# 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.

• Please dont downvote members,I want to learn these type of problems.I have not done them before.
– diya
Oct 23, 2015 at 16:32
• Counterpoint: please don't post zero-effort questions. Oct 23, 2015 at 16:34
– 5xum
Oct 23, 2015 at 16:37

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?

• Is $120(=8 \times 15)$ the correct answer?@5xum
– diya
Oct 23, 2015 at 16:48
• @diya Without justification, none of the numbers is the right answer if you ask me. You should be able to explain why $8\cdot 15$ is the right equation, or why some other equation is correct.
– 5xum
Oct 23, 2015 at 16:51
• @diya This hint is already very substantial. To give it a bit more depth, you should know about the rules of sum and product. Break it into cases based on where the mine in the center is. Oct 23, 2015 at 17:00

This is a supplement to 5xum’s hint. I’ve marked each blank cell with a colored dot; cells with the same color dot have something important in common. Seeing what they have in common, and how many of each kind there are, should help you with your analysis and calculation.

$$\begin{array}{|c|c|c|c|c|c|} \hline \color{purple}\bullet&\color{purple}\bullet&\color{green}\bullet&\bullet&\color{orange}\bullet&\color{brown}\bullet\\ \hline \color{purple}\bullet&2&\color{green}\bullet&1&\color{orange}\bullet&2\\ \hline \color{purple}\bullet&\color{purple}\bullet&\color{green}\bullet&\bullet&\color{orange}\bullet&\color{brown}\bullet\\ \hline \end{array}$$

From Brian M. Scott's hint: Answer $$= \text{C}^3_1\times\text{C}^5_1\times\text{C}^2_2 + \text{C}^3_1\times\text{C}^5_2\times\text{C}^2_1 + \text{C}^2_1\times\text{C}^5_2\times\text{C}^2_2$$

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

• Consider using LaTeX/Mathjax Mar 20, 2020 at 2:28