We'll define a type-~i~ scene as one excluding actor ~i~ ~(1 \le i \le 3)~. Let ~T_i~ be the total number of type-~i~ scenes, and ~C_i~ be the number of type-~i~ scenes included in the chosen subset. A subset is valid if and only if the following hold:

• ~0 \le C_i \le T_i~, for each ~i~,
• ~C_1 < C_2~ (equivalent to ~S_1 > S_2~), and
• ~C_1 < C_3~ (equivalent to ~S_1 > S_3~).

The number of different subsets with a given set of ~C~ values is equal to the product ~\prod_{i=1}^3 \binom{T_i}{C_i}~. Before going further, let's consider how we can efficiently evaluate choose values in general (modulo a prime ~P~ such as ~1\,000\,000\,007~). We'll first let ~\text{ModInv}(v, p)~ be the modular inverse of ~v~ for prime modulus ~p~, which is equal to ~v^{p-2} \bmod p~.

~\text{ModInv}(v, p)~ may be evaluated in ~\mathcal O(\log p)~ time using either exponentiation by squaring or the extended Euclidean algorithm. Then, we have:

$$\displaystyle \binom n k \bmod P = \frac{n!}{k!(n-k)!} \bmod P = \left[(n! \bmod P) \times \text{ModInv}(k!, P) \times \text{ModInv}((n-k)!, P)\right] \bmod P$$

If we precompute ~x! \bmod P~ and ~\text{ModInv}(x!, P)~ for each ~x~ between ~0~ and ~N~, inclusive, then we'll be able to evaluate any necessary choose value in constant time. Now, let ~Ways[i][j] = \left[\sum_{k=j}^{T_i} \binom{T_i}{k} \right] \bmod P~ – in other words, the number of size-~j~-or-greater subsets of type-~i~ scenes.

Note that ~Ways[i][j] = \left[Ways[i][j+1]+\binom{T_i}{j}\right] \bmod P~, meaning that we can precompute ~Ways[i][j]~ for all possible pairs ~(i, j)~ in ~\mathcal O(N)~ by iterating downwards through the ~j~ values.

Let's consider fixing the number of type-~1~ scenes in the chosen subset. For a given ~C_1~, there are ~Ways[2][C_1+1]~ subsets of type-~2~ scenes such that ~C_2 > C_1~ is satisfied, and similarly ~Ways[3][C_1+1]~ subsets of type-~3~ scenes. Therefore, the total number of valid subsets which can be chosen with a given ~C_1~ is ~\left[\binom{T_1}{C_1} \times Ways[2][C_1+1] \times Ways[3][C_1+1]\right] \bmod P~, which can be added up over all ~\mathcal O(N)~ possible choices of ~C_1~ to yield the final answer.