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?

(In reply to Off the cuff by Bob Smith)

1) Assume that your scheme works. This scheme does not allow for the real possibility that the first two families will intermarry in both cases, leaving the boys in the third family with no one to marry but their sisters. Since out of the 288 combinations you claimed, there are 8 ways that the third family could get "stuck," there are 288 - 8 = 280 ways left.

2) But then again the scheme does not work to begin with. Giving the first son in the second family 4 choices supposes that both sons in the first family chose his sisters.

 

Posted by TomM on 2005-07-18 17:11:34