To determine the number of different ways the instructor can choose 3 students from a class of 13, we use the concept of combinations. Combinations are calculated using the formula:

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

where:

• $n$ is the total number of students (13),
• $r$ is the number of students to be chosen (3).

Substituting the values:

$C(13, 3) = \frac{13!}{3!(13 - 3)!} = \frac{13!}{3! \cdot 10!}$

Simplify this expression by canceling out the $10!$ in the numerator and denominator:

$C(13, 3) = \frac{13 \cdot 12 \cdot 11}{3 \cdot 2 \cdot 1}$

Perform the calculations:

1. $13 \cdot 12 = 156$,
2. $156 \cdot 11 = 1716$,
3. $3 \cdot 2 \cdot 1 = 6$,
4. $\frac{1716}{6} = 286$.

Thus, the instructor can choose 3 students in 286 different ways.

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Do you want a detailed explanation or have any questions?

Here are 5 related questions for further exploration:

1. How would the answer change if the class had 20 students instead of 13?
2. What is the formula for permutations, and how does it differ from combinations?
3. What if the instructor also wanted to select a specific student as one of the 3? How many choices remain?
4. How would you calculate the number of ways to choose 2 groups of 3 students from the class of 13?
5. How does the concept of combinations apply in real-life scenarios beyond classroom settings?

Tip: Always verify whether the problem involves "combinations" (order doesn't matter) or "permutations" (order matters). This distinction is crucial!