Final answer:
There are 10 different ways for a teacher to split 6 students into two teams of three. This is calculated by using the combinations formula 6 choose 3, which gives 20, and then dividing by 2 because each team selection also determines the other team.
Step-by-step explanation:
The question asks in how many ways a teacher can split 6 students into two teams of three students each. This is a combinatorial problem, specifically a question of combinations without repetition. To solve this, we consider one team first. From six students, we choose three students to form the first team; this can be done in combination manners. The number of combinations is given by the binomial coefficient formula, which in this case is 6 choose 3:
C(6, 3) = \( \frac{6!}{3!(6-3)!} \) = \( \frac{6 \times 5 \times 4}{3 \times 2 \times 1} \) = 20
However, forming one team automatically forms the other team with the remaining students and these do not represent distinct groupings. Since each grouping is counted twice in this calculation, we divide by 2 to find the distinct groupings.
Total distinct groupings = \( \frac{20}{2} \) = 10.
Therefore, there are 10 different ways for the teacher to split the class into two teams of three.