In how many ways can three groups of two tennis players be seated in a row if no two members of the same group are adjacent? My teacher asked me a question about $3$ groups of tennis players, each consisting of two players. They wanted to take a picture together by sitting on chairs in a row. He asked, how many possible sequences for them if they are not allowed to sit next to their teammates? My teacher answered it with
$6!-3!2!2!2!$
with $6!$ as all possible sequences and $3!2!2!2!$ as all of them sitting next to each other. But that mean the answer max $2$ teams able sit next to their teammate right?
I'm quite sure he answered it wrong. Because he said none of them allowed sit next to their team. But with that answer, I'm quite convince they still can sit next to each other.
My answer is:
All Possible sequences
$6!$
3 teams with members sitting next to their teammate:
$3!2!2!2!$
2 teams with members sitting next to their teammate:
$_3C_2(4!2!2!-3!2!2!2!)$
1 Team with members sitting next to their teammate:
$_3C_1(5!2!-(_3C_2(4!2!2!-3!2!2!2!))$
So the answer is
$6!-(3!2!2!2!+(_3C_2(4!2!2!-3!2!2!2!))+(_3C_1(5!2!-(_3C_2(4!2!2!-3!2!2!2!))))$
But I'm not sure, can someone help me?
 A: Let us count; team A with players a and a', B with players b and b', and C with players c and c'.
Here is an example of a possible sequence; abcb'c'a', obviously we can put any player in the first chair, so 6 possibilities. Then we can put only 4 in the second chair, so 4 possibilities. Then also just 3 in the third chair as we can put the teammate (a' or the first chair's teammate) here. 
Now we can have two configurations :

*

*$6.4.2= 48$ beginning like abc , this type of sequence continue with 2 choices then 2 choices so in total $6.4.2.2.2= 192$ sequences.

*or $6.4= 24$ beginning like aba', this type of sequence continue with 2 choices (why?) then the last two positions are determined. So in total $6.4.2= 48$ sequences.

Giving you the total number of possible sequences to be :  $192+ 48= 240$
A: This amounts to seating $AABBCC$ so that no two adjacent letters are alike, and is easily tackled by inclusion-exclusion.
All arrangements - at least one pair together + at least two pairs together - all three pairs together
To get any pair together, you glue the pair, and arrange them in two possible ways
$= 6! - \binom31*2*5! + \binom32*2^2*4! - \binom33*2^3*3! = 240$
