Solving 5-Stone Counting Problem w/ Restrictions

AI Thread Summary
To solve the 5-stone counting problem with the restriction that the green stone cannot be at the front or back, it can only occupy the second, third, or fourth positions in the sequence. Placing the green stone in any of these positions allows for the remaining four stones to be arranged in 4! (24) different ways. Therefore, if the green stone is in the second position, there are 24 arrangements; the same applies for the third and fourth positions. This results in a total of 3 positions multiplied by 24 arrangements, equaling 72 possible arrangements. The final answer confirms the back of the book's solution.
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Hi, I wasn't sure how to approach this problem:

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.

BTW the back of the book says that the answer is 72.

Help please! thanks.
 
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How many ways can you place the green one, then place the rest?

P.S. please don't multiple post.
 
um 3? lol I don't get it...
 
The DO it.

Suppose you place the green stone in the second place. How many different ways are there to place the other 4 stones?

Suppose you place the green stone in the third place. How many different ways are there to place the other 4 stones?

Suppose you place the green stone in the fourth place. How many different ways are there to place the other 4 stones?

Okay, now how many ways is that all together?
 

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