Combinations Question regarding placement in multiple teams with restrictions? I've been having trouble with the following question:


  Nine players are to be divided into two teams of four, and one umpire. If two particular people cannot be on the same team, how many different combinations are possible?


The answer is 210, but I keep getting 70. The way I've been doing it is:

Let the three groups: the two teams and the umpire be represented as:

A _ _ _

B _ _ _

_

I have placed the three people in A's team: $7C3$ and B's team: $4C3$, then dividing by $2!$ due to the identical group sizes. What am I doing wrong?

Thanks.
Let's call the particular ones $A$ and $B$.

If none of the particular ones is chosen as umpire then there are
$7\times\binom{6}{3}=140$ possibilities ($1$ out of $7$ is picked
to be the umpire and then $3$ out of $6$ are chosen to join $A$)

If $A$ is chosen as umpire then there are $\binom{7}{3}=35$ possibilities
($3$ out of $8$ are picked to join $B$)

If $B$ is chosen as umpire then there are $\binom{7}{3}=35$ possibilities
($3$ out of $8$ are picked to join $A$)
This exact problem is explained in 2 parts.
Part 1: https://www.youtube.com/watch?v=LNX4Ya6Bg94 and
Part 2: https://www.youtube.com/watch?v=1qoX7RrMSNw&amp;t=459s 
The two players can be separated by making 1 an umpire.

So you have 140 (2 players on separate teams) +70 (player as umpire)