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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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    @DouglasS.Stones.Dear Douglas the sequence you are mentioning has two repetitions.Only when we add and subtract 1,2 we obtain we obtain no repetitions of numbers of the original sequence of the 8 natural numbers. The plus and mines signs always follow the above order.2012-11-04

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