We are tasked with finding how many ways we can select 4 students for a committee from a group of 9 students, where there are 4 nurses and 5 doctors, and the committee must have at least 3 nurses.

Case Breakdown:

We need to count how many ways we can select the committee of 4 students, ensuring there are at least 3 nurses. This means we have two cases to consider:

1. Case 1: The committee has 3 nurses and 1 doctor.
2. Case 2: The committee has 4 nurses and 0 doctors.

Step 1: Case 1 – 3 Nurses and 1 Doctor

• To select 3 nurses out of the 4 nurses, the number of ways is given by the combination formula: $\binom{4}{3} = 4$
• To select 1 doctor out of the 5 doctors, the number of ways is: $\binom{5}{1} = 5$
• Therefore, the number of selections in Case 1 is: $\binom{4}{3} \times \binom{5}{1} = 4 \times 5 = 20$

Step 2: Case 2 – 4 Nurses and 0 Doctors

• To select 4 nurses out of the 4 nurses, the number of ways is: $\binom{4}{4} = 1$
• Since no doctors are selected, there are no further choices for doctors. Thus, the number of selections in Case 2 is: $\binom{4}{4} \times \binom{5}{0} = 1 \times 1 = 1$

Step 3: Total Selections

Now, we add the results from both cases: $\text{Total selections} = 20 + 1 = 21$

Conclusion:

There are 21 ways to form a committee of 4 students with at least 3 nurses.

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Would you like to go over any specific details? Here are some additional questions to explore:

1. What is the formula for combinations and how does it differ from permutations?
2. How would the solution change if the committee had to consist of at least 3 doctors instead?
3. How do you calculate combinations when items are indistinguishable (like when selecting nurses and doctors)?
4. If the committee had to have exactly 2 nurses and 2 doctors, how many selections would there be?
5. Can you explain the difference between combinations and subsets in set theory?

Tip: Always make sure to check the constraints (like "at least 3 nurses") before breaking down the problem into cases.