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I am sorry i realized my mistake how about this approach. Total number of ways they can be arranged = 10! Always together = 9! Never together = 10! - 9! in never together case then can switch positions among themselves which would be = 2. If we fix their order of apparance then never together event will reduce to half. Therefore total number of way based on given condition = [9!+(10!-9!)/2] [9!+9!(10-1)/2] [9!+9!*9/2] 9!*5.5
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This looks very plausible. However I was puzzled why the final answer is different from the correct solution. I finally figured out. You have mixed oranges and apples in your deduction.
Let's see.
Total number of unrestricted outcomes: 10!
If the two people have to stay together and one must be in front of the other: 9!
NOTE here, number of outcomes that they do not stay together is NOT 10!-9!, because the first number you have not restricted one has to be in front of the other, and the second number you have.
So the number of outcomes that they do not stay together is 10!-2*9!. Then follow your logic you'll get the same result as the correct answer.
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