To solve this problem, we need to find the number of ways to select at least 4 members from a group of 7 members. We can approach this by calculating the combinations of selecting 4, 5, 6, and 7 members, then summing these values.

Step 1: Use the Combination Formula

The combination formula is given by:

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

where:

• $n$ is the total number of items.
• $r$ is the number of items to choose.

Step 2: Calculate Combinations for Each Case

1. Selecting 4 members:

   $\binom{7}{4} = \frac{7!}{4!(7-4)!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35$

2. Selecting 5 members:

   $\binom{7}{5} = \frac{7!}{5!(7-5)!} = \frac{7 \times 6}{2 \times 1} = 21$

3. Selecting 6 members:

   $\binom{7}{6} = \frac{7!}{6!(7-6)!} = \frac{7}{1} = 7$

4. Selecting 7 members:

   $\binom{7}{7} = \frac{7!}{7!(7-7)!} = 1$

Step 3: Sum All Combinations

$\text{Total} = \binom{7}{4} + \binom{7}{5} + \binom{7}{6} + \binom{7}{7} = 35 + 21 + 7 + 1 = 64$

Final Answer

There are 64 ways to select at least 4 members from the 7-member math club.

Would you like to see more details or have any questions? Here are some related questions to consider:

1. How would the answer change if the club had 8 members instead of 7?
2. What if the advisor wanted to select at least 3 members instead of 4?
3. How do combinations differ from permutations in this context?
4. Can you explain the role of factorials in the combination formula?
5. How would you compute combinations using a calculator?

Tip: When calculating combinations, remember that the order of selection does not matter, which is why the combination formula is used instead of permutations!