How many ways can we get a number by addition if each part of the addition has to be smaller or equal to a set value? For example, if we need to get 5 with the largest number we can use being 3, we can use:


*

*3 + 2

*3 + 1 + 1

*2 + 2 + 1

*2 + 1 + 1 + 1

*1 + 1 + 1 + 1 + 1


Is there any way to find out the solution with any two numbers without calculating every one of them?
 A: We can calculate  the number of ways with the help of generating functions. We encode the usage of 


*

*zero or more $1$s as $1+x+x^2+x^3+\cdots=\frac{1}{1-x}$

*zero or more $2$s as $1+x^2+x^4+x^6+\cdots=\frac{1}{1-x^2}$

*zero of more $3$s as $1+x^3+x^6+x^9+\cdots=\frac{1}{1-x^3}$
and we look for the coefficient of $x^5$ denoted with $[x^5]$ of the product $\frac{1}{(1-x)(1-x^2)(1-x^5)}$.
This needs a little algebra, but we can keep it simple since we can skip powers greater than $5$ when multiplying out.

We obtain
  \begin{align*}
\color{blue}{[x^5]}&\color{blue}{\frac{1}{(1-x)(1-x^2)(1-x^3)}}\\
&=[x^5]\left(1+x+x^2+x^3+x^4+x^5\right)\left(1+x^2+x^4\right)\left(1+x^3\right)\tag{1}\\
&=[x^5]\left(1+x+x^2+x^3+x^4+x^5\right)\left(1+x^2+x^3+x^4+x^5\right)\tag{2}\\
&=[x^5]\left(1\cdot x^5+x\cdot x^4+x^2\cdot x^3+x^4\cdot x+x^5\cdot 1\right)\tag{3}\\
&=[x^5]5x^5\\
&\,\,\color{blue}{=5}
\end{align*}

Comment:


*

*In (1) we  expand $\frac{1}{1-x}$, $\frac{1}{1-x^2}$ and  $\frac{1}{1-x^3}$  but only up to powers raised to $x^5$ since higher powers do not contribute to $[x^5]$.

*In (2) we multiply out the two right-most terms again skipping powers greater than $5$.

*In (3) we multiply out skipping all factors which do not give $x^5$.
A: It is a simple function call.  In Mathematica, for example:
IntegerPartitions[5, {1, 5}, {1, 2, 3}]

(*
{{3, 2}, {3, 1, 1}, {2, 2, 1}, {2, 1, 1, 1}, {1, 1, 1, 1, 1}}
*)
If you merely seek the number of such partitions:
Length@IntegerPartitions[5, {1, 5}, {1, 2, 3}]

(* 5 *)
[This function asks for all integer partitions of the number 5, of length 1 through 5, using only the numbers 1, 2 and 3.  Then take the "length" of this list, i.e., the number of partitions.]
Another example:
IntegerPartitions[10, {1, 10}, {1, 2, 3}]

(*
{{3, 3, 3, 1}, {3, 3, 2, 2}, {3, 3, 2, 1, 1}, {3, 3, 1, 1, 1, 1}, {3, 2, 2, 2, 1}, {3, 2, 2, 1, 1, 1}, {3, 2, 1, 1, 1, 1, 1}, {3, 1, 1, 1, 1, 1, 1, 1}, {2, 2, 2, 2, 2}, {2, 2, 2, 2, 1, 1}, {2, 2, 2, 1, 1, 1, 1}, {2, 2, 1, 1, 1, 1, 1, 1}, {2, 1, 1, 1, 1, 1, 1, 1, 1}, {1, 1, 1, 1, 1, 1, 1, 1, 1, 1}}
*)
Length@IntegerPartitions[10, {1, 10}, {1, 2, 3}]

(* 14 *)
A: If you wish to find the number of ways to express a positive integer $n$ as the sum of $m$ positive integers ($m\le n$) and each of the positive integers must be less than or equal to a certain value $\alpha$ then the number of such integers will be the number of positive integral solutions to these equations:
$x_1+x_2=n$ 
$x_1+x_2+x_3=n$ 
$x_1+x_2+x_3+\cdots+x_m=n$
such that $1\le x_i \le \alpha$
for the equation $x_1+x_2=n$ 
the number of positive integral solutions is the same as the coefficient of $x^n$ in the expansion $(x+x^2+x^3+\cdots+x^\alpha)^2$
for the equation $x_1+x_2+x_3=n$ 
the number of positive integral solutions is the same as the coefficient of $x^n$ in the expansion $(x+x^2+x^3+\cdots+x^\alpha)^3$
for the equation $x_1+x_2+x_3+\cdots+x_m=n$
the number of positive integral solutions is the same as the coefficient of $x^n$ in the expansion $(x+x^2+x^3+\cdots+x^\alpha)^m$
So the total number of the solutions of all of these equations is the desired number.
This method of finding the coefficient in the expansion works because the term $x^n$ is formed only if in the multiplication, the powers of some terms get added to give the specified $n$.
For example in your question, $5$ has to be expressed as the sum of positive integers not greater than $3$
$x_1+x_2=5$
Coefficient of $x^5$ in the expansion of $(x+x^2+x^3)^2$
$=(x+x^2+x^3)(x+x^2+x^3)$
$=x^2+2x^3+3x^4+2x^5+x^6$
Thus,$[x^5]=2$
Hence, 5 can be expressed in 2 ways as the sum of 2 positive integers lesser than or equal 3, which are
$(2,3)$ and $(3,2)$ ( yes, this method gives us the ordered pairs)
You can do the same for rest of the equations.
In general as well, if you allow integers from a certain interval (not only positive) say, $ \alpha \le x_i \le \beta$ then the number of ways to express $n$ as sum of integers (order is considered) is
($[x^n]$ is the coefficient of $x^n$)
$$\sum_{m=2}^n [x^n]  (x^ \alpha+x^{\alpha+1}+x^{\alpha+2}+\cdots+x^ \beta)^m$$ 
