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!!

Hi Brunel

I did not inderstand the concept behind writing 4c3 it means selecting 3 person out of 4 but it is 4 team so it should be 8c3

Am i Correct

Regards]

Archit

Bunuel - What do you mean by selecting "3 teams" out of 4? The question states that 3 people form a committee (or team) and need to be selected from 4 teams of 2 people each. Archit is right. Shouldn't it be 8C3?

However the slotting method you mention is correct. 8*6*4/3!

i think it should be only 4c3 * 2c1
3 teams out of four and 1 person out of 2
am i correct or missing sthn

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

My Method

First we find out the amount of combinations of 3 person teams from a pool of 8. C(3/8) = 8!/5!*3! = 56 ways

Next I find the number of ways we CAN make a 3 person team using two from the same group, so

We take 2 people from a 2 person group C(2/2) and mutiply that by taking any 1 person from the remaining 6 C(1/6)

C(2/2)*C(1/6) = 6

This is the number of combinations by taking both parties from pair A. As we have a four pairs we must multiply this by 4 (4x6 = 24)

So there are 24 ways in which two people from the same pair can work together.

Finally we subtract this from the total number of combinations to find the number of groups where pairs do NOT work together....

56 - 24 = 32!
_________________

It is an arrangement question similar to finding arrangements of AABBB = 5!/ (2!*3!) where 2! and 3! are done to remove the duplication of A and B as all As and all Bs are the same. Had all As or all Bs be different (A1 A2) or (B1 B2 B3) , then the answer would have been = 5!

Hope this helps.

How I Solved it:

Teams with no couple chosen = Total no. of possible teams - Teams with one couple chosen
= 8C3 - 4C1x6C1
= 56 - 24 = 32

where:
4C1 --> From 4 couples choose any one
6C1 --> From the 6 remaining candidates, choose any one