Let's consider a much smaller value and work our way up to the factorial 2009!
Start with a+b = 2. The value 2 is the smallest right hand side possible since a > 0 and b > 0, so a = b = 1 is the smallest a,b possible.
The only solution to a+b = 2 is a = b = 1 as mentioned.
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Now move onto a+b = 3. We have two solutions:
a = 1, b = 2
a = 2, b = 1
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Now move onto a+b = 4. We have three solutions
a = 1, b = 3
a = 2, b = 2
a = 3, b = 1
Note how 'a' counts up while b counts down. I have 'a' starting at the smallest value, and b is set to add with 'a', getting to the right hand side.
So 'a' counts up to 3, while b counts down from 3.
We cannot reach 4 since 0 is not allowed for a or b.
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The conjecture is that a+b = k has k-1 positive integer solutions. A simple proof of this is to list out all solutions like so
a = 1, b = k-1
a = 2, b = k-2
a = 3, b = k-3
....
a = k-3, b = 3
a = k-2, b = 2
a = k-1, b = 1
Where k > 2
Going from 1 to k-1 is exactly k-1 items. It's similar to how {1,2,3,..,m} has m different integers in it. Replace m with k-1 and you have the same idea.
This confirms there are k-1 solutions listed above.
The last step is to simply replace k with 2009! and we have 2009! - 1 different ordered pair solutions
Since the factorial 2009! is so massive, it's best to not expand this out using a calculator. Many calculators would produce "overflow", or similar, if you tried to compute the factorial 2009!