How many ways to make a score of N? In a game, there are three possible shots making points 3,5, and 10 respectively.
A player has scored exactly N points. What are the scoring sequences possible? This was the original problem I was given.
For instance, for N = 15, the four possible scoring sequences are:


*

*3,3,3,3,3

*5,5,5

*10,5

*5,10


This is a programming question and I could come up with a solution to figure out the exact sequences(using dynamic programming). The trouble I am having is with estimating the time complexity of this approach. More precisely, I am trying to figure out the number of solutions possible for a given N. For example, for N=15, the number of scoring sequences is 4. How can we mathematically come up with a bound for the number of scoring sequences of N if the three possible scores are 3, 5, and 10? 
 A: You can define your function recursively:
$$f(n)=f(n-3)+f(n-5)+f(n-10),\\\text{for all }n>0$$
With initial conditions:
\begin{align}
f(n)&=0,&\text{for all }n<0\\
f(0)&=1
\end{align}
(There's only one sequence that gets us $0$ points: the empty sequence. Note that $n=0$ does not satisfy the recurrence relation.)
Proof for the recurrence relation: there are $f(n-3)$ sequences that begin with $3$, $f(n-5)$ sequences that begin with $5$, and $f(n-10)$ sequences that begin with $10$.
A: If you are just interested in the number of ways without giving importance to the order, you would essentially need the number of non-negative integer solutions to
$$3x+5y+10z = N$$
Equivalently, we would need the coefficient of $a^N$ in
$$\dfrac1{1-a^3}\cdot\dfrac1{1-a^5}\cdot\dfrac1{1-a^{10}} = \sum_{k=0}^{\infty} c(k)a^k$$
A: If the order of the terms didn't matter, you could use ordinary generating functions to answer this question.  However, since the ordering matters, you need exponential generating functions.  The coefficient of $\frac{t^m}{m!}x^n$ in
$$f(x,t) = \exp(t(x^3+x^5+x^{10}))$$
gives the number of sequences of $m$ shots whose total score is $n$.  Put another way, expand
$$f(x,t) = \sum_{n\ge0} p_n(t) x^n$$
where $p_n(t)$ is a polynomial in $t$. Then replace each $t^m$ in $p_n(t)$ by $m!$ to get the total number of sequences of shots, of all possible lengths, whose total score is $n$.
As an example, if you ask Wolfram Alpha for the Taylor series of $f(x,t)$, you get
$$1+t x^3+t x^5+(t^2 x^6)/2+t^2 x^8+(t^3 x^9)/6+1/2\, t (t+2) x^{10}+(t^3 x^{11})/2+(t^4 x^{12})/24+1/2\, t^2 (t+2) x^{13}+(t^4 x^{14})/6+1/120\, t^2 (t^3+20 t+120) x^{15}+ O(x^{16}).$$
To recover the count for $n=15$, we have
$$p_{15}(t)=1/120\, t^2(t^3+20t+120) = (t^5+20t^3+120t^2)/120.$$
Replacing $t^m$ with $m!$ gives $(120 + 20(6) + 120(2))/120 = 4$. (In fact, breaking this apart, this tells you there's one sequence with $5$ terms, one sequence with $3$ terms, and two sequences with $2$ terms.)
