+0

+1
120
21
+10

There are 20 chairs numbered from 1 through  20 around a circular table. How many ways can three people be seated, so that no two people are adjacent?

Oct 14, 2019

#5
+1808
+3

Solution:

$$\text{The chairs are in a circle, identified by a numeric sequence, and chair one (1) follows chair twenty (20).}\\\text{ }\\ \text{There are 20 ways to choose a sequence of three (3) consecutive chairs.}\\ \text{There are 20 ways to choose a sequence of two (2) consecutive chairs.}\\ \text{There are 16 ways to choose a third chair that is \bf{not} in sequence with the two (2) consecutive chairs.}\\ \text{There are nCr(20,3) ways to choose a set of three (3) chairs.}\\ \text{So ... }\\ \dbinom {20}{3} - 20 - (20*16) = 800\\ \text{There are 800 ways to place three (3) persons in a circle of twenty chairs, such that no two persons are adjacent.}\\$$

GA

Oct 16, 2019
#11
+105683
0

Ok Ginger, I get your logic.  Seems reasonable.

If the three people are all different, different people in the same seats,  then I think you would multiply this by 3! = 6

800*6=4800

Do you agree with that?

Melody  Oct 16, 2019
#13
+1808
+2

Yes, I agree. The counts increase by a factor of (3!) when the persons are distinguishable.

Other points of interest:

The set counts remain the same whether or not the chairs are distinguishable by label.  Though the chairs are distinguishable, they are “fixed” in place and there is no permutation or rotation of the chairs themselves. The chairs are chosen by sets such that that no two persons are adjacent when occupying them and rotations of the sets through the 20 chairs occur in the selection process.

However, If the chairs are not labeled, then the distinguishable set counts of 800 drop to 40 (by factoring out the 20 rotations), because the positions of the persons are no longer discernable with respect to the chairs. Only the relative space (empty chairs) between the persons is discernable.

This is probably the solution for this question: https://web2.0calc.com/questions/help_4800

GA

GingerAle  Oct 17, 2019
#7
+105683
+1

Deleted

Oct 16, 2019
edited by Melody  Oct 17, 2019
#8
+6045
0

....

Rom  Oct 16, 2019
edited by Rom  Oct 18, 2019
#9
+105683
0

Thanks Ginger and Thanks Rom for confirming her answer.

I guess I have some homework to do.

Rom, how do you know that Ginger is correct?

Melody  Oct 16, 2019
edited by Melody  Oct 16, 2019
#12
+6045
+1

....

Rom  Oct 16, 2019
edited by Rom  Oct 18, 2019
#14
+105683
0

Thanks Rom :)

Melody  Oct 17, 2019
#15
+1

n chairs around a circular table, m people, at least k chairs between them (n>=m(k+1))

If chairs are numbered, people are distinguishable:

$$n*(m-1)!*{n-m*(k+1)+(m-1) \choose m-1}=a(n,m,k)$$

If chairs are numbered, people are indistinguishable:

$$\frac{n*(m-1)!*{n-m*(k+1)+(m-1) \choose m-1}}{m!}=\frac{n*{n-m*(k+1)+(m-1) \choose m-1}}{m}=b(n,m,k)$$

If chairs are not numbered (indistinguishable), people are distinguishable:

$$\frac{n*(m-1)!*{n-m*(k+1)+(m-1) \choose m-1}}{n}=(m-1)!*{n-m*(k+1)+(m-1) \choose m-1}=c(n,m,k)$$

However if the chairs are indistinguishable AND the people are indistinguishable, we can't always divide the third formula by m factorial (or alternatively, the second one by n the number of chairs) to get the number of arrangements where both chairs and people are indistinguishable.

