There are three families each with two sons and two daughters. In how many ways can these young people be married so that each couple includes one member of each gender, no couple is brother and sister, and there are no singles left over?
First take the older boy in Family 1. He has a choice of 4 brides. Whichever one he chooses, call her older brother also has a choice of 4 brides, 2 from Family 1 and 2from the remaining family. [(4 · 2) + (4 · 2) = 16]
Now it is the turn of the older brother in the remaining family. If the second boy chose one of his sisters he has 3 choices. If the second boy chose a girl from Family 1, he has 2 choices. [(4 · 2 · 2) + (4 · 2 · 3) = 40.
Next it is the younger boys' turns. At this point, either there is one girl left from each family, or there is one family with 2 girls left, one with 1 left and one with none left. In either case, there is a boy with one married and one unmarried sister. He has 2 choices. After he has made his choice, each of the other two has only 1 choice. [40 · 2 · 1 · 1 = 80]
Posted by TomM
on 2005-07-23 04:28:02