/sci/ - Science & Math

11520

736

/sci/ is not a homework board

>>7088945
Does this look like homework to you?

/sci/ is always having an identity crisis

2 * 7!

8 * 6! = 5760 if I'm not mistaken

>>7088922
if they're in the same order but rotated 1 position around the circle does that count

16 * 6!

=

11520

>>7088946
It's certainly easy enough to be homework

'bout three fiddy

>>7088967
This is probably correct.

>>7088960
Wait, take that times 2 as you can switch the two members of the couple.
>>7088967 is right

>>7088962
You just need to divide by the number of possible rotations since rotations that aren't identity don't fix any dispositions .That would make 1440 possibilities.

>>7088960
it's 16*6!, because the two sitting together can reverse order.

28

just think 7 untis: 7!
but the 7th unit has two people in so x 2!
7! x 2= 10080 ways

>The first person of the gay couple can sit in 8 different seats
>The second person of the gay couple can then sit in 2 different seats
>The others can sit in every seat

So 8 * 2 * 6 * 5 * 4 * 3 * 2 * 1 possibilities.

(6*5*4*3*2*1)*4*2=5760
First seat the other 6 relative to the couple= 6*5*4*3*2*1
The decide where the couple will sit=*4 possibilities
then switch the couple from places=*2

>>7089050

When the gay couple inverts, you count the positions of the other gay loners again, so multiply your answer by 2.

>>7088922
Okay faggots here we go:

First we break down unecessary covolution by pretending the circle is a line.

Then we know there are eight people and that two of them always sit together.

Thus we can treat the two as one package or person. Then we can permutate each person in the new model giving us: 7! permutations.

But wait Faggots there's more, because these cunts are sitting in a circle we must account for the special case where the two people are sitting next to each other by sitting each on one end of the line. Thus we have to only focus on the faggots permuting between / or around depending on how you perceive it. So 8-2 = 6 people so 6! permutations.

In total, we have 7! + 6! ways to permute for a total of 5760 seating combinations.

>>7089118
You considered the the two who sit together as a pair, but neglected the fact they they could still switch seats with eachother.

Multiply your answer by 2 and you are good to go.

>>7089187
You're right. My apologies, and thanks.

The answer is 1440, guys.

>>7088967
This is what I got. 16 ways for the couple to sit together, then the 6 remaining can sit anywhere.

http://fileserver.net-texts.com/asset.aspx?dl=yes&id=9908

Page 2, question 11. The answers are a few pages later.

>people not considering rotated seating arrangements to be equivalent
Did you not do elementary combinatorics problems in high school or something? Anyway it's the same except divided by 8.

What does !7 actually mean?

>>7088958
/thread

I'm gay

>>7089604
>What does !7 actually mean?
Seven's really excited to see you.

Srsly, it's 7 * 6 * 5 * 4 * 3 * 2 * 1

>>7089609
Just to be an asshole (because that's what /sci/ is about), !7 is actually 0. 7! is 7*6*5*4*3*2*1. But !n is 0 if n isn't 0 (and non-zero, most likely 1, if n=0).

>>7089607
I'm trans

>>7088922
Choose a seat, call it A. There are two seat adjacent to A, and choosing one leaves us with six free seats. Choosing A, and a seat adjacent, gives 6! arrangements. There are 8 choices for A, and each has two choices of adjacent seating. This yields the formula 2*8*6! configurations.

>>7089619
!7 = 1854
http://mathworld.wolfram.com/Subfactorial.html

>>7088922
This was actually an SAT challenge question I had one time. I remember Brian (the math instructor) solved it very elegantly. It's quite difficult for a 15 year old though.

I would think that since the chairs are otherwise identical, it makes no difference where the first of the couple sits, so the answer should be 2*6! otherwise you're counting the same orientation 8 damn times.

Why am I wrong?

>>7090242
Because the fact of the matter is that the chairs are each unique. They have their own place in the room, and sitting, say closer to the door, is different than sitting over by the window.

>>7090258
unless they aren't. I feel like that should be specified. Put numbers on the chairs or something.

Wouldn't it be two times 7! considering the pair are always together and treated as one for each flip?

>>7090455
I thought the same.

6!x2=1440 is this correct?

>>7090455
This would be correct for a situation with 7 chairs, where the couple occupies the same chair (in 2 possible ways). If you treat the couple as 1 entity, there are 7 entities you have to seat, but not on just 7 chairs.

>>7090485
you do treat it as one entity, and if there were 8 chairs and 8 people it would be 7! not 8!, so when you consider that two people must sit together you then have 7 places and it's 6! and since they can be swaped it's 2x. Although I am not complete sure if it's correct, it was a long time since i've last done this.

