Let's solve the two parts of this problem step by step:

Part 1: Total Number of Ways to Choose the Committee

The total number of people we can choose from is 19 (18 teachers + 1 principal). We need to choose a committee of 5 people from these 19.

The number of ways to choose 5 people from 19 is calculated using the combination formula:

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

where:

• $n = 19$ (total number of people),
• $r = 5$ (number of people to choose).

Thus, the total number of ways is:

$\binom{19}{5} = \frac{19!}{5!(19 - 5)!} = \frac{19!}{5!14!}$

Using a calculator:

$\binom{19}{5} = \frac{19 \times 18 \times 17 \times 16 \times 15}{5 \times 4 \times 3 \times 2 \times 1} = 11628$

So, there are 11,628 ways to choose the committee.

Part 2: Percentage of Committees Where the Principal Is Not Included

If the principal is not included, we are choosing all 5 people from only the 18 teachers. The number of ways to choose 5 people from 18 teachers is:

$\binom{18}{5} = \frac{18!}{5!(18 - 5)!} = \frac{18 \times 17 \times 16 \times 15 \times 14}{5 \times 4 \times 3 \times 2 \times 1} = 8568$

Now, to find the percentage of committees where the principal is not included:

$\text{Percentage} = \left( \frac{\binom{18}{5}}{\binom{19}{5}} \right) \times 100 = \left( \frac{8568}{11628} \right) \times 100 \approx 73.7\%$

Thus, approximately 73.7% of the committees will not include the principal.

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Would you like more details or have any questions? Here are 5 related questions:

1. How many ways can a committee of 4 teachers and the principal be chosen?
2. What if there are 20 teachers instead of 18—how does that affect the result?
3. How many ways can a committee of 3 teachers and 2 principals be chosen?
4. What is the probability that a randomly chosen committee includes exactly 2 teachers?
5. How many committees can be formed where at least one teacher is included?

Tip: The combination formula $\binom{n}{r}$ is useful whenever you are selecting objects without regard to order!