SEE EDIT BELOW: Lets say that I have the set {$ 2,3,4,5,6,7 $} and that I want to add multiples $\geq0$ of these elements in different orders (sequences) to get $100$. How many such sequences are there?
At first I thought about the number of non-negative integer solutions to an equation on this form: $$x_1\ +\ x_2\ +\ ...\ +\ x_n\ =\ k$$ but doesn´t that just count the number of times each element occurs? Can I use this, and if so, how, and then somehow order them and count the number of sequences? How can I do this?
EDIT: Im sorry for the confusion but I need to correct some things, my bad. The goal is to partition the number $100$ into EXACTLY $15$ integers-parts where all these parts must be multiples of elements of the set {$2,3,4,5,6,7$}. I know that the number of partitions of $100$ into exactly $15$ parts is $p(100,15)$, but these include all the partitions using all the numbers from $1$ to $100$, right?
So is there a way to just get the partitions of the set above? Thank you for great answers so far even though I messed the question up!
Let $a_n$ be the number of sequences that add up to $n$. Then $a_n=0$ for $n<0$, and $a_0=1$. Conditioning on the first element of the sequence yields recurrence $$a_n = \sum_{i=2}^7 a_{n-i},$$ which you can readily use to compute $a_{100}$.
To obtain the generating function $A(z)=\sum_{n=0}^\infty a_n z^n$, multiply both sides of the recurrence by $z^n$ and add them up, yielding \begin{align} A(z) &= a_0 z^0 + \sum_{n=1}^\infty \sum_{i=2}^7 a_{n-i} z^n \\ &= 1+\sum_{n=1}^\infty \sum_{i=2}^{\min(7,n)} a_{n-i} z^n \\ &= 1+\sum_{i=2}^7 \sum_{n=i}^\infty a_{n-i} z^n \\ &= 1+\sum_{i=2}^7 z^i \sum_{n=i}^\infty a_{n-i} z^{n-i} \\ &= 1+\sum_{i=2}^7 z^i A(z) \\ &= 1+A(z) \frac{z^2-z^8}{1-z}, \end{align} so $$A(z) = \frac{1}{1-\frac{z^2-z^8}{1-z}} = \frac{1-z}{1-z-z^2+z^8}.$$