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?
Each son may marry one of four girls. Let's say the eldest son gets first choice, then the second son only gets to choose from three. Repeat for next family. Last family's sons only get to choose from remaining 2 girls. So the knee jerk answer is 4 * 3 * 4 * 3 * 2 * 1 = 288.
Off the cuff
