The question I'm looking at is:
Andy, Bill, Carl and Dave are 4 students on a team of 10. 5 must be chosen for a tournament, how many teams can be picked if Andy or Bill or Carl or Dave must be on the team.
Using the inclusion-exclusion principle:
Let $A_1 =$ teams with Andy, $A_2 =$ teams with Bill, ect.
$$|A_i| = {9 \choose 4}= 126$$
$$|A_i \cap A_j| = {8 \choose 3} = 56\text{ for }i \neq j$$
$$|A_i \cap A_j \cap A_k| = {7 \choose 2} =21\text{ for }i \neq j \neq k$$
$$|A_1 \cap A_2 \cap A_3 \cap A_4| = {6 \choose 1} = 6$$
So then $|A_1 \cup A_2 \cup A_3 \cup A_4| = 4(126) - 6(56) + 3(21) - 6 = 225$
But when I use the complements principle to subtract all teams without Andy, Bill, Carl, and Dave from all teams I get:
$${10 \choose 5} - \displaystyle{6 \choose 5} = 252 - 6 = 246$$
which is not the same. So I'm clearly doing something wrong with one of these but I don't know which one is wrong.