If n is an integer such that the sum of the digits of n is 2, and $10^{10} < n < 10^{11}$, the number of different values of n is

This is tough! I initially thought about combinations with repetition but the lower bound 10^10 < n is tricky. Can someone explain why the number of digits matters here?

I got 10 but the answer says 11? I must have missed something. The sum of digits is 2, so the digits could be a 2 and ten 0s, or two 1s and nine 0s. But the 2 can't be in the first position if the rest are zeros? Actually it can, since 2 followed by zeros is 20000000000, which is >10^10 and <10^11. So that's valid. Then for two 1s, the first digit must be 1, and the other 1 can be in any of the remaining 10 positions. That gives 10 numbers. Plus the one with a 2 gives 11 total. But wait, is 20000000000 allowed? Yes, because 10^10 = 10000000000, so 2*10^10 is bigger. So why did I think 10? I forgot the single 2 case. So answer should be 11. But I chose 10. Oops.

I think the answer is 11. The digit sum is 2. Possible digit multisets: {2,0,0,...,0} and {1,1,0,...,0}. For the first, the 2 must be the first digit (since if it's not, the number would start with 0, which is not allowed for an 11-digit number). So 1 number. For the second, the first digit must be 1 (can't be 0), and the other 1 can be in any of the remaining 10 positions. So 10 numbers. Total 11. But is there any other combination? What about {1,0,1,0,...}? That's the same as two 1s. So 11.