Math Is Fun Forum

The table is perfectly symmetrical about the 'x' diagonal.

eg,
Row "A" has 1 @ "x", then 30 @ "2", then 1 @ "3", and row "f" has 1 @ "2", then 30 @ "3", then 1 @ "x";
Column "A" has 1 @ "x", then 31 @ "2", and column "f" has 31 @ "3" then 1 @ "x".

Also, starting with column A = 62, the A.P. of +1 requires the following handshake totals:
A=62, B=63, C=64...d=91, e=92, f=93.

In your first line under "f" the term should be 2;Please see the symmetrical 1st point on the last line for "f" under "A" It is 2. The sheet is symmetrical albout the main diagonal.The proposal is "A" will have 2 handshakes with all others including "f".

From all of the new formulas, the updated general formula for the answer in terms of n seats is:

The updated answer for n=15:

(about 51.19% of the total arrangements)

And we are as far as ever from an intuitive explanation

Hi anna_gg,

That is strange. The numbers I have given are not in doubt assuming the truth of your formulas. I calculated them using the combinations and permutations calculator from this site with the help of the "pattern" rule. Perhaps you could give it a try here https://www.mathsisfun.com/combinatorics/combinations-permutations-calculator.html
I suppose it is possible that I made a logical error

I can only tentatively reason that the answer is probably between 8,460,000 - 8,490,000 and might be close to 8,482,000. But I am no good at sorting out complex sequences.

Let's say you have five dots in a line and another five dots in a line some distance away.
The total number of ways of drawing five lines, connecting the dots on one side with the dots on the other and each line connecting two dots that aren't connected by any other lines, will give the total number of ways of selecting the two questions, but we can't have the lines going straight across because that would be equivalent to selecting the same question for the same student.
Here's an equation using the inclusion-exclusion principle:

48,120 represents the total number of ways of drawing the five lines with at least one line going straight across.
That isn't what we want.  We want the total number of ways of drawing five lines with none going straight across. So...

Since each two questions for each student can be selected in two different ways we divide by 2^5.

I'm not sure exactly why my earlier answer was wrong.
Anyways, good job anonimnystefy!  This is a hard problem.