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Let's play mini-Sudoku!

 size(3cm); draw((0,0)--(4,0)--(4,4)--(0,4)--cycle,black+2bp); draw((0,2)--(4,2),black+2bp); draw((2,0)--(2,4),black+2bp); draw((0,1)--(4,1)); draw((0,3)--(4,3)); draw((1,0)--(1,4)); draw((3,0)--(3,4));

We wish to place an "X" in four boxes, such that there is exactly one "X" in each row, column, and 2x2 outlined box. For example:

 size(3cm); draw((0,0)--(4,0)--(4,4)--(0,4)--cycle,black+2bp); draw((0,2)--(4,2),black+2bp); draw((2,0)--(2,4),black+2bp); draw((0,1)--(4,1)); draw((0,3)--(4,3)); draw((1,0)--(1,4)); draw((3,0)--(3,4)); label("X",(0.5,1.5)); label("X",(1.5,3.5)); label("X",(2.5,2.5)); label("X",(3.5,0.5));

In how many ways can we do this?

 

Please explain very well in this quesiton.

 Mar 18, 2015

Best Answer 

 #1
avatar+99352 
+10

I think this is it

 

1)  there are 4 choices for the first row, the X can go anywhere.    Just put your X in

2)  There are only 2 choices for the next row because it cannot go in the same box as the first one did.

3)  There are 2 choices for the next row because it can go in either box but it cannot go under either of the 2 x's that are already there.

4)  There is only one possibility left for the last row

 

so the number of possibilities is        4*2*2*1=16

 Mar 19, 2015
 #1
avatar+99352 
+10
Best Answer

I think this is it

 

1)  there are 4 choices for the first row, the X can go anywhere.    Just put your X in

2)  There are only 2 choices for the next row because it cannot go in the same box as the first one did.

3)  There are 2 choices for the next row because it can go in either box but it cannot go under either of the 2 x's that are already there.

4)  There is only one possibility left for the last row

 

so the number of possibilities is        4*2*2*1=16

Melody Mar 19, 2015

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