To determine how many different groups of 3 trainees can be selected from 11 applicants, we use the combination formula, which is used when the order of selection does not matter. The formula for combinations is:

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

Where:

• $n$ is the total number of applicants (11 in this case),
• $r$ is the number of trainees to be selected (3 here),
• $!$ denotes factorial, which is the product of all positive integers up to that number.

Let's plug in the values:

$C(11, 3) = \frac{11!}{3!(11 - 3)!} = \frac{11!}{3! \cdot 8!}$

We can simplify this since the $8!$ in both the numerator and denominator cancel out:

$C(11, 3) = \frac{11 \times 10 \times 9}{3!} = \frac{11 \times 10 \times 9}{3 \times 2 \times 1}$

Now, calculate the numerator and denominator:

$C(11, 3) = \frac{990}{6} = 165$

So, there are 165 different groups of trainees that can be selected.

Would you like any more details or further clarifications?

Here are 5 related questions:

1. How does the combination formula change when the order of selection matters?
2. What is the difference between combinations and permutations?
3. How would you calculate the number of ways to select 4 applicants out of 11?
4. What are some real-world scenarios where you would use combinations?
5. Can you calculate the number of ways to select a committee of 5 people from 12 candidates?

Tip:

When solving problems involving combinations or permutations, always remember to carefully identify whether the order matters (permutations) or not (combinations).