Partitions of 12 into exactly three parts

Text

In how many ways can $12$ be written as a sum of exactly three positive integers, disregarding order (for example $10+1+1$ and $1+10+1$ are the same partition)?

Solution

Count the triples $a\ge b\ge c\ge 1$ with $a+b+c=12$, ordered by the smallest term $c$:

• $c=1$: $a+b=11$, with $b\le 5$: $\ (10,1),(9,2),(8,3),(7,4),(6,5)$ → $5$
• $c=2$: $a+b=10$, with $2\le b\le 5$: $\ (8,2),(7,3),(6,4),(5,5)$ → $4$
• $c=3$: $a+b=9$, with $3\le b\le 4$: $\ (6,3),(5,4)$ → $2$
• $c=4$: $a+b=8$, with $b=4$: $\ (4,4)$ → $1$

In total $5+4+2+1=\boxed{12}$ partitions.