the formula is (using burnside's lemma):

$$\frac{1}{n}*\sum_{i,\ 0\leqslant i\leqslant n-1\wedge\frac{n}{gcd(i, n)} \mid m }^{} b(gcd(i,n), \frac{m}{\frac{n}{gcd(i,n)}}, k) = \frac{1}{n}*\sum_{i,\ 0\leqslant i\leqslant n-1\wedge\frac{n}{gcd(i, n)} \mid m }^{} \frac{gcd(i,n)*{gcd(i,n)- \frac{m}{\frac{n}{gcd(i,n)}}(k+1)+(\frac{m}{\frac{n}{gcd(i,n)}}-1)\choose (\frac{m}{\frac{n}{gcd(i,n)}}-1)}}{\frac{m}{\frac{n}{gcd(i,n)}}} \\=\frac{1}{n}*\sum_{i,\ 0\leqslant i\leqslant n-1\wedge\frac{n}{gcd(i, n)} \mid m }^{} \frac{n*{gcd(i,n)- \frac{gcd(i,n)*m}{n}(k+1)+ \frac{gcd(i,n)*m}{n}-1)\choose ( \frac{gcd(i,n)*m}{n}-1)}}{m}$$

so we can divide the second formula to get the last one if and only if gcd(n,m)=1

Oct 17, 2019
#16
+1

Ok for some reason I can't edit my answer so i'll edit here:

so we can divide the second formula by n to get the last one if and only if gcd(n,m)=1

also we can further simplify the last equation by canceling out n:

$$\frac{1}{n}*\sum_{i,\ 0\leqslant i\leqslant n-1\wedge\frac{n}{gcd(i, n)} \mid m }^{} \frac{n*{gcd(i,n)- \frac{gcd(i,n)*m}{n}(k+1)+ \frac{gcd(i,n)*m}{n}-1)\choose ( \frac{gcd(i,n)*m}{n}-1)}}{m}= \\ \sum_{i,\ 0\leqslant i\leqslant n-1\wedge\frac{n}{gcd(i, n)} \mid m }^{} \frac{{gcd(i,n)- \frac{gcd(i,n)*m}{n}(k+1)+ \frac{gcd(i,n)*m}{n}-1)\choose ( \frac{gcd(i,n)*m}{n}-1)}}{m}$$

Guest Oct 18, 2019
#17
+105683
+1

Formulas like yours are probably very useful if you are feeding them into a computer program BUT when we are doing the logic as a one off, Ginger's way is easy for any mathematician to follow so therefore for our purposes it has more relevance.

Melody  Oct 18, 2019
#18
+1

I agree, the last formula is annoying, I was trying to convey that ginger's way of dividing by 20 to get the number for a question where the chairs aren't numbered won't always work (here it works because gcd(20, 3)=1). Keep in mind her way of getting to the answer of 800 will become harder as the number of people seated around the table grows

Guest Oct 18, 2019
#19
0

When I try to submit a comment It tells me I didn't check the "i'm not a robot" box, and after I check It again and submit it again, It submits the answer but It doesn't let me edit it (shows the comment as if I didnt write it)

Guest Oct 18, 2019
#20
+105683
0

I have not had direct touble editing but

I am  having trouble with LaTex at the moment.

It is not always rendering properly for me.

Melody  Oct 18, 2019
#21
+1808
+3

TPM, I was wondering when you were going to show up again. Your appearances have a definite correlation to the mushrooms that suddenly appear on cow manure. ...And the psychotropic effects are similar too.   This time your appearance is related to math instead of advocacy for morons and brain-dead vandals.   ... It looks like you’ve continued to improve your math skills.  Your final equation is impressive –I’m still playing with it. (It could stand to have more annotation.)  It’s much better than your delusional, chaotic, bullshit infused presentation here.

LancelotLink sent me similar –well annotated, equations showing the relationships between Stirling Numbers of the First and Second kinds to Lah numbers (aka Stirling Numbers of the Third kind), and to Bell numbers. The annotations included footnotes showing applications in physics where in certain mediums, electromagnet resonance produces harmonic convergences that become soliton waves reflecting in some time-domains while passing through others.  A similar phenomenon also occurs for acoustical resonance in certain mediums. All of these are related to the geometry and hypergeometry of the medium. Even though most of the math was and still is beyond my skills, I thought it quite cool that that these relationships casually occur in nature and that someone could actually discover them.

Your equations are not anything I could casually produce, but I can understand them if I invest the time. I’m assuming they are your equations –the wagging head and jointless gestures are readily apparent, but it lacks the long monologues (aka Blarney), and... Perhaps the computer knows something, because it ... (shows the comment as if I didnt write it). ... so ...if not, could you give a source?

GA

GingerAle  Oct 18, 2019
edited by GingerAle  Oct 18, 2019