A minimum of 3 voting members is required at all meetings of some association. If there is a total of 9 voting members in the group, find the number of ways this group can be formed.

To solve this problem, we need to find the number of ways to choose at least 3 voting members out of a total of 9. This is a combination problem, where the order of selection does not matter.

### Step-by-Step Solution:

1. **Number of Voting Members:** We have a total of \( n = 9 \) voting members.
2. **Minimum Required Members:** We need to select at least 3 members, so the possible group sizes can be 3, 4, 5, ..., 9.
3. **Using Combinations:** The number of ways to choose \( r \) members from \( n \) members is given by the binomial coefficient:
   \[
   \binom{n}{r} = \frac{n!}{r! \cdot (n-r)!}.
   \]
4. **Total Number of Ways:** We need to sum up the combinations for all group sizes from 3 to 9:
   \[
   \sum_{r=3}^{9} \binom{9}{r}.
   \]

   Let's compute each of these:

   - \( \binom{9}{3} = \frac{9!}{3! \cdot (9-3)!} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84 \)
   - \( \binom{9}{4} = \frac{9!}{4! \cdot (9-4)!} = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 126 \)
   - \( \binom{9}{5} = \frac{9!}{5! \cdot (9-5)!} = \frac{9 \times 8 \times 7 \times 6 \times 5}{5 \times 4 \times 3 \times 2 \times 1} = 126 \)
   - \( \binom{9}{6} = \frac{9!}{6! \cdot (9-6)!} = \frac{9 \times 8 \times 7 \times 6 \times 5 \times 4}{6 \times 5 \times 4 \times 3 \times 2 \times 1} = 84 \)
   - \( \binom{9}{7} = \frac{9!}{7! \cdot (9-7)!} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 36 \)
   - \( \binom{9}{8} = \frac{9!}{8! \cdot (9-8)!} = \frac{9 \times 8}{2 \times 1} = 9 \)
   - \( \binom{9}{9} = \frac{9!}{9! \cdot (9-9)!} = 1 \)

5. **Sum the Combinations:** Add all these values together to find the total number of ways:
   \[
   84 + 126 + 126 + 84 + 36 + 9 + 1 = 466.
   \]

### Final Answer:
There are **466** ways to form a group of at least 3 voting members from a total of 9 members.

Would you like further details or have any questions about this solution?

### Related Questions:
1. How would the solution change if the group size required at least 4 members?
2. What if the group size could not exceed 5 members?
3. How many ways can you select exactly 3 members from 9 members?
4. If there are 10 members instead of 9, how many ways can you form a group of at least 3?
5. How does the formula for combinations work in different scenarios like selecting groups with conditions?

### Tip:
For combination problems, always ensure to identify whether the problem involves combinations or permutations by checking if the order of selection matters.

Explore how to calculate the number of ways to form a group of at least 3 voting members from a total of 9 using combinations. This problem highlights the use of the combination formula, covering concepts from combinatorics and binomial coefficients. Ideal for students in Grades 10-12.

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