# A Basic Permutations Problem the order matters or doesn't confusion

I just decided to refresh my basic combinatorics problem-solving skills and picked up this fun book by Hollos "Combinatorics Problems and Solutions." I am having a severe brain freeze over the solution to this problem, my question is, why is this a permutation?

There are 10 jobs that need to be filled, 4 of which must be filled by men, 3 by women, and the remaining 3 by either. If there are 20 men and 6 women available, in how many ways can we fill the jobs?

The solution says:"The 20 men can fill the 4 jobs in $\frac{20!}{(20-4)!}$ ways, the 6 women can fill the 3 jobs in $\frac{6!}{(6-3)!}$ ways. There are 16 men and 3 women still available to fill the 3 remaining jobs, which can be done in $\frac{19!}{(19-3)!}$ ways. So the total number of ways to fill the 10 jobs is $$\frac{20!}{(20-4)!} \times \frac{6!}{(6-3)!} \times \frac{19!}{(19-3)!}$$

So my question is, but isn't the order irrelevant here, i.e. shouldn't we use combinations instead of of permutations because we'd want to avoid overcounting and divide, for example, 20! men by 16! and also 4! to avoid overcounting?

• If the jobs are distinguishable, then it matters who receives which job. – N. F. Taussig Feb 14 '18 at 20:50