Total number of ways to select a squad of $15$ from $30$ probables $ =^{30}C_{15} $
Since each of the $5$ zones must be represented by their captains, it leaves out $5$ players and we can select the remaining $10$ players from any of the $25$ players. So, our required probability $ = \dfrac{{}^{25}C_{10}}{{}^{30}C_{15}}$
- None of these is correct
Supposing instead of “captain” if any player can represent a zone, the prolem becomes really difficult.
If we have to represent each zone, we have to select one player each from all the five zones which can be done in $5 \times 7 \times 8 \times 6 \times 4$ ways.
Remaining $10$ people can be selected from $25$ remaining probables in $^{25}C_{10}$ ways.
So, required answer $=\frac{5 \times 7 \times 8 \times 6 \times 4 \times^{25}C_{10} }{^{30}C_{15}}$
The above approach is wrong because we are doing over counting and thus is only an upper bound. For example we count the selection of Player A from Zone North in the mandatory single player case and player B from North Zone in the $^{25}C_{10}$ case repeatedly when Player B is chosen as a mandatory player.
This problem is similar to distinct balls in distinct bins where no bin is empty (solution for this is given by $S(n,k)\times k!$ where $S(n,k)$ is Stirlings number of second kind). But here we have $5$ distinct bins (zones) and $15$ distinct balls (players) with the added constraint of an upper bound on each bin. One can try solving 😊
