Combinatorics with percentage constraints Say I have a $K$-component alloy system, where the $K$ components are chosen from $N$ possible metal elements ($N>K$), and the concentration of each component always ranges from $a$% to $b$% (only consider integer percentages). Two alloys are considered to be different if any metal content differs by 1%. If we don't distinguish the arrangement of the components (i.e. the order doesn't matter), what is the total number of unique alloy compositions?
For now I only know if we assume the concentrations of all compoenents are always equal, the total number of unique alloy compositions is $\displaystyle{N \choose K} = \frac{N!}{(N-K)!K!}$. But how should we take into account the percentage constraint?
Edit: To make a component considered as "exists" in the alloy, obviously it has to be at least having a percentage of 1%, thus we always have $K−1 \le a \le b \le 100−K+1$.
To give you some idea: let's say $K=3,a=2,b=98$, the first component can take a percentage equal to $2,3,\ldots,98$, but when the first component takes a percentage of 98% for example, the other two components have to both take a percentage of 1%.
 A: Too long for a comment. I suspect that your question does not  have a nice closed form answer.
There are ${N}\choose{K}$ ways to choose the metals. Then you want to multiply that by the number of compositions of $100$ into $K$ parts, where the parts are restricted to the integers in $[a,b]$.
Start with the wikipedia page on compositions.
For the restriction to $[a,b]$ this article linked there might help:
Compositions of $n$ with parts in a set.
A: You want to find the number of ways to distribute $100$ percentage points into $N$ boxes, of which $N-K$ are empty and the remaining $K$ have between $a$ and $b$ percentage points.
First, fix $K$. We can use a generating function $G(z)$ to count the number of possibilities.
$$\tag{1}
G(z)=(z^{a}+z^{a+1}+\cdots+z^{b})^K=z^{aK}(1-z^{1-a+b})^K(1-z)^{-K}
$$
By appealing to the binomial theorem, we may write
$$\tag{2}
G(z)=z^{aK}\left[\sum\limits_{m=0}^K\left(\begin{matrix}K \\ m \end{matrix}\right) (-z^{1-a+b})^{m}\right]\left[\sum\limits_{n=0}^\infty \left(\begin{matrix}-K \\ n\end{matrix}\right) (-z)^n\right]
$$
The coefficient of $z^{100}$ in the Maclaurin expansion of $G$ corresponds to the number of possibilities $\Omega_K$. By setting $aK+m(1-a+b)+n=100$ we can collapse one sum to find
$$\tag{3}
\Omega_K=\sum\limits_{m=0}^M(-1)^{aK+m(b-a)} \left(\begin{matrix}K \\ m \end{matrix}\right) \left(\begin{matrix}-K \\ 100-aK-m(1-a+b)\end{matrix}\right)
$$
Where $M=\operatorname{floor}(\frac{100-aK}{1-a+b})$. The total number of combinations $\Omega(N,K)$ is then $\Omega_K$ weighted by the number of ways to choose $K$ boxes out of $N$, which is again a binomial coefficient.
$$\tag{4}
\Omega(N,K)=\left(\begin{matrix}N \\ K\end{matrix}\right)\Omega_K
$$
A: Only integer percentages are allowed to each component which must have a minimum of $a$% and maximum of $b$ %, and the %ages for the $K$ components must sum to $100$%, then the best way I see is to use generating functions.
Firstly, we put $a$ into each of the $K$ components, then $100-Ka$ are left to distribute
then  in the most rudimentary form of generating functions, we want to find the coefficient of $(100-Ka)$ in the expansion of $(1 + x + x^2 + .... + x^(b-a))^K$
and multiply by the $\binom{N}{K}$ varieties of components possible in the alloy.
PS
OP has remarked that the term for the base is called the "host" metal, and also clarified that  all percebtages are integers, and that they must add up to 100% So I have simplified the answer.
