<s>[INST] Find the total number of partitions of $12$ having unequal positive parts. $P(k, n) = P(k - 1, n - 1) + P(k - n, n)$
Let $P^\star(k,n)$ denote the number of partitions of $k$ having exactly $n$ positive parts, all of which are unequal. For example, $P(8, 3)$ implies $8$ can be partitioned as $8 = 2 + 2 + 4,$ but $P^\star(8,3)$ does not.
$P^\star(k, n) = P(k - \binom n2, n)$ 
We need the $\sum_{i=1}^4P^\star(12,i)$ because $P(k, n)$ is not defined for $P^\star(12,n)$ where $5 \le n \le 12.$
Now,
$P^\star(12,1) = P(12, 1)$
$P^\star(12,2) = P(11, 2)$
$P^\star(12,3) = P(9, 3)$
$P^\star(12,4) = P(6, 4)$
Now we calculate the four $P(k, n)$ above.
$P(12, 1) = 1.$
$P(11, 2) \\ = P(10, 1) + P(9, 2) \\ = 1 + P(8, 1) + P(7, 2) \\ = 1 + 1 + P(6, 1) + P(5, 2) \\ = 1 + 1 + 1 + P(4, 1) + P(3, 2) \\ = 1 + 1 + 1 + 1 + P(2, 1) + P(1, 2) \\ = 1 + 1 + 1 + 1 + 1 \\ = 5.$
$P(9, 3) \\ = P(8, 2) + P(6, 3) \\ = P(7, 1) + P(6, 2) + P(5, 2) + P(3, 3) \\ = 1 + P(5, 1) + P(4, 2) + P(4, 1) + P(3, 2) + 1 \\ =1 + 1 + P(3, 1) + P(2, 2) + 1 + P(2, 1) + P(1, 2) + 1 \\ = 1 + 1 + 1 + 1 + 1 + 1 + 1 \\ = 7.$
$P(6, 4) \\ = P(5, 3) + P(2, 4) \\ = P(4, 2) + P(2, 3) \\ = P(3, 1) + P(2, 2) \\ = 1 + 1 \\ = 2.$
So, the answer to this problem is $1 + 5 + 7 + 2 = 15.$
Does the calculation above make sense to you? Thanks.
 [/INST] Your answer is correct.  As a check, here is a list of the $15$ partitions of $12$ into unequal positive parts:
\begin{align*}
12 & = 12\\
   & = 1 + 11\\
   & = 2 + 10\\
   & = 3 + 9\\
   & = 4 + 8\\
   & = 5 + 7\\
   & = 1 + 2 + 9\\
   & = 1 + 3 + 8\\
   & = 1 + 4 + 7\\
   & = 1 + 5 + 6\\
   & = 2 + 3 + 7\\
   & = 2 + 4 + 6\\
   & = 3 + 4 + 5\\
   & = 1 + 2 + 3 + 6\\
   & = 1 + 2 + 4 + 5
\end{align*}
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