Different seven-digit numbers could Sid have meant to type 
Sid intended to type a seven-digit number, but the two 3's he meant to
  type did not appear. What appeared instead was the five-digit number
  $52115$. How many different seven-digit numbers could Sid have meant to type?

First time, I attempted this problem as follows-
I considered, there are 12 places available for two 3's (as they can appear together)
_ _ 5 _ _ 2 _ _ 1 _ _ 1 _ _ 5 _ _
So, we can select 2 places from 12 places in ${12\choose 2}$ = $66$ ways.
But, I found the correct approach is simply choosing 2 places from available 7 place in ${7\choose 2}$ = 21 ways. However, I can't find my mistake in previous one. Can anyone explain what am I missing?

The answer 21 is got as follows: If the threes are together then we have 6 numbers (choosing one of the 6 gaps in 52115. If the threes are not together, we need to pick two places out of 6 and this can be done in $\binom{6}{2} = 15$ ways. Thus a total of 21 numbers.
