Answer is C) 21

7!/2(5!)

5040/240

The answer is C (21) - nice work.
Your solution of 7!/2(5!) looks considerably different from mine, but it obviously seems to work. I'd love to hear how you arrived at it.

Here's my approach to the question:

We need to place the two missing 3’s into the number 52115. There are two cases to consider:
(a) The two 3’s are separated by other numbers (e.g., 532135 or 3521135)
(b) the two 3’s appear together (e.g., 5332115 or 5211533)
Case (a): look at the typed number as: _5_2_1_1_5_
The placeholders shown are potential locations to place 3’s. To meet the case (a) criterion of having separated 3’s we need to select 2 of the 6 placeholders and place a 3 in each location. We can do this 6C2 (15) ways
Case (b): Use the same setup: _5_2_1_1_5_
To meet the case (b) criterion of having 3’s together, we need to select 1 of the 6 placeholders and place both 3’s there. We can select 1 of the 6 placeholders 6 ways.
So there are 21 ways (15+6) to place our missing 3’s

Mr2Bits' approach works because you can think of the problem as follows:

Say you have seven chairs in a row. There are 21 (7C2) pairs of chairs in which you could place the two 3's. Now fill the five remaining empty chairs in order with the digits 52115. In this way, for each of the 21 different choices we could make for where to place the 3's, we get a different possible seven-digit number, and we clearly get all of them this way.

Ian Stewart wrote:Mr2Bits' approach works because you can think of the problem as follows:

Say you have seven chairs in a row. There are 21 (7C2) pairs of chairs in which you could place the two 3's. Now fill the five remaining empty chairs in order with the digits 52115. In this way, for each of the 21 different choices we could make for where to place the 3's, we get a different possible seven-digit number, and we clearly get all of them this way.

Very nice! That's what I love about math - lots of ways to reach the same conclusion. This approach is the most eloquent so far!