5. Eight strangers are preparing to play bridge. How many ways can they be grouped into two bridge games - that is, into unordered pairs of unordered pairs of people?
Solution:
315
Putting 8 people into 4 pairs and putting those 4 pairs into 2 pairs of pairs are independent. If the people are numbered from 1 to 8 , there are 7 ways to choose the person to pair with person 1 . Then there are 5 ways to choose the person to pair with the person who has the lowest remaining number, 3 ways to choose the next, and 1 way to choose the last (because there are only 2 people remaining). Thus, there are $7 \cdot 5 \cdot 3 \cdot 1$ ways to assign 8 people to pairs and similarly there are $3 \cdot 1$ ways to assign 4 pairs to 2 pairs of pairs, so there are $7 \cdot 5 \cdot 3 \cdot 3=315$ ways.