in the previous case , we get $\binom{n}{4}\binom{4}{2}\binom{2}{1}=\binom{n}{4}\binom{4}{2}2$ .

so , the sum of these three cases must be equal to $\boxed{2x}$ As we got in the first method for division.