Order matters, because you can think of the n workers all lined up
waiting to be assigned a type of work, and it certainly matters
in what order you assign the types of work in.  So we know to use 
permutations, not combinations.

There are k = P(k,1) ways to assign a type of work to worker #1.

For each of those k ways to assign a type of job to worker #1, 
there remain k-1 types of work to assign to worker #2.

So there are k(k-1) = P(k,2) ways to assign a type of work to each of 
workers 1 and 2.

For each of those k(k-1) = P(k,2) ways to assign a type of job to workers
#1 and #2, there remain k-2 types of work to assign to worker 3.

So there are k(k-1)(k-2) = P(k,3) ways to assign a type of work to each of 
workers 1, 2 and 3.

...

So when we get down to n workers, we'll have

k(k-1)(k-2)···[x-(n-1)] = P(k,n).

That's the answer.

Edwin