Since the order of selection is not important, we are dealing with a combination problem. The formula for combinations is:

$C(n, r) = \frac{n!}{r!(n - r)!}$

In this case:

• $n = 30$ (the total number of students),
• $r = 2$ (the number of students to be chosen).

Substitute these values into the formula:

$C(30, 2) = \frac{30!}{2!(30 - 2)!} = \frac{30!}{2! \cdot 28!}$

Simplify this expression by canceling $28!$ from the numerator and denominator:

$C(30, 2) = \frac{30 \times 29}{2 \times 1} = \frac{870}{2} = 435$

So, there are 435 ways to choose 2 students from a class of 30.

Answer: 435 ways

Would you like a further breakdown or explanation?

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1. How would the calculation change if we had to select 3 students instead of 2?
2. How many ways can we select a group of 5 students from the same class?
3. What if the order of selection mattered? How many ways could 2 students be chosen then?
4. Can you calculate the number of ways to choose 2 students from a class of 50?
5. How would the approach differ if we had 10 students and needed to form a team of 4?

Tip: When the order of selection doesn't matter, always use combinations, not permutations.