To solve this problem, we need to calculate the number of ways to form a 4-member group from a total of 9 students (5 girls + 4 boys).

The formula for combinations, which gives the number of ways to choose a subset of $k$ elements from a set of $n$ elements, is given by:

$\text{Number of combinations} = \binom{n}{k} = \frac{n!}{k!(n-k)!}$

Here, $n = 9$ (total students) and $k = 4$ (group size). So, we have:

$\binom{9}{4} = \frac{9!}{4!(9-4)!} = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = \frac{3024}{24} = 126$

So, the correct answer is:

c. 126

Would you like more details or have any questions about this solution? Here are some additional related questions you might find useful:

1. How many 3-member groups can be made from the same set of 9 students?
2. What is the probability of randomly selecting a group of 4 students that includes exactly 2 girls?
3. How many ways can we form a group of 4 students with at least 1 girl?
4. How many different ways can we select 2 boys and 2 girls from the group?
5. If we add 1 more girl to the classroom, how many 4-member groups can be made?

Tip: When calculating combinations, the order of selection doesn't matter. This distinguishes combinations from permutations, where order is important.