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?
 A: Consider the following possibilities:
_ 3 5 _ _ 2 _ _ 1 _ _ 1_ _ 5 _ 3
_ 3 5 _ _ 2 _ _ 1 _ _ 1_ _ 5 3 _
3 _ 5 _ _ 2 _ _ 1 _ _ 1_ _ 5 _ 3
3 _ 5 _ _ 2 _ _ 1 _ _ 1_ _ 5 3 _
All of these would give the same result of 3521153, but are counted separately by your first method.
A: An intuitive explanation is:
Consider 6 _ 5 _ _ 2 _ _ 1 _ _ 1 _ _ 5 _ _, we have $6$ places to put the next $6$:
6 6 5 _ _ 2 _ _ 1 _ _ 1 _ _ 5 _ _
6 _ 5 6 _ 2 _ _ 1 _ _ 1 _ _ 5 _ _
6 _ 5 _ _ 2 6 _ 1 _ _ 1 _ _ 5 _ _
6 _ 5 _ _ 2 _ _ 1 6 _ 1 _ _ 5 _ _
6 _ 5 _ _ 2 _ _ 1 _ _ 1 6 _ 5 _ _
6 _ 5 _ _ 2 _ _ 1 _ _ 1 _ _ 5 6 _
Continuing:
_ _ 5 6 _ 2 _ _ 1 _ _ 1 _ _ 5 _ _ gives $5$ (different from before) places
_ _ 5 _ _ 2 6 _ 1 _ _ 1 _ _ 5 _ _ gives $4$
_ _ 5 _ _ 2 _ _ 1 6 _ 1 _ _ 5 _ _ gives $3$
_ _ 5 _ _ 2 _ _ 1 _ _ 1 6 _ 5 _ _ gives $2$
_ _ 5 _ _ 2 _ _ 1 _ _ 1 _ _ 5 6 _ gives $1$
Adding these, $6+5+4+3+2+1=21$.
A: 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.
