The problem involves choosing 6 students from a class of 22 students. Since the order of selection doesn't matter, we are dealing with a combination.

The formula for combinations is:

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

where:

• $n$ is the total number of students (22),
• $r$ is the number of students to be chosen (6).

So, we need to calculate $C(22, 6)$:

$C(22, 6) = \frac{22!}{6!(22 - 6)!} = \frac{22!}{6!16!}$

We can simplify this by calculating only the relevant terms from $22!$ and $16!$:

$C(22, 6) = \frac{22 \times 21 \times 20 \times 19 \times 18 \times 17}{6 \times 5 \times 4 \times 3 \times 2 \times 1}$

Now, let's compute the value:

$C(22, 6) = \frac{22 \times 21 \times 20 \times 19 \times 18 \times 17}{720}$

I'll calculate this.The number of different ways the instructor can choose 6 students from a class of 22 students is $74,613$.

Would you like a detailed breakdown of the steps?

Here are 5 related questions for practice:

1. How many ways can the instructor choose 5 students from a class of 30?
2. How many ways can a team of 4 students be selected from a group of 15?
3. What if the instructor wants to select both a group leader and 5 students from a class of 22? How many ways are there to do this?
4. How many ways can the instructor choose 6 students from 22 if order matters?
5. In how many ways can 7 students be chosen from a class of 25?

Tip: In combinations, the order of selection does not matter, but in permutations, the order does!