Seven boys and three girls are playing basketball. I how many different ways can they make two teams of five players so that both teams have at least one girl?
1. **Total number of ways to choose 10 players into two teams of 5:**
   The total number of ways to divide 10 players into two teams of 5 is given by:
   \[
   \binom{10}{5} = \frac{10!}{5!5!} = 252
   \]
   However, since the teams are indistinguishable (i.e., Team A vs. Team B is the same as Team B vs. Team A), we need to divide by 2:
   \[
   \frac{252}{2} = 126
   \]

2. **Calculate the number of ways to divide the players such that both teams have at least one girl:**
   - **Case 1: One girl in one team and two girls in the other team:**
     - Choose 1 girl out of 3 for the first team:
       \[
       \binom{3}{1} = 3
       \]
     - Choose 2 girls out of the remaining 2 for the second team:
       \[
       \binom{2}{2} = 1
       \]
     - Choose 4 boys out of 7 for the first team:
       \[
       \binom{7}{4} = 35
       \]
     - The remaining 3 boys automatically go to the second team.
     - Total for this case:
       \[
       3 \times 1 \times 35 = 105
       \]

3. **Summing up the cases:**
   Since there is only one valid case where both teams have at least one girl, the total number of ways is:
   \[
   105
   \]

The final answer is \(\boxed{105}\).