Does your method give the same results with mine, for n=2 to n=11? By the way, these are kind of "confirmed" because I also did the calculations in Excel, but I would like to see a more "scientific" way, with arrangements, permutations etc.
Do you agree with the answer given?
Hmmm, then allow me to make a nice ansatz...
n = 12 = 79200
n = 13 = 506880
n = 14 = 2745600
n = 12 = 79200
n = 13 = 506880
n = 14 = 2745600
The method and logic can be really simple if you can immediately answer a few more questions:
What do you get for n = 1,2,3,4?
What do you get for n = 1,2,3,4?
]]>Of course all the possible permutations for n=14 are 14! and also number 1 can only be in 1st position. I will leave the rest for you, as I am a novice at combinatorics
Can you please show the method to calculate this and the logic behind it?
I started by setting all the possible positions for each number:
1 only goes to 1st position.
2 can be in any position of 2, 3, 4 or 5.
3 in any position of 2, 3, 4, 5, 6, 7, 8 and so on.
Actually, not in "any" position - these are only the starting positions, to which then we must also apply the restriction of [n/2] preceding n in the arrangement.
But then I don't know how to continue
The rest is yours
FYI floor(n/2) = integer part (n/2)
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