"A Math Club has 12 members. They have to choose euclid contest team of 3 members and a National Math Competition team with 4 members. Students can be on both teams, but James agrees to only be on at most one team. In how many ways can both teams be chosen?"
Here's what I thought should be the answer but doesn't make much sense:
Case 1: James is in Euclid team:
$\therefore $ $11\choose2$ for Euclid Team
This would be because James is in team Euclid assumed so we only have to get 2 other players and we have 11 students left.
Also : $11\choose4$ for Math team
so total ways would be $11\choose2$ . $11\choose4$ (Product Rule)
Case 2: James is in Math team:
$\therefore $ $11\choose3$ for Math team
This would be because james is in team Math assumed so we only have to get 3 other players and we have 11 students left.
Also : $11\choose3$ for Euclid team , as the 11 students can form a team of 3 for Euclid competition here.
so total ways would be $11\choose3$ . $11\choose3$ (Product Rule)
Case 3: James in no team
$\therefore $ $11\choose3$ . $11\choose4$ (Product Rule)
as James is in no team so we are left with 11 people for both teams
Now the question is did I do it correctly? Because I have never seen cases being used in a permutation questions?
Therfore, The total number of ways would be to add all the cases meaning
Case 1 + Case 2 + Case 3 :
($11\choose2$ . $11\choose4$) + ($11\choose3$ . $11\choose3$) +($11\choose3$ . $11\choose4$)
Is that correct?
Disregarding the wish of James there are $\binom{12}4\binom{12}3$ possibilities.
Among them are $\binom{11}3\binom{11}2$ possibilies with James is in both teams.
Subtracting we find: $$\binom{12}4\binom{12}3-\binom{11}3\binom{11}2$$ possibilities that do not disregard the wish of James.