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.