>>7090465

It's like the number of different guesses are turning into a polinmial ITT,

>>7090485

Except that you'd treat the pair of chairs as one chair since they're always together is what I thought of it as. They're inextricably bound together because of the pair of people sitting on them and thus become one unit. With one difference, that each position can be duplicated because of their switching.

>>7090485

But I thought we were just considering the people as the entities, with all chairs being out of the equation because firstly, they're fixed in place and not marked and secondly like I said the question asks how many ways the people, not the chairs + the people, can be rearranged.

It would be just like if someone painted 8 circles and asked how many different ways the 8 people could position themselves standing on the circles with two of the people always holding hands next to each other.

>>7090476
7!/7 x 2 = 1440

IT'S 11520 YOU FUCKING FAGGOTS

I'M 15 AND I SOLVED THIS

>>7090506

>>7090490
>and if there were 8 chairs and 8 people it would be 7! not 8!,

Are you sure?

>and if there were 8 chairs and 8 people it would be 7! not 8!,
>and if there were 7 chairs and 7 people it would be 6! not 7!,
.....
>and if there were 3 chairs and 3 people it would be 2! not 3!,

That doesn't seem likely. And I'm postulating this simply by whittling down to a computable polynomial.

>>7090505
>7!/7
>/7

>which=6!

uwotm8?

>>7090510
He's not wrong.
Sure, it's 1440 if you count all "rotationally similar" solutions as identical, but they're not actually identical.

>>7090511
you have to that into account that it's in a circle, so 123 and 231 are the same, and that way if there were 3 people it would be 2!

>>7090520
Thank you!

>>7090510
You're the retard, retard. The problem with the
>hurr treat them as a unit and put seven on a line
approach is that it under-counts. Try that same method with 3 people and a triangular table.

>>7090526
And to clarify, in case it hasn't been done enough, the correct answer assuming the chairs are fixed is
>2 * 8 * 6!
whereas without that assumption we merely have
>2 * 6!
which is the answer in >>7089240.

>>7090535
>assuming the chairs are fixed
you don't count in how many ways the fucking chairs go, but the fucking people sitting on them.

>>7090495
>>7090485
But because there are 8 chairs, the couple has 8 possible positions in the circle, not 7. For every position of the couple there are 6 empty seats left for 6 people, which is 6!, and you multiply it by 2 for when the two friends are swapped.

>>7090539
meant to quote
>>7090490

>>7090538
Yes but you can (if you like) disregard the 8 possible rotations of each positioning.

Why are you all putting emphasis on the couple thing? The "2 of them always sit together" has no relevance to the question; which is how many different seating setups can you have. Fuck the couple information it's not fucking relevant.

>>7090538
I gave two answers, one accounting for graph isomorphisms, meaning we can rotate the chairs around the table (or rotate the table, whatever your preference). The other supposes that the chairs are fixed relative to the table. Not sure what your problem is.

Hello, OP here
In the picture it says:
"How many combinations(people sitting in DIFFERENT places) are there"
Which means that if for example every person switches seats with the person on the right we have a new combination.

The question is not clear. Is the couple line supposed to explain what sitting in a circle means? Or does it suggest that 2 of the 8 people are always sitting next to each other no matter what?

If that is the case, we got 4 couples, so all you need to do if calculate the order amount in which 4 couples can sit.

The "multiply that by 2" is incorrect because we got 4 couples who can each switch places. E.g. the two people in couple 1 can switch places, and the two in couple 2 can. Or couple 1 doesn't, but couple 2 does.

>>7090548
>>7090546
sry than, my bad, in that case you dont do n!/n, but just n!. It would be 7!x2 = 10080, 7! cuz ttwo go together.

>>7090545
What do you mean? If 2 always sit next to each other that limits the number of possible permutations.

4 couples means (4*3*2*1) - 1 combos, is 23. (-1 because it's a circle and the last couple with interhook with a previous combo).

Then among the 4 couples the two people can switch, so (4*2)^4 is 4096

4096 * 23 = 94140 combinations.

>>7090554

Because we don't know which two form a couple. Do we have to randomize that as well?

>>7090551
This is false, see
>>7090485

It's obvious really (though I got it wrong too), cuz if you first explain 7! to people you start by saying "Okay the first thing can fit in 7 places, then this place is occupied and the next can fit in 6 places" &c, but in this case the first thing (the couple) can fit in EIGHT places. So instead of 7*6! it's 8*6!.

>>7090521

Then so would 312 be the same! No. No, you're factoring the actual chairs in as a fixed point of reference in your model. Forget the chairs. The chairs are not part of the permutation so all rearrangements of the people are from the POV of the people themselves, NOT any of the chairs OR any fixed point of reference in the room NOR any cartesian coordinates.

Take one of the 8 people as a fixed point of reference if you want. Now what?

