Here I would like to ask how do you count the number of partitions of an integer, e.g. 12, given the partition size can only be 5 and $ 1\leq \text{the value of each partition} \leq5$.
I can only count each individual case by hand through writing each case, but I sometimes forget to include other possible cases. Is there any quick and useful technique to count each individual case of the partition of an integer by hand only?
For a small number such as 12, the simplest way is to manually make cases. This is an example:
Case 1: all the partitions are equal. Only one partition i.e. $(4,4,4)$ exists
Case 2: 2 partitions are equal. if the numbers are $(a,b,b)$ then $a,b$ are solutions of $2b+a=12$, $1\le a,b \le 5, a \ne b$. Only $(2,5,5)$ satisfies this.
Case 3: all the partitions are different. If the numbers are $(a,b,c)$ then $a,b,c$ are solutions of $a+b+c=12$, $1\le a,b,c \le 5, a \ne b \ne c$. $(3,4,5)$ satisfies this condition.
The only partitions possible are $(2,5,5)$, $(3,4,5)$ and $(4,4,4)$.
Note that the techniques I used above are similar to Multinomial theorem, but not exactly same. This is because you mentioned that $(2,5,5)$ is the same as $(5,2,5)$ or $(5,5,2)$. If these cases are different, then multinomial theorem can be used. For larger numbers with more partitions, this technique gets complicated and you should use the generating function described in the comments.