1. First, calculate $15C_{10}$ using the formula for combinations: $$15C_{10} = \frac{15 \times 14 \times 13 \times 12 \times 11 \times 10!}{10! \times 5 \times 4 \times 3 \times 2 \times 1} = 3003$$ 2. Consider the problem involving 3 sisters treated as a single unit: - If the 3 sisters are together in the team, treat them as 1 unit. - Total players excluding sisters is $15 - 3 = 12$. - Choose the remaining 7 players from the 12: $$12C_7 = \frac{12 \times 11 \times 10 \times 9 \times 8 \times 7!}{7! \times 5 \times 4 \times 3 \times 2 \times 1} = 792$$ 3. Now consider the team with none of the sisters: - All sisters are excluded, so choose all 10 players from the remaining 12. - Calculate $$12C_{10} = \frac{12 \times 11}{2 \times 1} = 66$$ 4. Total number of teams fulfilling the sister conditions: $$792 + 66 = 858$$ 5. Next, for the condition of forming numbers: - Numbers must be even and cannot start with zero. - Possible last digits (even numbers) are 0, 2, 4, 6, 8. 6. For the case (i) where the last digit is 0: - This part likely leads to counting how many numbers meet these criteria; problem context is given but further calculation depends on additional info. Final answers: - $15C_{10} = 3003$ - Total teams with sister conditions = $858$ The problem about even numbers with restrictions is partially stated without a full request for solution at this stage.