>>7090559
I don't think so, because then it doesn't make a difference since there's always 8 couples, or you might as well say "8 of them always sit together". I mean if they're being nazi's and adding irrelevant information in a tricky way that could be the case but I'm assuming they're not.

>>7090556
>(-1 because it's a circle and the last couple with interhook with a previous combo).

>>7090565
(not the guy ur replying to)
Let's take the couple then, they can be LR or RL, and the remaining six places next to them is 6!, so 2*6!.

>>7090559

I assumed it's always the same two people. If you randomize two of them sitting together that makes absolutely no sense. Because someone is always sitting next to at least two random people.

It has to be the same two people always sitting together for all the permutations.

Take the answer you would get if you didn't have the couples and divide that by two.

8*7*6*5*4*3*2*1 = 40320, / 2 = 20160

>>7090570
>and the remaining six places next to them is 6!
>next to them
>them

They are still part of the permutations. As a single entity/couple that can be treated as a permutation unit identical to all the others except for the fact that in every single permutation as a single entity, they have exactly two ways of arranging themselves.

You're thinking like they're sitting in the corner of the room all the time while the other 6 do all the shuffling in the middle.

>>7090570

it's 2 * 6!, minus the recurring because it's a circle

so (2 * 6!) * 25%, because as we have seen, for every combo in 1234 3412, 2341 and 4123 we have the same, so you take a quarter of that.

>>7090578
>minus the recurring because it's a circle
>25%

That doesn't sound right. 1/4th off a permutation of four. But we have eight.

Also we're not told whether to take the central point of reference as the circle center, or 0,0 in cartesian coordinates. The question itself is deficient in it's full description.

>>7090576
I was specifically talking about the case where you disregard the rotations inside the circle. Not sure what you're getting at.

>>7090578
Nigger what the fuck.
If you take the couple as reference, every permutation of the couple and the 6 people on the 6 seats around them is unique.

>>7090586

We've got four couples dude

>>7090591
>disregard the rotations inside the circle.
Even if you disregard their LR RL rotations(if that's what you meant) and they're always joined at the hip and don't rotate positions with each other(the two) you STILL have to account for THEM shuffling as a single unit with the other 6 single people.

>Not sure what you're getting at.

See above. But I've just realized something else, we're not told whether to take the fixed, central point of reference as the circle center of the 8 chairs, or 0,0 in cartesian coordinates where each chair is at 100% unique coordinates. The question itself is "wrong" because it's not precise about the preconditions. So there's actually two correct answers. One for cartesian coordinates and the other for relative positioning with no fixed point of reference.

>>7088922
Which high school is it from, OP? Doesn't look hard enough to be from Nada HS.

>>7090595

I thought we had one couple composed of two people who always sat together?

>>7090597
>and the other for relative positioning with no fixed point of reference.
This is what I was talking about. In this case, only the position of the persons in relation to each other matters. You can take 1 person (or couple) as a reference and count how many permutations there are of the other persons in relation to this reference person, since the absolute position of the reference dude doesn't matter.
This is 2*6!, don't you think?

>>7090604
>This is 2*6!, don't you think?

Yes, sounds right for relative positioning. My calcs were first based on purely fixed, unique positions of the chairs in unchanging coordinates, hence almost linear.

Maybe the translator left something out and made the question more ambiguous than it should have been. We can say there are two correct answers because of the wording(or lack of) of the question.

Speaking of wording and translation, another anon suggested that there are actually two couples, not one.

>>7090600
No idea, it is probably made up. Pretty hard on my opinion for a highschool entrance test. I did not even know what x! was when i was in highschool.

>>7090609
It's not really harder than HS entry level stuff in Europe or US (I think), it's just a matter of whether or not you have had a lesson in combinatorics in elementary school.

>>7090609
http://school.js88.com/scl_h/22018140/100/id-5690.html

They take their entrance exams seriously in Japan.

Look, this is how it's done. As my beautiful illustration shows, there are 8 possible positions for the pair to sit in. Since they are 2 people, there are a total of 16 positions in which they can be seated, due to the fact that they can switch places with one another. This leaves us with 16 possible combinations for these people. Then, with the pair seated, we have 6 open seats and 6 people to seat. One person sits in one of the six chairs. Then one of the remaining 5 sit in one of the five remaining chairs and so on. Then the chairs are all filled. This leaves us with:
16 (number of possible couple positions)*6 (one person sitting in one of the six empty chairs)* 5(one person sitting in one of the five empty chairs)*4*3*2*1 (for the same reason as the ones prior)
16*6*5*4*3*2*1=11520
Anyone willing to prove me wrong?

>>7091071
I hope you are just trolling.

>>7091079
Care to explain?

>>7091079
He's correct though.

