How many ways are there to select $3$ arrangements out of objects m,m,n,n,p?
I tried the partial permutation formula first $n!/(n-r)!$ and got 60, but then realized this had repeats in it. Such as counting $m,m$. So I then tried to account for $m,m$ by finding 3 ways to organize $m,m$ and multiplying this by $2!$ (What's left from n,n,p where since n,n repeats we count it as $2!$). I did the same with $n,n$ getting 6. Then I add this together and get a total of $12$ partial permutations. Any insights into of this is correct and of there's a better way when there are repeating elements. Thanks
I count six arrangements that use both $m$s. As you note, the "odd" letter can be either an $n$ or a $p$ (two choices) and it can go in any of three positions.
Similarly, there are six more arrangements that use both $n$s.
Finally, there are $3!=6$ arrangements of $m, n, p$ (the only way to avoid using a pair of letters), for a total of $18$.