7 people
5 Doctors taken 2 at a time
C(5,2) = 5! / [ 2! (5-2)! ] = (5*4* 3) / 3*2 = 10 combinations of doctors
4 nurses, 2 at a time
C(4,2) = 4!/ [ 2! * 2!]= 4*3*2 / 4 = 6
6*10 = 60
How many possible combinations of doctors and nurses could be chosen for the committee????
5 Doctors taken 2 at a time
C(5,2) = 5! / [ 2! (5-2)! ] = (5*4* 3) / 3*2 = 10 combinations of doctors
4 nurses, 2 at a time
C(4,2) = 4!/ [ 2! * 2!]= 4*3*2 / 4 = 6
6*10 = 60
nCr = n! / r!(n - r)!
Where n is the total number of items, r is the number of items to be selected, and "!" denotes the factorial operation.
For selecting 2 doctors from 5, the calculation would be:
5C2 = 5! / 2!(5-2)!
= (5 * 4 * 3!) / (2 * 1 * 3!)
= (5 * 4) / (2 * 1)
= 10
For selecting 2 nurses from 4, the calculation would be:
4C2 = 4! / 2!(4-2)!
= (4 * 3 * 2!) / (2 * 1 * 2!)
= (4 * 3) / (2 * 1)
= 6
To find the total number of possible combinations, we can multiply the two results:
Total combinations = 10 * 6
= 60
Therefore, there are 60 possible combinations of doctors and nurses that could be chosen for the committee.
The number of combinations of selecting k items from a set of n items is given by the formula:
C(n, k) = n! / (k!(n-k)!),
where n! represents the factorial of n, which is the product of all positive integers less than or equal to n.
In this case, we need to select 2 doctors from a group of 5 doctors, and 2 nurses from a group of 4 nurses.
So, the number of possible combinations can be calculated as follows:
C(5, 2) * C(4, 2) = (5! / (2!(5-2)!) * (4! / (2!(4-2)!))
= (5! / (2!3!)) * (4! / (2!2!))
= (5 * 4 * 3!) / (2!3!) * (4 * 3 * 2!) / (2!2!)
= (5 * 4) / 2! * (4 * 3) / 2!
= (20 / 2) * (12 / 2)
= 10 * 6
= 60.
Therefore, there are 60 possible combinations of doctors and nurses that could be chosen for the committee.