# Why are is partitions counting technique wrong?

I recently heard about partitions. I tried to count them using the following technique:
1) Ways to write $5$ as a sum of five positive integers:
$$1+1+1+1+1$$ 2) Number of ways to write $5$ a sum of four positive integer:
$$2+1+1+1$$ 3) Ways to write $5$ as a sum of three positive integers:
$$3+1+1$$ $$2+2+1$$ 4) Ways to write $5$ as a sum of two positive integers: $$4+1$$ $$3+2$$ 5) Finally, ways to write $5$ as a sum of one number:
$$5$$

So, there are seven different ways to write $5$ as a sum of positive integers. Which is correct. But if I continue to use this technique for higher numbers I do not get the correct answer. For example if I do it for 6.

1) Ways to write $6$ as a sum of six positive integers:
$$1+1+1+1+1+1$$ 2) Ways to write $6$ as a sum of five positive integers:
$$2+1+1+1+1$$ 3) Ways to write $6$ a sum of four positive integer:
$$3+1+1+1$$ $$2+2+1+1$$ 3) Ways to write $6$ as a sum of three positive integers:
$$4+1+1$$ $$3+2+1$$ 4) Ways to write $6$ as a sum of two positive integers: $$5+1$$ $$4+2$$ $$3+3$$ 5) Finally, ways to write $6$ as a sum of one number:
$$6$$

The answer that I got is $10$ but the actual partitions according to wolfram alpha is $11$. My question is, what did I miss? Because as I increase the number the error becomes larger and larger.

• You missed $2+2+2$. – Brian M. Scott May 1 '16 at 5:30
• @BrianM.Scott Is there any formula for counting them? – Anonymous May 1 '16 at 5:33
• Nothing very nice; you can get an idea of what’s known here and here. – Brian M. Scott May 1 '16 at 5:35
• @BrianM.Scott So, basically you need a computer for larger numbers. Right? – Anonymous May 1 '16 at 5:39
• Pretty much, yes, though there are ways to avoid pure brute force, e.g., using recurrences. – Brian M. Scott May 1 '16 at 5:41