Given 20 pairs how many diff. 3 member groups can be formed that don't contain any of the original 20 couples?

I don't know how to solve so please help!!!Given 20 pairs how many diff. 3 member groups can be formed that don't contain any of the original 20 couples?
There are 20 couples or 40 people.





Pick the first person from the 40 people --%26gt; 40 ways


There are then 38 people to pick from (exclude the person and his/her partner) --%26gt; 38 ways.


Then there are 36 ways to pick the next person (exclude the 4 people that are either chosen or partners) --%26gt; 36 ways.





If you were to multiply that out you'd get:


40 x 38 x 36 = 54,720 ways





However, realize that the order of picking the people doesn't matter. If you pick Amy, Cindy and Edward, that's the same as picking Edward, Amy and then Cindy. So divide by the ways to arrange 3 people --%26gt; 3! = 3 x 2 x 1 = 6 ways.





54,720 / 6 = 9,120 ways





Answer:


9,120 waysGiven 20 pairs how many diff. 3 member groups can be formed that don't contain any of the original 20 couples?
Ok - 20 pairs = 40 people





member groups that don't contain an original pair means that the 1st member can be chosen from any of the 40 people.





2nd person can not be the 1st person's pair, leaving 38 possibilities.





3rd person can't be either pair for the first 2 chosen, leaving 36 possibilities.





The total is the product of the possibilities 40x38x36 = 54,720





The trap here is to identify that there are 40, not 20, people.
It is 6,840....because 20 X 19 X 18 = 6,840....