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?
Choose the correct choice from the following answers.
A) 20
B) 22
C) 26
D) 30
E) 32
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= 
<gadget id="calculator">56 - 24</gadget>
<output>32</output>
32!   ANSW32R:32
<result>E</result>
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?