Find the possible combinations

[Mentallic]

Homework Statement

This is something I want to know regarding a game series I am a big fan of.
If we are given a certain amount of points that can be distributed among a certain number of different skills, with the added restriction that each skill has a maximum number of points that can be put into it which is less than the total points we are given, then find the possible combinations.

Homework Equations

$$P(n,r)=\frac{n!}{(n-r)!}$$

$$C(n,r)=\frac{n!}{r!(n-r)!}$$

The Attempt at a Solution

I was able to find that if we ignore the maximum limit for each skill, then if
S = number of skills
P = number of points

The number of combinations C, should be equal to

$$C=\frac{(S+P-1)!}{P!(S-1)!} = C(S+P-1,P)$$

Now to add the limit in each skill to the mix, I'm not feeling comfortable about. Any ideas/hints?

 

[Xerxes1986]

$$C=\frac{(S+P-1)!}{P!(S-1)!}$$

Is exactly the answer if there were no limits. This problem reminds me of the Einstein solid where you have N oscillators (S skills) and q energy units (P points). You have S buckets and P balls to place in the buckets.

What I would do next is to calculate the number of ways you can Put (P - (P_max+1)) units of energy into (S - 1) skills. This is the situation where you have put P_max + 1 points into 1 bucket and are distributing the rest among the other buckets. Then put 1 more in the filled bucket and calculate the number of arrangements all the way up to 1 bucket having P skill points. This is assuming you can't completely fill 2 buckets (P < 2*P_max)

So, the number of states where you have overfilled 1 bucket is:

$$\sum^{P-P_m}_{n=1} \frac{(P-P_m-n+S-2)!}{(P-P_m-n)!(S-2)!}$$

You can probably use the same or a similar argument for P>2*p_max