Let's solve the problem in parts based on the image:

Problem:

A committee of 4 is to be chosen from:

• 5 science teachers
• 3 language teachers
• 4 mathematics teachers

We are asked to calculate how many committees can be selected in three different scenarios:

a) No restrictions

We need to choose 4 members from a pool of $5 + 3 + 4 = 12$ teachers. The number of ways to choose 4 members from 12 is given by the combination formula:

$\text{Total Ways} = \binom{12}{4} = \frac{12!}{4!(12-4)!}$

Let me compute that.For part a, the number of ways to choose 4 committee members from 12 teachers with no restrictions is $\binom{12}{4} = 495$.

b) At least one from each group (science, language, and mathematics)

We need to ensure that the committee has:

• At least 1 science teacher
• At least 1 language teacher
• At least 1 mathematics teacher

This means we have to select 1 teacher from each of these groups, and the 4th member can come from any of the remaining teachers.

1. First, we choose 1 science teacher from 5: $\binom{5}{1}$
2. Then, we choose 1 language teacher from 3: $\binom{3}{1}$
3. Then, we choose 1 mathematics teacher from 4: $\binom{4}{1}$
4. For the 4th member, we choose from the remaining 9 teachers (those not yet chosen): $\binom{9}{1}$

The total number of ways for this case is the product of these combinations:

$\text{Total Ways} = \binom{5}{1} \times \binom{3}{1} \times \binom{4}{1} \times \binom{9}{1}$

Let me calculate this.For part b, the number of ways to choose the committee with at least one teacher from each group (science, language, and mathematics) is 540.

c) Two science teachers $X$ and $Y$ must be chosen, and language teacher $Z$ is unavailable

In this case:

• Two specific science teachers, $X$ and $Y$, must be on the committee. This leaves us with 2 remaining spots to fill.
• Language teacher $Z$ cannot be selected, so we can only choose from the 2 remaining language teachers.
• We can still choose from the remaining 3 science teachers, 2 language teachers (since $Z$ is unavailable), and 4 mathematics teachers.

Thus, we need to choose 2 more members from:

• 3 remaining science teachers
• 2 remaining language teachers
• 4 mathematics teachers

This gives a total of $3 + 2 + 4 = 9$ teachers to choose from, and we need to select 2.

$\text{Total Ways} = \binom{9}{2}$

Let me calculate that.For part c, the number of ways to select the committee when two specific science teachers $X$ and $Y$ must be chosen, and language teacher $Z$ is unavailable, is 36.

Summary of Results:

• a) No restrictions: $\mathbf{495}$
• b) At least one from each group: $\mathbf{540}$
• c) Two specific science teachers must be chosen, and a specific language teacher is unavailable: $\mathbf{36}$

Do you want more details on any of these solutions? Here are five related questions to further your understanding:

1. How does the combination formula differ from permutations in committee selections?
2. What happens if we are required to have exactly two mathematics teachers on the committee?
3. How would the total change if we only had 2 language teachers instead of 3?
4. What if a specific mathematics teacher must also be included in part c?
5. How does changing the size of the committee (e.g., from 4 to 5) affect the number of possible selections?

Tip: When dealing with selection problems, always start by determining the total pool of candidates and then apply any restrictions carefully to avoid miscounting.