Integer partitions of $N$ into three distinct parts How many ways are there to partition 10 into 3 distinct parts?

For example, $\{7,2,1 \}$, $\{6,3,1 \}$ and $\{5,4,1\}$ are valid partitions, but $\{4,5,1\}$ would not be an additional partition because it has the same elements as $\{5,4,1\}$. $\{8,1,1\}$ is not admissible because it has two 1's.
The Wikipedia article on partitions indicates how to see that the number of partitions into at most three parts is the same as the number of partitions with greatest part at most three.  It gives the number of partitions of $n$ into at most three parts as the nearest integer to $\frac {(n+3)^2}{12}$, which for $n=10$ is $14$.  For exactly three parts, we subtract the $5$ partitions into $2$ parts and the $1$ partition into $1$ part, leaving $8$.  They are:
$$1,1,8\\1,2,7\\1,3,6\\1,4,5\\2,2,6\\2,3,5\\2,4,4\\3,3,4$$
Of these, four are into distinct parts.  There is also equality between all partitions into distinct parts and all partitions into odd parts, but I don't know how to equate that to your problem with both distinct parts and a given number.
enter image description hereThe formula &lt;C(n+2,2)/6-0.75*(n+1)+89/72&gt; is the answer of the number of solutions of a+b+c=n, where a,b,c>0 and &lt;x&gt; is the integer closest to x