given that zero is not allowed in either the leftmost place or the rightmost place.

The given multiset is

MSJ{0:3,1:1,3:1,5:1,8:1,9:1}

or alphabet

S={0,0,0,1,3,5,8,9}

I started the problem by dividing the S into two subsets

S={0,0,0,1,3,5,8,9} cardinality = 8

F={1,3,5,8,9} cardinality = 5

Now, my professor is a stickler for showing proof when showing work. Absolutely no English words allowed in the proof.

I got Ocp Îµ SxF {a1,a2,a3,a4,a5,a6,a7,a8|F Ocp a1, F Ocp a8, S Ocp a2…7}

=(F^2)(S^6)/3!

=(5^2)(8^6)/3! .

Does the logic of this proof track?

Also, Ocp = occupies.