I will assume that we are looking for the number of ordered $6$-tuples with sum $<10$. So for example, the solution $x_1=6$, $x_2=x_3=x_4=0$, $x_5=2$, $x_6=0$ is to be considered different from the solution $x_1=2$, $x_2=x_3=x_4=0$, $x_5=6$, $x_6=0$.
You will find a very clear discussion of the ideas for a combinatorial solution in Wikipedia under Stars and Bars.
Let $k$ and $n$ be fixed.
The particular problems discussed there are (i) the number of solutions of $x_1+x_2+\cdots +x_k=n$ in positive integers and (ii) the number of solutions of $x_1+x_2+\cdots +x_k=n$ in non-negative integers.
There is no sense in my repeating here what is well done in an easily accessible source. However, your problem is somewhat different from the ones discussed in the Stars and Bars article.
Your condition is
$x_1+x_2+\cdots +x_6<10$, so you are dealing with an inequality rather than an equation. But the inequality can be rewritten as $x_1+x_2+\cdots +x_6\le 9$. If you want to count the non-negative solutions of this inequality, that is the same as the number of non-negative solutions of the equation $x_1+x_2+\cdots +x_6+x_7=9$.
The above trick of dealing with $<$ by adding an extra variable works in general.
The number of positive solutions of $x_1+x_2+\cdots+x_k<n$ is the same as the number of positive solutions of $x_1+x_2+\cdots +x_k+x_{k+1}=n$.
We prove that this is the case by showing that there is a one-to-one correspondence between the positive solutions of the inequality and the positive solutions of the equation.
Given a solution $(a_1,a_2,\dots,a_k)$ of the inequality $x_1+x_2+\cdots+x_k<n$, let
$$a_{k+1}=n-(a_1+a_2+\cdots+a_k).\qquad\qquad(\ast)$$
Then $a_{k+1}$ is positive, and is completely determined by $(a_1,a_2,\dots, a_k)$. It is clear from $(\ast)$ that $a_1+a_2+\cdots+a_k+a_{k+1}=n$, so $(a_1,a_2,\dots, a_k,a_{k+1})$ is a positive solution of $x_1+x_2+\cdots +x_k+x_{k+1}=n$.
All positive solutions of $x_1+x_2+\cdots+x_k+x_{k+1}=n$ are obtainable in this way from a solution of the inequality $x_1+x_2+\cdots +x_k<n$. For if $(a_1,a_2,\dots,a_k, a_{k+1})$ is a positive solution of $x_1+x_2+\cdots+x_k+x_{k+1}=n$, then $a_1+a_2+\cdots+a_k<n$, so $(a_1,a_2,\dots,a_k)$ is a positive solution of the inequality $x_1+x_2+\cdots+x_k<n$.
We have exhibited an explicit bijection between the set of positive solutions of the inequality $x_1+x_2+\cdots+x_k<n$ and the set of positive solutions of the equation $x_1+x_2+\cdots+x_k+x_{k+1}=n$. Thus the two sets have the same number of elements.
In a similar way, one can show that the number of non-negative solutions of $x_1+x_2+\cdots+x_k<n$ is the same as the number of non-negative solutions of $x_1+x_2+\cdots +x_k+x_{k+1}=n-1$.
Best Answer
By setting $x_i=5-y_i$, we can instead solve $$(5-y_1)+(5-y_2)+\cdots+(5-y_7)=30$$ for $y_1,\ldots,y_7 \in \{0,1,2,3,5\}$. This is equivalent to $$y_1+y_2+\cdots+y_7=5.$$
From here, the possible multisets of $y$-values are:
This gives $420$ possibilities.