From a panel of 4 doctors, 4 officers, and one doctor who is an officer, how many committees of 3 can be made if it contains at least one doctor and one officer?





The total number of committees of 3 that can be formed from the given panel of 9 people is:

9C3 = (9 * 8 * 7) / (3 * 2 * 1) = 84

Therefore, there are 84 possible committees of 3 that can be formed from the panel of 9 people.



The number of committees without a doctor can be found by selecting 3 officers from the panel of 5 officers (excluding the doctor who is also an officer). This can be calculated as:

5C3 = (5 * 4 * 3) / (3 * 2 * 1) = 10

Therefore, there are 10 committees of 3 that can be formed without a doctor.



The number of committees without an officer can be found by selecting 3 doctors from the panel of 4 doctors. This can be calculated as:

4C3 = (4 * 3 * 2) / (3 * 2 * 1) = 4

Therefore, there are 4 committees of 3 that can be formed without an officer.



The number of committees with both a doctor and an officer can be found by subtracting the number of committees without a doctor or without an officer from the total number of committees. This can be calculated as:

84 - 10 - 4 = 70

Therefore, there are 70 committees of 3 that can be formed with both a doctor and an officer.



Thus, the number of committees of 3 that can be formed from the panel of 9 people containing at least one doctor and one officer is 70.