>>7091083
Not him but that was a 5 hour necropost to a 'you should be able to solve this' troll thread.

>>7088958
ALMOST there
it should be 2*6!
you thought there were 7 "groups" (6 individual and the pair of two) but they are seated around a circular desk. this means (example from a three people scenario)

abc bca cab are the same seating pattern, because they all connect in a circle anyway

so you have to fix the position of a single person to make the circular seating patter be as though it were a regular pattern, to do this you subtract 1 from the 7 so 6!

this is a fact

>All these idiots on my /sci/
Get out. There are two acceptable answers (with justification):
2 * 6! and 2 * 8 * 6!

>>7091355
>All these idiots on my /sci/
yup, shit like these show really how many highschoolers and whatnot is on /sci/

>>7091071
>Anyone willing to prove me wrong?

You're half wrong and it's because the question is ambiguously worded. You're assuming that the chair positions are fixed in the model and defined by cartesian coordinates and each chair is a uniquely located entity.

See >>7090597 >>7090604 and >>7090607

>>7091361
You know, it's fine when they're excited. It becomes agitating when they're smug (and wrong).

Starting with the fags, one gaylord can choose any of the 8 chairs and his bf wants to sit next to him giving him 2 possibilities. Now the fags have 16 possible ways to sit on 8 chairs. Because the other 6 people don't care, they can sit on the remaining 6 chairs in any order. Thus we have 6! different ways for the others to sit per each gaycombo, ie. 16*6! = 11520.

Are weabos generally as good at math as regular japs are or are they total opposites; artistic types?

>>7091367
>because the question is ambiguously worded
It's really not.

>You're assuming that the chair positions are fixed
No, he isn't.
No matter how you scramble the chairs, the question clearly states they are arranged in a circle.
Clearly one of the chairs is further north than the rest, or closer to the water fountain, or whatever.
Rotationally similar solutions exist, and there's nothing in the question to suggest they should be "folded together".

>>7093002
but every chair also has a unique position in space and every time someone new sits on it it is changed in a way that it is statistically impossible that any exact configuration will ever occur

same seating could certainly be interpreted as everyone has the same person on his right and left as before

2! * 6!

come on, this is literally middle school/high school shit

>>7093011
>same seating could certainly be interpreted as everyone has the same person on his right and left as before
...unless you actually read the question.
Then you would know that it asks about: "people sitting in different places", NOT "people sitting in different relative positions".

I've never been able to do these "how many possible combinations" problems. Fuck them.

>>7088922

28 * 7!

Count the probability of the couple, times combination

>>7093074
It also said "combinations" and not "permutations" so these two statements conflict.

>>7093117
But the later, parenthetical phrase is obviously intended to clarify the earlier wording.

>>7093002
>the question clearly states they are arranged in a circle.
>Clearly one of the chairs is further north than the rest, or closer to the water fountain, or whatever.

And there's the ambiguity. It wants you to ASSUME. Not that big of a deal though, it just means there's two answers instead of just one.

>>7093515
>It wants you to ASSUME. Not that big of a deal though, it just means there's two answers instead of just one.
Nobody's being asked to ASSUME anything.
The question clearly states:"people sitting in different places", NOT "people sitting in different relative positions".
No assumptions required.
You can now go back to /b/ and download pictures of underage boys in drag.

>>7093551
angsty highschooler detected

>>7093551
>The question clearly states

The question doesn't clearly state the chairs are to be disregarded. It doesn't matter if they're in a circle because like the different people in the question the chairs are also different unique objects. which are differenciated by their coordinate locations.

Also, it's a translation from Japanese which makes it more likely that ambiguity has been added as would be the case if an artistic weabo wanted to troll /sci/ but didn't himself understand the entire question to begin with given his rudimentary understanding of Japanese and/or permutations.

>>7093681
You would say anything in order to avoid admitting you cannot answer this wouldnt you?

>>7093754

I already answered it, then realized there were two answers. Others ITT also seem to agree with me about that.

>>7089050
there are 8 places the couple can sit.

Mathematicians haven't done high school math since high school. They probably don't even know how to carry the one anymore.

36 right?

>>7088922
Take the couple and put them in two seats, multiply by every permutation of the remaining 6, switch the two people. Then rotate the couple around the table while repeating the same process.

6!*2*8 = 11520

Which is less than every possible permutation of 8 people (8! = 40320).

3.141593 roughly.


>Ask a serious question if you want a serious answer on a math board

it´s 16 :

you can place person A on 8 seats and person B right or left of person A. So 8*2 = 16 to cover all possibilities.

>>7095690
I´m sorry I thought i was just about the two of them my bad.
actually >>7095674
is right

how do you even do this
it's not required to live, watch anime fap and post on 4chan

>>7088970
nice burn m8 lol xd #rekt