How many ways are there to write $64$ as a sum of $10$ natural numbers (without $0$) such that each number is $\le 12$ (here, the order of these natural numbers doesn‘t matter)?
My first idea would be to fill up to any sum between $52$ and $63$ using only $9$ numbers. Afterwards, the tenth number is fixed. However, I don‘t know how to count the fill-up and there might be duplicate counts.
(This is problem #15 from Arnold‘s Problems for children from 5 to 15, so I don‘t expect that a solution involves complex results about partitions.)
Note that we can deal with the order not mattering by ordering in non-decreasing order.
In other words, we choose each number by choosing how much larger it should be than the previous number.
So, we are picking numbers $b_0,...,a_9$ such that $b_0>0$, $a_1,...,a_9\geq0$, and such that $b_0+(b_0+a_1)+(b_0+a_1+a_2)+...+(b_0+...+a_9)=10b_0+9a_1+...+a_9=64$. For the sake of symmetry, we let $a_0=b_0-1$, giving us the equation $$10a_0+9a_1+...+a_9=54$$where $a_0,...a_9\geq0$ and $a_0+a_1+...+a_9<12$.
Can you solve this from here? It's a bunch of casework (I promise it is not as bad as it looks). The answer should be 9673.