Suppose that $A, B$ and $C$ are the game abilities, being $C$ the special one. As $C$ must be upgraded in level 6 or after, the levels in which this is possible are 6, 7, 8, 9, 10, 11, 12, 13, 14 and 15 (it is impossible in level 16 because otherwise it would be impossible to upgrade $C$ for the second time), so there are 10 possibilities. If $C$ is upgraded in level 6, 7, 8, 9 or 10, there are 6 possibilities to upgrade the ability for the second time. But if it is upgraded in level 11, 12, 13, 14 or 15, there are 5, 4, 3, 2 and 1 possibilities respectively. So there are $5\cdot6+5+4+3+2+1 = 45$ different ways to upgrade $C$ exactly two times. In the other levels there are three possible choices: upgrade $A$, upgrade $B$ or don't upgrade any ability. As there are fourteen levels in which this decision must be made, there are $3^{14}$ ways to upgrade abilities $A$ and $B$. So the answer to the problem is $45\cdot3^{14} = 5\cdot3^{16}$.

Fix two valid positions for $S$. The total number of ways to do this is $45$. The number of ways to place $A$ or $B$ is invariant of the positions for $S$. For that, this number is the binary strings possible with length $16-2=14$. Then we get $2^{14}$ configurations. Thus the total is $45 \cdot 2^{14}$.

For the first 5 levels , You have 2 options - Either one ability or none (2C1)^5 ways For the next 5 levels , You have 3 options - 1st ability or 2nd ability or none (3C1)^5 ways For the next 5 levels : Fix the special ability in ay 2 levels : (5C2)=20 ways then you get 3 options -1st or 2nd or none in other 4 levels: ---->(3C1)^3 ways Thus the final answer is (2^7)*(3^8)*5 ways

We will denote $\alpha, \beta, \zeta$ the three options, with $\zeta$ being the special one. $\alpha$ and $\beta$ make up $14$ of the choices, giving $2^14.$ We multiply this factor by the number of ways to choose positions for the two $\zeta.$ The first is at least $6,$ second at least $11.$ There are $6\cdot 5+\binom{6}{2}=45$ ways to do this, thus the answer is $2^14\cdot 45.$