Thread: Combinations

Hi there, I have done the following questions, but I am not sure my logic is strong enough to give me confidence in these answers.



A small manufacturing company will start operating a night shift. There are 20 machinists employed by the company.

a. If a night crew consists of 3 machinists, how many different crews are possible?



b. If the machinists are ranked 1,2,…,20 in order of competence, how many of these crews would not have the best machinist?

My thought here is to find how many crews have the best machinist and subtract that from the previous answer. There are 3 spots to fill so given one spot has the best machinist therefore 19 x 18 = 342 crews have the best machinist.

Finally 1140 – 342 = 798 crews do not have the best machinist.


c. How many of the crews would have at least 1 of the 10 best machinists?

Here for “at least one” means find the sum of crews that have one, two and three of the best 10 machinists.

For crews with one of the best 10 machinists 10 x 10 x 9 = 900

For crews with one of the best 10 machinists 10 x 10 x 9 = 900

For crews with one of the best 10 machinists 10 x 9 x 8 = 720

But the sum of these are greater than 1140


d. If one of these crews is selected at random to work on a particular night, what is the probability that the best machinist will not work that night?

Not too sure how to start this one, maybe if I knew the method to the previous part it would be clearer.

I agree with your approach in part a.

Think of part b as asking how many 3 man crews there are if you choose from the 19 other mechanics.

Part c is more easily answered by asking the question - how many crews are there that have none of the top 10 mechanics, i.e., 10 choose 3.

Part d should just be part b divided by part a.

I understand your logic, thanks for the reply.

I have made this problem harder than it is.