That would be 5! - 2*4!
You have 5 differently colored stones-red, orange, blue, green, purple. If the green stone cannot be placed at the front or the back of the sequence, how many possible arrangements can you make?
I know that without the above restriction, the amount would be 5!=120.
But I don't get how to use the restriction.
The back of the book says that the answer is 72.
Thanks for any help
There are 5 stones. Since you CANNOT place the green stone at the front, there are 4 stones you CAN place there. Assuming you have placed that stone, there are 4 stones left. Since you CANNOT place the green stone at the end (and one of the 4 left is green), there are 3 stones left that you CAN place there. Now you have 3 stones left to put in the remaining 3 places- of course there are 3! ways of doing that.
The total number of ways you can place the 5 stones WITHOUT the green stone being first or last, is (4)(3)(3!)= 4*3*6= 72. That is, of course, exactly the same as 5!- 2(4!).