In a game of Minesweeper, a number on a square denotes the number of mines that share atleast one vertex with that square. In a game of Minesweeper, a number on a square denotes the number of mines that share atleast one vertex with that square.A square with a number may not have a mine,and the blank squares are undetermined.In how many ways can the mines be placed in the given configuration on the blank squares.
 

$(A)120\hspace{1cm}(B)105\hspace{1cm}(C)95\hspace{1cm}(D)100$



Number on the middle square is $1$,so there is one mine which shares atleast one vertex with this square.So there are 8 ways to put the mine adjacent to square numbered $1$.But then i got stuck.Please help me.
Hint: First, think about where the mine causing the middle number $1$ can be. How many possible positions are there?

Then, given the position of that mine how many mines does the left $2$ need, and in how many places can you place them?
From Brian M. Scott's hint: Answer $= \text{C}^3_1\times\text{C}^5_1\times\text{C}^2_2 + \text{C}^3_1\times\text{C}^5_2\times\text{C}^2_1 + \text{C}^2_1\times\text{C}^5_2\times\text{C}^2_2$

$= 3\times5\times1 + 3\times10\times2 + 2\times10\times1 = 15 + 60 + 20 = 95$