Your answer to part a) is correct.
This is part b)
Now you do part c). Your answer is not correct.
I am just seeking confirmation to my answers. Thanks in advance!
Here is the question.
A company employs 8 persons in department A, 5 in department B and 3 in department C. A project team comprising of 6 people is required. How many ways can this team be assembled if:
a) From every department, 2 persons are picked.
b) A minimum of 2 are picked from department A and
c) A representative for every department is picked.
Am I on the right track or I veered off somewhere along the way?
Question c is a bit tricky for me because there is no number given. You are just told picking a representative for every department. It could be picking 3 for one department and 1 for the remaining departments or 2 for each or or ... I thought adding all the numbers (8+5+3) and choosing 6 from the total will be right.
There are seventeen employees in all, eight in department A and then nine not in department A.
If we choose only two from department A, , then we must choose four not in department A, .
You do that for each of ,
Now you add them up.
Part c) is difficult. There are nine combinations you have to calculate.
I want to say that the coefficient of in
gives the answer.
...r varies from 1 to 8
hence this expression contains
similarly for other terms of the product given above.
Now consider the first case of the table given by
It has the no. of ways
now this case is covered in the product by the multiplication of first term, of second and third term.
Hence, all the cases in the table are summed up in the coefficient of in the product.