Solution:

We observe that all the operations commute with one another, and that applying an operation twice is the same as not applying it. Hence we can associate to each operation an ON/OFF button (corresponding to whether or not that given operation has been carried out), and to each sequence of operations one and only one state of the buttons.

The combinations obtainable are then at most as many as all the possible states of the buttons, which are $2^{7}$.

Suppose now, for contradiction, that there are two different states leading the 8 cells to contain the same numbers. This means that there exists a state of the buttons (not all set to OFF) whose result is to transform the combination 00000000 into itself.

Suppose that operation $T$ has been carried out. Then necessarily $B_{1}$ and $B_{6}$ have also been carried out (in order for the first and last cell to be 0). This means that $B_{2}$ and $B_{5}$ are OFF, otherwise the second and second-to-last cell would have a 1; for the same reason $B_{3}$ and $B_{4}$ are also OFF, but this would mean that cells 4 and 5 contain a 1, which is impossible.

Hence operation $T$ has not been carried out, so neither have $B_{1}$ and $B_{6}$, otherwise the outer cells would have a 1. But then $B_{2}$ and $B_{5}$ are also OFF; similarly $B_{3}$ and $B_{4}$ must not have been carried out either, which is absurd because we had assumed that at least one operation had been performed.

Therefore each state of the buttons corresponds to a different combination, hence the reachable combinations are exactly $2^{7}$.