Source:700-800 Math: Combination and Permulation

[sajibfin06]

Source:700-800 Math, suppose a 4 member team is to be formed from 7 members, if 2 of the 7 member refuse to be on the same team, how many arrangement are possible ?

[MartyMurray]

Source:700-800 Math, suppose a 4 member team is to be formed from 7 members, if 2 of the 7 member refuse to be on the same team, how many arrangement are possible ?

I would break this down into a question about three sets of teams.

One set would include teams made up of one of the two who refuse to be on the same team, along with three of the remaining five who are ok with being on a team with anyone

The second set would include teams made up of the second of the two who refuse to be on the same team, along with three of the remaining five.

The third set would include teams made up of four of the remaining five.

The number of teams in the first set = 1 * (5*4*3)/3! = 10

The number of teams in the second set = 1 * (5*4*3)/3! = 10

The number of teams in the third set = (5*4*3*2)/4! = 5

The total number of teams = [spoiler]10 + 10 + 5 = 25[/spoiler]

[DavidG@VeritasPrep]

We can find the total possibilities without restriction, and then simply subtract out the undesirable scenarios when those two folks who don't like each other are on the team.

Total without restriction: 7C4 or (7*6*5*4)/(4*3*2*1) = 35

For our undesired outcome, imagine that we have four positions and that two of those positions are already occupied by the people who don't want to be on the same team. Well, there are only two positions left to fill, and for those positions, we'll be selecting from the 5 people left, so we want 5C2 to see how many ways undesirable outcomes we can make: 5C2 = 5*4/2 = 10

Total outcomes - undesired outcomes = 35 - 10 = 25.