In how many arrangements can you arrange 3 green cones and 3 blue cones in a line along the boundary of football ground if no two cones of the same color are to be adjacent?  Choose the correct choice from the following choices:
A) 6
B) 12
C) 36
D) 72
E) 144
BGBGBG or GBGBGB --&gt; 2 ways,   Both 3 B's and 3 G's can be arranged in 3! ways.   Thus total 2*3!*3!=72.   Answer: 72.
<result>D</result>
In how many arrangements can you arrange 3 green cones and 3 blue cones in a line along the boundary of football ground if no two cones of the same color are to be adjacent?