8-element permutations of a multiset {3:0,1:1,1:3,1:5,1:8,1:9} with the restriction 0 is not allowed in left or rightmost position

Question

I am lost in how to approach this problem due to the wording:

Count the number of distinct 8-digit numbers that may be made by permuting the multi-set:

$$MS:=\{0:3,1:1,3:1,5:1,8:1,9:1\}$$

Given that $0$ is not allowed in either the leftmost place or the rightmost place

If possible, could someone please attempt this problem?

Counting, specifically balls-in-bins, is driving me crazy.

Answer 1

We wish to count the number of permutations of the multiset $0, 0, 0, 1, 3, 5, 8, 9$ in which $0$ does not appear in the leftmost or rightmost position.

As stated in the comments, we may use each element in the multiset the number of times it appears in the multiset. Two permutations are distinguishable if there is at least one position they do not have the same digit in that position.

Strategy:

1. Choose three of the middle six positions for the zeros.
2. Arrange the remaining five distinct numbers in the remaining five positions.

We can choose three of the middle six positions for the zeros in $\binom{6}{3}$ ways and arrange the remaining five distinct numbers in the remaining five positions in $5!$ ways. Hence, the number of arrangements of the multiset in which a zero does not occupy the leftmost or rightmost position is $$\binom{6}{3}5!$$