M&S->Permutations and combinations

Let the players be P1, P2, P3, ... P15 with P1 and P2 being the brothers.
Also, as mentioned in the question, let the teams be 1st, 2nd and 3rd.
(a) When assigning players for 1st team, there are 15C5 = 3003 possible combinations and for each specific combination of players in 1st team, there are 10C5 = 252 possible combinations for 2nd team and 3rd team together.
Therefore total no. of ways = 3003 x 252 = 756756
(b) We count the no. of possible ways that P1 and P2 are in the same team:
No. of possible ways that they are in the first team = 13C3 x 10C5 = 72072
Similarly, No. of possible ways that they are in the second and third team are 72072 for each.
So total no. of possible ways that P1 and P2 are in the same team = 72072 x 3 = 216216
Therefore, no. of possible ways that they are in different team = 756756 - 216216 = 540540
(c) Suppose P1 must be in the first team:
No. of possible ways that P2 in the second team = 13C4 x 10C5 = 180180
No. of possible ways that P2 in the third team is also 180180.
So total no. of possible ways = 360360.

give me numerical answer please