How many ways can 5 people be assigned to 3 distinct teams, with at least one team having 2 members?
How to approach this problem? Would appreciate a step by step explanation of the same
Uh just do casework or inclusion/exclusion. Have u tried casework?
Well, I am not being to wrap my head around this problem.
When it says in how many ways can 5 people be assigned in 3 teams with at least one team having 2 members…do we need to ensure that all 5 people are assigned? Because if that is the case then, the only arrangement that will work is:
Team 1: 2 people
Team 2: 2 people
Team 3: 1 person
Also different ways the teams can have two members is (2,2,1), (2,1,2), (1,2,2)
So using the inclusion exclusion principle:
If there were no restrictions, then the total no of ways the 5 people could have been arranged in 3 groups is simply: 3^5 = 243 (as any of the member can be assigned to any team, since no restrictions have been mentioned!)
from this we then exclude the cases where the condition of “at least one team” having 2 members is is not met. This happens when teams are arranged in the following fashion (one team has 3 members and rest of the two teams have 1 member each)
So for assigning 3 out of the 5 people in one team: 5c3 = 10 ways.
Assigning remaining 2 people into 1 person each in of the other two teams = 2! = 2
we also know such an arrangement can be possible in any one of three ways (3,1,1) (1,1,3) (1,3,1)
So the total number of assignments where no team has exactly 2 members is:
3 * 10 *2 = 60 ways
Thus, the total number of ways to assign 5 people to 3 teams such that at least one team has exactly 2 members is 243-60 = 183
- is this even correct?