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...must be something with mod(3).

I also read that these types of problems are solved by using Hamming codes

No idea how!!

No limitation in number of weightings, right?

Earplugs and lots of puzzles to keep busy!!!

Hey guys you are both so clever!

Now it makes more sense!! Despite the rain

Last one for f with f should be x, not 4.

thickhead wrote:

When we have finished with "P" and are going to "Q" we find another 3 can not be added as the diagonal joining (f,B) to (B,f) falls into a ditch ("x" of the main diagonal )and the total for Q remains as that of "P" viz. 77. this difficulty may be overcome by revising Q's handshake with "f" to 4 instead of 3.That should be continued to all the persons from "R" to "e".

I don't get it. If Row "A" has 1 @ "x", 30 @ "2" and 1 @ "3", then shouldn't Column A have also the same? How come it has 31 @ "2" (instead of 30) and none @ "3"?

If the intersection of, say, A2 with A31 is 3, then also A31 with A2 should be 3 (not 2), right??

phrontister wrote:

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.

And also last A17 with A16 is 3, while A16 with A17 is 2.

anna_gg wrote:

Yes but then the total will be 92 instead of 93, which will be the same with the previous one!

thickhead wrote: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".

Yes but then the total will be 92 instead of 93, which will be the same with the previous one!

thickhead wrote:

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".

Why not?

He could possibly have 10 handshakes with someone else, as long as his total number of handshakes with everyone is different from everybody else's.

thickhead wrote:

what I assumed is that a must have different number of handshakes from others but these numbers could be used by others also. e.g. if A had 10 handshakes with P ,R can have 10 handshakes with C ;there is no bar for that but obviously P can not have 10 handshakes with anybody other than A.

So finally what is the correct number for X?

We are not told that each people has different number of handshakes with each of the others; only that each one "participates" in a different number of handshakes in total. Therefore, if A participates in 2*31=62 (because we are told that each pair exchanges at least two handshakes), then all others from B to f participate in a different number of handshakes (but I don't know how to find them). I guess their sum must be 496*X*2.

thickhead wrote:

A had 2 handshakes with B,3 handshakes with C, 4 handshakes with D .... 32 handshakes with f (32nd person).So total handshakes by A=2+3+3+4+.......+31+32=527 handshakes. Add handshakes of all people and divide by 2.

62+63+...+93 = 2480.

Are we dividing this by 496 (i.e. the total possible number of pairs)?

You mean, the number of handshakes each of them had was different?

Are we tied to use all 3 options?

Ohh yes, you are right! I am stupid

thickhead wrote:

Yes but you must add also the 50 initial handshakes, since we know that each invitee shook hands with at least one other. So we start with 50 and then the first person that shook hands with EVERY other, had 48 more (since we have already counted the first one), the second 47 etc.

thickhead wrote:

It must be a combination of pieces with number of blocks divisible by 3.

I think the correct answer is 2291229 if my calculations are correct.

Relentless wrote:

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

Sorry, {w,k,w,w} and {w,m,w,w} - if not mistaken (after 11 hours of work, this is normal!!).

Relentless wrote:

Hi (:

The list does include {w,w,w,m} and {w,w,w,w} in the 9th and 10th rows.

I used "pattern *,w,w,*" for this.

"pattern *,m,m,m,*" for the males,

and both for both.

I had tried the calculator myself but the results I get, when compared with some tests I did through programming (of course, for numbers smaller than 15), do not match! No idea why!! I have made all possible combinations for the syntax of "pattern" rule, with * and ?, in case I was missing something!

Relentless wrote:

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

Apparently your list is missing {w,w,w,m} and {w,w,w,w}.

Relentless wrote:

19, right?

man, woman, kid

{m,m,w,w}

{m,w,w,m}

{m,w,w,w}

{m,w,w,k}

{m,k,w,w}

{w,w,m,m}

{w,w,m,w}

{w,w,m,k}

{w,w,w,m}

{w,w,w,w}

{w,w,w,k}

{w,w,k,m}

{w,w,k,w}

{w,w,k,k}

{k,m,w,w}

{k,w,w,m}

{k,w,w,w}

{k,w,w,k}

{k,k,w,w}

Also,

Number of combinations with two adjacent females to 4 seats: I get 21 instead of 19 (and of course I haven't yet checked the bigger numbers). So this is where our difference of 57 vs 55 comes from (yours is 81-19-5 while mine is 81-21-5).

Relentless wrote:

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.

Answers up to 12 seats: For up to 3, I agree (have tried listing all of them). For 4 seats, I get 55 instead of 57 - and of course I haven't tried the rest of them.

Relentless wrote:

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.

Very well!

Fruityloop wrote:

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.