The problem:
A deck is missing a number of cards. It has S spades, C clubs, and E red cards, for a total of M cards. A side point is that C = S + 1. A hand of N cards is dealt.
A) What is the probability of the player receiving more clubs than spades?
B) What is the probability of the player NOT receiving more spades than clubs?
And now for what I believe to be the solution:
The probability of a hand of N cards containing R spades is
$$ \begin{pmatrix} S \\ R\end{pmatrix} / \begin{pmatrix} M \\ N\end{pmatrix} $$
Thus, the probability of a hand of N cards containing exactly R spades AND B clubs is in fact the probability of the hand containing R spades, B clubs, and D reds, where D = N - R - B, and so is equal to
$$ \begin{pmatrix} S \\ R\end{pmatrix} \begin{pmatrix} C \\ B\end{pmatrix} \begin{pmatrix} E \\ D\end{pmatrix}/ \begin{pmatrix} M \\ N\end{pmatrix}^3 $$
I would then need to sum it across all values of B and R where
A) B is greater than R
$$ \sum_{R=1}^S \sum_{B=R+1}^C (\begin{pmatrix} S \\ R\end{pmatrix} \begin{pmatrix} C \\ B\end{pmatrix} \begin{pmatrix} E \\ D\end{pmatrix}/ \begin{pmatrix} M \\ N\end{pmatrix}^3) $$
B) B is greater than or equal to R
$$ \sum_{R=1}^S \sum_{B=R}^C (\begin{pmatrix} S \\ R\end{pmatrix} \begin{pmatrix} C \\ B\end{pmatrix} \begin{pmatrix} E \\ D\end{pmatrix}/ \begin{pmatrix} M \\ N\end{pmatrix}^3) $$
All that would be left would be to make sure to select meaningful values of R and B (e.g. R + B cannot be greater than N).
Do you agree?
What you have here is a discrete uniform distribution over the set of possible hands. In such a distribution the probability of a set $S$ is the size of $S$ divided by the total number of hands $\binom{M}{N}$ (Not $\binom{M}{N}^3$!). The number of hands with exactly $B$ clubs and $R$ spades is $\binom{C}{B}\binom{S}{R}\binom{E}{N-B-R}$, and so you can indeed get the answer by summing over legal values of $R$ and $B$.
To summarize, your answers are almost correct - you just need to lose the power 3 in the denominator.