In a class of $30$ students, how many pairs of teams, one football ($11$ players) and one basketball ($5$ players) can form with the condition:
a) no student can participate in both teams
b) at most one student participates in both teams
c) any student can participate in both teams
My attempt:
a) I think in first ask we do this $C_{11}^{30}C_{5}^{19}$, but one of my friends made like on photograph.
b) $C_{11}^{20}C_{5}^{20}$?
c) $C_{11}^{30}C_{5}^{30}$?
Your answer $$\binom{30}{11}\binom{19}{5}$$ is correct.
Since at most one students participates on both teams, either no student participates on both teams or one student does.
We calculated the number of ways teams could be formed if no student participates on both teams above.
For the case in which one student participates on both teams, there are $30$ ways to select the student who participates on both teams. We now need to select $10$ of the remaining $29$ students to complete the football team and $4$ of the remaining $19$ students to complete the basketball team. Hence, there are $$\binom{30}{1}\binom{29}{10}\binom{19}{4}$$ ways to select the teams if one student participates on both teams.
Therefore, there are $$\binom{30}{11}\binom{19}{5} + \binom{30}{1}\binom{29}{10}\binom{19}{4}$$ ways to select the teams if at most one student participates on both teams.
Follow the model in the preceding question. Choose the students who participate on both teams, choose the remaining members of the football team from the remaining students, then choose the remaining members of the basketball team from the remaining students.