All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > General
No Incest, Please (Posted on 2005-07-18) Difficulty: 3 of 5
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?

No Solution Yet Submitted by Erik O.    
Rating: 3.6000 (5 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution Off the cuff, revisited | Comment 8 of 13 |

80 ways

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
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (3)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2017 by Animus Pactum Consulting. All rights reserved. Privacy Information