Problem 1: How many ways are there to make a committee of 15 board members?

To solve this, you need to choose 15 members from a total of 12 junior board members and 18 senior board members. The total number of ways to do this is given by the combination formula:

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

Where:

• $n$ is the total number of members (12 juniors + 18 seniors = 30),
• $k$ is the number of members to select (15).

The total number of ways to select 15 members from 30 is:

$\binom{30}{15} = \frac{30!}{15!(30 - 15)!}$

Problem 2: How many ways are there to make a committee of 15 board members if exactly 6 must be juniors?

Here, you must select 6 junior board members from the 12 juniors, and the remaining 9 members from the 18 senior board members. This involves two separate combinations:

1. Choose 6 juniors from 12: $\binom{12}{6} = \frac{12!}{6!(12 - 6)!}$
2. Choose 9 seniors from 18: $\binom{18}{9} = \frac{18!}{9!(18 - 9)!}$

The total number of ways to form the committee is the product of these two combinations: $\binom{12}{6} \times \binom{18}{9}$

I can calculate these values for you if you'd like the exact number.

Would you like me to compute the final results?

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Questions to expand understanding:

1. How does the combination formula change if we add restrictions like gender or rank?
2. How does this problem change if we had to form a committee with at least 6 juniors?
3. How many ways are there to select exactly 10 senior board members from the 18?
4. What if the committee size was 20 instead of 15?
5. How would the formula change if we had to select all the juniors and seniors?

Tip: When dealing with large numbers in combinatorics, factorials can grow very quickly, so using software or calculators is recommended for precise results.