We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
0
128
3
avatar

Alice, Andrew, and six other students are to be divided into two teams of four people. How many ways are there to form the teams so that Alice and Andrew are on opposite teams?

 

So if you choose between Alice and Andrew you have 2. You multiply that by 6*5*4 because those are her other teamates. I then choose the others team, the leftover between Alice and Andrew is 1 so I multiply that by the remaining teamates he can have 3*2*1. So wouldn't the answer be 1440. But this answer is wrong so I don't know what to do.

 Mar 23, 2019
 #1
avatar+5662 
+4

we put Alice and Andrew on opposite teams.

We then have \(\dbinom{6}{3}=20\) ways to select 3 players from the remaining six for one of the teams.

 

Once those 3 are selected the other 3 are automatically on the other team.

 

In your working you treated permutations of players within a team as important.  They aren't.

 Mar 23, 2019
edited by Rom  Mar 23, 2019
 #2
avatar+721 
+2

I believe Rom is correct

CoolStuffYT  Mar 23, 2019
 #3
avatar+533 
+2

Because there are only two teams, Alice and Andrew are on opposite teams. Now, we can choose 3 people out of the other 6 to be on Alice's team, and the other 3 to be on Andrew's. So, the answer is 6C3 = 6! / (3!)^2 = 720 / 36 = 20.

 Mar 23, 2019

6 Online Users

avatar