Problem:
In how many ways can 6 juniors and 6 seniors form 3 disjoint teams of 4 people so that each team has 2 juniors and 2 seniors?
Answer Choices:
A.
B.
C.
D.
E.
Solution:
Select the first junior and call this person . There are 5 other juniors and pairs of seniors who could team up with . Select the next junior not yet on a team, say . There are 3 other juniors and pairs of seniors who could team up with . The third team consists of the people not yet chosen. Thus there are ways to form the teams.
There are ways to assign the seniors to teams 1,2 , and 3 , and, similarly, 90 ways to assign juniors to those teams. But there are ways for the three teams to be ordered, so the number of ways to form the teams is .
The problems on this page are the property of the MAA's American Mathematics Competitions