Math Is Fun Forum

Correct.

Solution please??

Solutions so far are not correct, so any further ideas are most welcome and appreciated!!

Can't think of anything so far!!

Any chance I understand the reasoning and solution? Am pretty novice
Only thing I understand is 3^15 for the total possible number of arrangements without the restrictions

We don't know.

It must be something binding the employees to go to work on the next day, and this for as many employees as possible.

Dear Bobby,
I agree that intuition is not the proper way of solving math problems

I trust your solution is correct. Can you please explain No 2 an in particular, how you got 80/3 from the total number of possible arrangements? If I understand well, this is the number of ways to select 5 out of 31.

Many thanks,
Sam

Unfortunately I don't know the answer and I have only performed the calculations up to n=11 - like I said, I did them using the https://www.mathsisfun.com/combinatorics/combinations-permutations-calculator.html but it can only handle numbers up to 10000000

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.

For n=4 I get 3 of 24
For n=3 I get 2 of 6
for n=2 I get 1 of 2
n=1 is out of scope, since floor(n/2)=0 and 0 does not belong to the set.

I don't know what you mean by "bunch"...
I pick one book at a time, check for the dedication and then continue. As soon as I find all 5, I stop.

If this is of any help, I got the answer (played with the "Combinations and Permutations Calculator" and with the "pattern" tools) for n = 3 to n = 11 but the tool cannot process such big numbers for n = 14
So for n=4 the arrangements that satisfy the requirement are 3 out of the total of 24
n=5 --> 8 of 120
n=6 --> 20 of 720
n=7 --> 80 of 5040
n=8 --> 210 of 40320
n=9 --> 896 of 362880
n=10 --> 3360 of 3628800
n=11 --> 19200 of 39916800

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

1,3,2,5,4,8,6,7,10,9,11,12,14,13 or
1,2,5,4,3,9,10,6,8,11,12,13,7,14.