So, I was making unordered combinations of natural numbers which add upto a certain natural number. I was able to go till 6 when I got to know about the partition function. I was pleased to see that my combinations were correct however i could only come up with 10 such combinations for 6 and it states that there exist 11 such combinations. Please tell me the eleventh combination. The combinations i have found are below: $$(6), (1,5), (2,4), (3,3), (1,1,4), (1,2,3), (1,1,1,3), (1,1,2,2), (1,1,1,1,2), (1,1,1,1,1,1,1)$$
Also is there a way to tell the combinations rather than just the number of combinations. I mean something other than hit and trial as I have already failed in that but also a bit simple if possible.
As lulu pointed out, the missing partition of 6 is (2, 2, 2).
Here's an algoirthm for generating all partitions from Knuth's The Art of Computer Programming (vol. 4A, section 7.2.1.4, p391). Given a partition $\lambda$ that isn't all 1s, it starts $1\dots 1(x+1)$ for some $x \ge 1$ (and the number of 1s could be 0). Replace the initial terms $1\dots 1(x+1)$ with $rx\dots x$ for the appropriate remainder $r \le x$ to get the "next" partition.
Here's how the algorithm runs for $n = 6$: \begin{align*} (\underline{6}) \text{ with } x = 5 & \mapsto (1,5), \\ (1,\underline{5}) \text{ with } x = 4 & \mapsto (2,4), \\ (\underline{2},4) \text{ with } x = 1 & \mapsto (1,1,4), \\ (1,1,\underline{4}) \text{ with } x = 3 & \mapsto (3,3), \\ (\underline{3},3) \text{ with } x = 2 & \mapsto (1,2,3), \\ (1,\underline{2},3) \text{ with } x = 1 & \mapsto (1,1,1,3), \\ (1,1,1,\underline{3}) \text{ with } x = 2 & \mapsto (2,2,2), \\ (\underline{2},2,2) \text{ with } x = 1 & \mapsto (1,1,2,2), \\ (1,1,\underline{2},2) \text{ with } x = 1 & \mapsto (1,1,1,1,2), \\ (1,1,1,1,\underline{2}) \text{ with } x = 1 & \mapsto (1,1,1,1,1,1). \end{align*}
In each partition, I've underlined the $(x+1)$ part used in the algorithm; any later parts stay the same.