permutations (1 Viewer)

[nrumble42]

Nine persons gather to play football by forming two teams of four to play each other, the remaining person acts as referee. if 2 particular people are not to be in the same team, how many ways are there to choose the teams?

 

[pikachu975]

Nine persons gather to play football by forming two teams of four to play each other, the remaining person acts as referee. if 2 particular people are not to be in the same team, how many ways are there to choose the teams?

Number of ways to pick team 1: 9C4
Number of ways to pick team 2: 5C4
So total of ways to pick two teams = 9C4*5C4 / 2! as the teams can be picked the exact same but in opposite order.

Now find the complementary,
Number of ways to pick team 1: 8C4 (since we're pretending the 2 people are 1 unit)
Number of ways to pick team 2: 5C4
So total of ways to pick two teams = 8C4*5C4 as the teams can be picked the exact same but in opposite order.

Therefore the required ways to pick the teams:
(9C4*5C4)/2 - (8C4*5C4)/2 = 140

Not sure if it's right...


Another option would just be:
Picking 3 people from 7 (don't include the 2 guys yet): 7C3
Then for this team pick 1 from the 2 guys: 2C1
Then pick team 2: 5C4
So total = 7C3 x 2C1 x 5C4 /2 (because order doesn't matter so dividing by 2 eliminates the option of picking the teams in reverse order)
= 175

I think this is the right one which tsoliman had first

 

[tsoliman1]

Nine persons gather to play football by forming two teams of four to play each other, the remaining person acts as referee. if 2 particular people are not to be in the same team, how many ways are there to choose the teams?

Selecting Team 1:
7C3 (1 of the 2 people is not selected in the same group as the other one and one of the 2 must be picked. Order doesnt matter so no need to divide by 2)

Selecting Team 2:
5C4 (The second person is included in here after the 4 (including one of the 2) have been selected from Team 1. order doesnt matter)

Therefore multiplying them gives:
=175

Not sure if it is right

 

[jathu123]

lol here's an (another!) alternate way

Case 1: Each team has 1 person it in. So the first team has person A and the other team has person B. We need to choose 3 people in the first from 7. So 7C3. For the other team, we need to choose 3 people from 4. So 4C3. So total = 4C3*7C3*1C1 = 140. We don't divide by 2 cause the teams are distinguishable (A on one, B on another)

Case 2: Now consider the case in which person A or person B is the referee. If person A is the referee, there (8C4*4C4)/2 = 35 ways of choosing the first team (divide by 2 cause they're indistinguishable). This is the same if person B was the referee. So total = 2*35 = 70

Overall: 140+70 = 210

 

[pikachu975]

So what was the answer?

 

[nrumble42]

the answer is 210. sorry for the late reply