How many cricket teams can the manager choose if one particular batsman refuses to play when one particular bowler does? 
A certain country has a cricket squad of 16 people, consisting of 7 batsmen, 5 bowlers, 2 all- rounders and 2 wicket-keepers. The manager chooses a team of 11 players consisting of 5 batsmen, 4 bowlers, 1 all-rounder and 1 wicket keeper.  Find the number of different teams the manager can choose if one particular batsman refuses to be in the team when one particular bowler is in the team. 

I did it 
$6C5 \cdot 5C4 \cdot 2 \cdot 2 + 7C5 \cdot 4C3 \cdot 2 \cdot 2$. It is completely wrong. How to do it?
 A: If you describe in words the logic behind your expression, you should see why it's wrong (not completely wrong, but still wrong). And correcting the logic could help you correct the answer.
Your first term $\binom{6}{5}\cdot\binom{5}{4}\cdot2\cdot2$ counts all possible teams that do not include that batsmen. Note that the $\binom{5}{4}$ factor here counts all possible choices of $4$ bowlers, both with or without the conflicting bowler in them. So far, so good.
What does your second term $\binom{7}{5}\cdot\binom{4}{3}\cdot2\cdot2$ represent? The $\binom{4}{3}$ factor probably says that you've included the conflicting bowler, and then count all ways to select three more bowlers. But the $\binom{7}{5}$ factor counts all choices of $5$ batsmen, even the ones that include the conflicting batsman. So this allows both of them to be on the team, which shouldn't happen.
State more carefully the different possible cases that need to be accounted for, and that should help you come up with the correct expression.

Here's a somewhat different approach, which I think may be a bit easier: first count all possible such teams, and then subtract the number of teams that we can't allow, i.e. the number of teams with both this batsman and this bowler included. (This is, essentially, an instance of the inclusion-exclusion principle.)
A: There are $5$ from $7 = 21$ ways to choose batsmen. For everyone of those there are $4$ from $5 = 5$ bowlers and for every one of those there are $2$ choices of wicket keeper and $2$ choices of all-rounders. Subtracting combinations that contain a certain bowler and batter involves counting the number of ways a batter can be in the group of $5$ which is $4$ from $6 = 15$, and for those, the choic of bowlers is $4$ from $4 = 1$. So the calculation looks like........
$([\binom{7}{5} - \binom{6}{4}] \cdot \binom{5}{4} + \binom{6}{4})\cdot 2 \cdot 2$ 
$=(6\cdot 5 + 15)\cdot 2 \cdot 2 =  180$
