A committee of three people is to be chosen from four teams

The easy way to solve this problem is to say there are 4C3 ways to choose a team. (2C1)(2C1)(2C1)(2C0) to choose the players. Hence the solution is (4C3)(2C1)(2C1)(2C1)(2C0)=32.

You can also do this longer way: Since each member on the committee must be from a different team lets look at the converse: How many ways can people of the same team be on the committe and subtract them out. We can pick 3 people from 8, 8C3=56, this is the total number of combinations. We can find the number of ways the a two people from the same team are on the committee by

1) The paired team on the committee is the first team
(2C2)(2C1)(2C0)(2C0)=2
(2C2)(2C0)(2C1)(2C0)=2
(2C2)(2C0)(2C0)(2C1)=2

2) The paired team on the committee is the second team

(2C1)(2C2)(2C0)(2C0)=2
(2C0)(2C2)(2C1)(2C0)=2
(2C0)(2C2)(2C0)(2C1)=2

3) The paired team on the committie is the third team

(2C1)(2C2)(2C0)(2C0)=2
(2C0)(2C2)(2C1)(2C0)=2
(2C0)(2C2)(2C0)(2C1)=2


4) The paired team on the committie is the fourth team
(2C1)(2C0)(2C0)(2C2)=2
(2C0)(2C1)(2C0)(2C2)=2
(2C0)(2C0)(2C1)(2C2)=2

All those two add up to 24. 56-24=32.

Here is my question, please help me answer below
TWO APPROACHES
COMBINATION==========================
Now if I wanted to do find the number of combinations of pair of the same team on the committee using combination.

(4C1)=# of ways to pick one team where we will grab two people from
(1C1)= # of ways to pick the two people from the same two people on the team
(3C1)= # of ways to pick the last person from the other three teams
(2C1)=# of ways to pick the person from the other group

Hence (4C1)(1C1)(3C1)(2C1)=24 . 56-24=32. Is my thinking correct here?

PERMUTATIONS================================

Now if I wanted to do find the number of combinations of pair of the same team on the committee using permutation.
I have three slots for the three committee members. I have 8 people. I have 8 to choose from for the first slot. 1 person for the second slot, as it has it be the first slots team member. The last person can be any of the 6. Hence 8*1*6=48. But I did not consider the different arrangement's of the Pair Pair Nopair committee. I have 3!/2!=3 to jumble them. So I have 48*3=144 ways of picking two people from the same team on the committe. 56-144 is negative! What am I doing wrong here !! HELP Bunuel!! Help Anyone!! Any help would be appreciated!!