A committee of three people is to be chosen from four teams of two. What is the number of different committees that can be chosen if no two people from the same team can be selected for the committee?
A. 20
B. 22
C. 26
D. 30
E. 32
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!!
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