In a basketball league, there are six teams of 5 players each. A set of 7 players is selected randomly out of the 30 players in the league. What is the probability that at least one player from each team is selected?
The number of ways to choose 7 players from 30 players is done using the choose formula ${\binom {n}{k}}={\frac {n!}{k!(n-k)!}} = \dbinom{30}{7}$
The probability at least one player from each team will be selected, $P(A) = \frac{ \binom{7}{5} \binom{7}{5} \binom{7}{5} \binom{7}{5} \binom{7}{5} }{\binom{30}{7}} $
Approaching directly:
Note that for every team to contribute at least one player to our selection, that requires that every team contributes exactly one player to the selection with the exception of one team which contributed exactly two.
If the first team contributed two people and every other team contributed one person, there are $\binom{5}{2}$ ways to pick which two people the first team contributes, and $\binom{5}{1}$ ways to pick which one person each other team contributes respectively.
There are then $\binom{5}{2}\binom{5}{1}\binom{5}{1}\binom{5}{1}\binom{5}{1}\binom{5}{1}$ ways in which we may select two people from the first team and one person from each other team.
The probability then that there are two people from the first team and one person from each other team is $\dfrac{\binom{5}{2}\binom{5}{1}\binom{5}{1}\binom{5}{1}\binom{5}{1}\binom{5}{1}}{\binom{30}{7}}$.
Remembering that it could have been any of the teams that was the team to contribute two people, and recognizing that the situation is symmetrical and the probability would be the same for each team, we get the probability that every team contributes at least one person (i.e. there is a team that contributes two people and each other team contributes one person) as being:
$6\times \dfrac{\binom{5}{2}\binom{5}{1}\binom{5}{1}\binom{5}{1}\binom{5}{1}\binom{5}{1}}{\binom{30}{7}}$
Approaching indirectly:
The probability that none from the first team are selected is $\dfrac{\binom{25}{7}}{\binom{30}{7}}$, that none from the first two teams are selected is $\dfrac{\binom{20}{7}}{\binom{30}{7}}$, etc...
Using inclusion exclusion on the events that a team is not included in the selection we get the probability that there is at least one team not included in the selection as:
$\dfrac{\binom{6}{1}\binom{25}{7} - \binom{6}{2}\binom{20}{7}+\binom{6}{3}\binom{15}{7}-\binom{6}{4}\binom{10}{7}+0-0}{\binom{30}{7}}$
Subtracting this away from $1$ gives the probability that every team was included at least once in our selection. You should be able to verify that it is indeed equal to the value calculated earlier.
"Which technique is 'better'?" They are both fine. For this specific problem the direct approach is less work and the indirect approach looks more tedious. That being said, if we were picking ten people instead and we had a great many more possible breakdowns such as 5-1-1-1-1-1, 4-2-1-1-1-1, 3-3-1-1-1-1, 3-2-2-1-1-1, 2-2-2-2-1-1,... it gets much much more tedious to do it directly and at that point it would have been less work to do it indirectly instead.