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Let's have the following sequence of natural numbers: 1, 2, 3, 4, 5, 6, 7, 8. The permutations of these 8 numbers are equal to 8!. We can obtain some of these permutations by adding and subtracting one or more numbers within this sequence e.g. 8-1=7, 7+1=8, 6-1=5, 5+1=6, 4-1=3, 3+1=4, 2-1=1, 1+1=2; also we have 8-3=5, 7+1=8, 6-3=3, 5+1=6, 7-3=4, 6+1=7, 4-3=1, 1+1=2, and so on. My question is: how many permutations we can obtain with this method of adding and subtracting numbers within the sequence?

addendum:

The question asks the following: If we have a sequence of natural numbers in ascending order and we add to or subtract from each number of this sequence one or more numbers within this sequence how many permutations of this sequence we can obtain?

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Every derangement of the ordered list $8,7,6,\cdots,2,1$ can be obtained by adding or subtracting one of the elements of $[8]:=\{1,\cdots,8\}$ to each entry of the list (with no restrictions on repetitions). This is because $(i,j)\mapsto |i-j|$ surjects from $[8]\times[8]\setminus\rm diag([8])$ onto $[8]$. If you disallow leaving a list entry fixed, these are all of the possible effects, whereas if you do allow entries to remain preserved, every possible permutation of the list is possible. It is unclear to me however what you are actually asking. –  anon Nov 4 '12 at 3:59
    
@anon.We know the number of permutations of these 8 elements.Now if you find another permutation except the three above will become clear to you what I am asking. –  Vassilis Parassidis Nov 4 '12 at 4:38
    
@anon.Can you tell me please what number can you add and subtract to obtain the permutation 3,7,1,8,4,6,2,5. –  Vassilis Parassidis Nov 4 '12 at 5:15
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@anon: If you look carefully at Vassilis’s examples, both follow the format $8-a$, $7+b$, $6-a$, $5+b$, $4-a$, $3+b$, $2-a$, $1+b$, where $a=b$ is permitted. He may be considering only modifications of this type. (And it seems perfectly clear that he’s using one-line notation with commas and without parentheses.) –  Brian M. Scott Nov 4 '12 at 6:25
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The sequence 8-3=5, 7+1=8, 6-3=3, 5+1=6, 7-3=4, 6+1=7, 4-3=1, 1+1=2 doesn't match the -a/+b description (e.g. 7-3=4). –  Douglas S. Stones Nov 4 '12 at 8:14

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