If every white marble is adjacent to at least one other white marble, then we have two options:
- We have two groups of two adjacent white marbles in our permutation
- We have one group of four adjacent white marbles in our permutation
where the second option is really a subset of the first one (two adjacent groups of two adjacent white marbles). So, in order to count all possibilities, we need to count the number of ways to put two groups of two white marbles in the gaps between our $n$ black marbles (the number of gaps here include the "gap" before the black marbles and after them). So we have $(n+1)$ gaps, in which we need to put our white marble groups. The number of ways to put the first group in gaps is $(n+1)$, and the number of ways to put the second group is $(n+2)$ (as we have a new gap after adding a new element to our permutation). The total number of ways is $\frac{(n+1)(n+2)}2$, because there is no distinction between the order, in which we place the first and the second group.
Answer: $\frac{(n+1)(n+2)}2$