To solve this problem, we need to consider the different combinations of students that include at least one Chinese and one European student. We will use combinations to calculate the number of ways to choose students.

Let C, E, and A represent the number of Chinese, European, and American students chosen, respectively. We have the following cases:

1. C = 1, E = 1, A = 2: \(7 \times 6 \times \binom{4}{2} = 252\)

2. C = 1, E = 2, A = 1: \(7 \times \binom{6}{2} \times 4 = 420\)

3. C = 1, E = 3, A = 0: \(7 \times \binom{6}{3} \times 1 = 140\)

4. C = 2, E = 1, A = 1: \(\binom{7}{2} \times 6 \times 4 = 504\)

5. C = 2, E = 2, A = 0: \(\binom{7}{2} \times \binom{6}{2} \times 1 = 315\)

6. C = 3, E = 1, A = 0: \(\binom{7}{3} \times 6 \times 1 = 210\)

Adding these cases together gives the total number of ways:

\(252 + 420 + 140 + 504 + 315 + 210 = 1841\)