Question

Eight machinists are currently registered with a temp agency. The agency has four different openings for machinists. Exactly one person can take each opening, but, since the openings are on different days, it is possible for a single person to take more than one opening.

a. The temp agency decides it will just choose four machinists to receive an opening without deciding which machinist will take which opening. How many possible ways can the agency make this selection?

b. As it turns out, every machinist wanted the same opening. Now the agency needs to assign a specific machinist to each specific opening, but they will not decide how many openings a single person can take. (For example, one person could get every opening or there could be a different person for each opening.) How many possible ways can the agency make this selection?

c. After widespread complaints, the agency now has a policy that each machinist can receive no more than one opening. Now how many ways can the agency assign machinists to openings?

d. Finally, the agency will assign one machinist to two openings, and every other opening must be assigned to a different machinist. How many ways can the agency make this selection?

Answer 1

a) The number of machinists=8, number of openings=4.

if we simplify the question it becomes: how many ways we can choose 4 out of 8 people. So the answer is  

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b) since it is given that one person could get every opening or there could be a different person for each opening so we can select 1 out of 8, 2 out of 8, 3 out of 8 or 4 out of 8 persons. So the number of ways is
(0) + () +3) + (%) = 162

c) since we have 4 openings and a machinist can have at most 1 opening so we need 4 machinists at first which can be done in

ways.

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Now we can assign them in 4 openings in 4! =24 ways. So the answer is  $24\times 70=1680$ ways.

d) since the agency will assign one machinist to two openings, and every other opening must be assigned to a different machinist we need total 3 machinists for 4 openings.

so we can choose 3 out of 8 people in $\binom{8}{3}$ ways. Then out of these 3 machinists 1 can be selected in $\binom{3}{1}$ ways who will be assigned for 2 openings. Now the other 2 machinists can be assigned to 2 remaining openings in 2! ways.

So total number of ways is = $\binom{8}{3}\times\binom{3}{1}\times 2! = 336$ .

Thank you.