Question

Problem 2 Six different jobs need to be done today at the restaurant where you work. You have a pool of nine available workers. If you can assign only one worker to each job and no more than one job to a worker, how many different worker/job assignments are possible

Answer 1

Answer:   60480

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Explanation:

The order matters because the jobs are different. This means we'll use a permutation.

We have n = 9 workers to pick from and r = 6 jobs to fill.

Plug those values into the nPr permutation formula.

$n P r = \frac{n!}{ (n-r)! }\\\\9 P 6 = \frac{9!}{ (9-6)! }\\\\9 P 6 = \frac{9!}{ 3! }\\\\9 P 6 = \frac{9*8*7*6*5*4*3*2*1}{ 3*2*1 }\\\\9 P 6 = 9*8*7*6*5*4 \ \ \text{.... Note: The 3*2*1 cancels}\\\\9 P 6 = 60480$

There are 60480 ways to assign the 9 workers to the 6 different jobs where order matters.

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Another approach:

Label the 6 jobs as A,B,C,D,E,F

For job A, we have 9 workers to pick from.

For job B, we have 8 workers to pick from. This is because it states that there is "no more than one job to a worker".

For job C, we have 7 workers to pick from. And so on. We have this countdown going on. We stop once we reach job F.

We end up with 9*8*7*6*5*4 = 60480 different permutations

The string "9*8*7*6*5*4" is found in the second to last step of the section above.