In how manys can a programming team and a hacking team be selected if Jack does not serve on both teams? A club has $12$ members. They need to choose a programming team of $3$ members and a hacking team of $4$ members. Members can be on both teams, but Jack only wants to be on at most one team. How many ways can both teams be chosen?
I did two different ways. I solved if Jack chooses programming, $C(12,3)=220$ $C(11,4)=330$. In total there are $550$ different ways both teams can be picked.
But, if Jack chooses hacking, I solved $C(11,3)=165$ and $C(12,4)=495$. In total there are $650$ ways both teams can be chosen.
I was wondering if this is the correct method.
 A: Saying that Jack is on at most one team means that he is either on one of the two teams or on neither team.
Method 1:  We consider cases.


*

*Jack is selected for the programming team:  We must two of the other eleven club members for the programming team and four of the other eleven club members for the hacking team.  This can be done in 
$$\binom{11}{2}\binom{11}{4}$$
ways.

*Jack is selected for the hacking team:  We must select three of the other eleven club members for the programming team and three of the other eleven club members for the hacking team.  This can be done in
$$\binom{11}{3}\binom{11}{3}$$
ways.

*Jack is selected for neither team:  We select three of the other eleven club members for the programming team and four of the other eleven club members for the hacking team.  This can be done in 
$$\binom{11}{3}\binom{11}{4}$$
ways.


Since these cases are mutually exclusive and exhaustive, the number of ways the teams can be selected without placing Jack on both teams is found by adding the above results.
Method 2:  We subtract the number of selections in which Jack is placed on both teams from the total number of possible selections.
If there were no restrictions, we could select three of the twelve club members to be on the programming team and four of the twelve club members to be on the hacking team in 
$$\binom{12}{3}\binom{12}{4}$$
ways.
If Jack is on both teams, we have to select two of the other eleven club members to be on the programming team and three of the other eleven club members to be on the hacking team.  Hence, there are 
$$\binom{11}{2}\binom{11}{3}$$
selections in which Jack is on both teams.  
Therefore, there are 
$$\binom{12}{3}\binom{12}{4} - \binom{11}{2}\binom{11}{3}$$
ways to select the teams without placing Jack on both teams.
Note:  In your attempt, you made the following errors.


*

*You did not consider the possibility that Jack is on neither team.

*If Jack is selected for the programming team, then we must select two of the other eleven members of the club to be on the team.  We cannot make $\binom{12}{3}$ selections since those include selections without Jack.  You made the analogous error when Jack is selected for the hacking team.

*When you computed the number of ways the two teams could be formed if Jack is selected for exactly one of them, you should have multiplied the number of ways the remaining members of the programming team can be selected by the number of ways the the remaining members of the hacking team can be selected rather than adding them.  You add when two events cannot occur at the same time.  In this case, once it is determined on which team, if any, Jack will serve, the choices of the remaining members of the hacking and programming teams can be made independently.

