# Counting eleven digit integers with the sum of the digits 2

Suppose n is an integer, such that the sum of the digits of n is 2, and $10^{10} \lt n \lt 10^{11}$. The number of different values for n is:

Let me try to list them :

(1) 11000000000
(2) 10100000000
(3) 10010000000
...
(10)10000000001


So I am getting 10 possible values but the answer is 11.What am I missing here ?

EDIT: I was missing 20000000000.

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20,000,000,000? – Jonas Meyer Oct 29 '10 at 5:24
Your new question doesn't make sense. Also, it seems better to start a new question instead of replacing an existing one. – Yuval Filmus Oct 29 '10 at 5:57
@ Yuval Filmus : Fixed. – Quixotic Oct 29 '10 at 6:07
The question should not be marked as unanswered. – TCL Nov 22 '10 at 0:17