Mathematics - 2013-05-03 (page 3 of 5)

5:00 PM

I want to count the number of $3$ digit numbers obtained from $A$ such the their digit sum is odd such that

$(i)$ There is no repetition of digits.
$(ii)$ There is repetition.

robjohn

@WillJagy I know, and if they are going to be hyper-sensitive, then perhaps they need to come back later.

Peter Tamaroff

Now, for $(i)$ I did this. We need $O,O,O$ or $O,E,E$ ($E$=even, $O$=odd)

robjohn

yes

Peter Tamaroff

In the first case, I have $4!/(4-3)!=24$ combinations.

Wait.

robjohn

@PeterTamaroff there are two evens and 4 odds, that leaves 4 choices times 3! arrangements

Peter Tamaroff

5:04 PM

I am counting $4\cdot 2!+4+4\cdot 2!$ by writing the second case as

$$OEE\\EOE\\EEO$$

robjohn

So we get 48 with no repetition

Peter Tamaroff

44?

24+8+8+4?

robjohn

I count 24 for both cases

Peter Tamaroff

@robjohn Oh, right.

robjohn

you have only two evens, so they are both there, then there are 4 choices for the odd. that times six arrangements

Peter Tamaroff

5:06 PM

I have $2!\cdot 4+2!\cdot 4+2!\cdot 4=3!\cdot 4$

robjohn

yes

Peter Tamaroff

OK, $48$.

Now I need to count repetition.

First, the case $OOO$.

J. M.

@WillJagy recently?

Peter Tamaroff

This is just the functions from $\{1,3,5,7\}$ to $\{1,2,3\}$ so $3^4$?

robjohn

@J.M. welcome back :-)

skullpatrol

5:08 PM

@Hayaku welcome back :-)

J. M.

@BabakS. If they don't understand at once, a picture of two great circles likely will.

robjohn

@PeterTamaroff you have the same two basic things OEE and OOO

J. M.

@robjohn What, I never left! ;)

Hayaku

@skullpatrol hello

Peter Tamaroff

@robjohn Yes. Now the case $O,E,E$

Well, in this case the $E$'s give me more options.

They give me $2^2$ instead of $2!$.

So it is $3\cdot 2^2 \cdot 4$

J. M.

5:10 PM

Damn, sometimes math outdoes the cafeteria. The cafeteria doesn't get asked to shovel lunch into students' mouths.

Peter Tamaroff

The total is then $3^4 +3\cdot 4^2$.

robjohn

@J.M. did you see my comment about the banner?

J. M.

@robjohn That's what I was alluding to. :)

Peter Tamaroff

@J.M. What are you alluding to?

robjohn

22 mins ago, by robjohn

Perhaps we need a banner... This is the Math chatroom; spoonfeeding is down the hall

skullpatrol

5:11 PM

Some people are born with a silver spoon...

Peter Tamaroff

@robjohn Why did you write that?

Hayaku

@WillJagy I don't think it has anything to do with background, it's just about learning to set problems up. Evaluation is a different skill than construction. It is not a "plug and chug" type of question

skullpatrol

Your Prof said it was for extra interest, so don't take it so seriously pal.

Hayaku

@skullpatrol I'd still like to learn it. None of my friends seem to know how to solve it either. Prof said it was "very tricky"

Peter Tamaroff

@Hayaku Solve what?

Hayaku

5:15 PM

@PeterTamaroff stats question

Peter Tamaroff

@Hayaku \meruns.

Tobias Kildetoft

nearing the daily cap from just a single short answer to a practically trivial question.

Peter Tamaroff

@TobiasKildetoft The $\sqrt x$ one?

Tobias Kildetoft

yeah

Peter Tamaroff

@TobiasKildetoft I got only five because of some nitpickers. Bleh.

Tobias Kildetoft

5:22 PM

I am a lot happier with the rep I got from the one about groups with 2 proper nontrivial subgroups

(though the other answer is a lot cleaner than mine)

J. M.

@PeterTamaroff Where is this?

(The front page is scaring me these days.)

Peter Tamaroff

@TobiasKildetoft I got some nice rep from this one

@J.M. Here

J. M.

@TobiasKildetoft That's the bikeshed. :)

Tobias Kildetoft

@J.M. Bikeshed?

J. M.

@PeterTamaroff Definitely a dupe.

Peter Tamaroff

5:23 PM

@J.M. Yeah.

Hayaku

can anyone help point me in the right direction? if I roll dice repeatedly, taking a sum as I go, until the sum is greater than 1, what is the probability that the final roll's value is between a and b?

Tobias Kildetoft

@Hayaku what are the possible outcomes of the dice?

Hayaku

1-6

Tobias Kildetoft

most of my dice have all values at least 1

but then you only roll once

Hayaku

i figure if the final roll is between a and b then we are going backwards from a-1..2..3..4..5..6 and b-1..2..3..4..5..6 until we hit some initial value

and then this range of initial values dictates the probability out of the outcome set

but i don't know if this is right

Tobias Kildetoft

5:26 PM

@Hayaku but if you only roll until the sum exceeds 1, you roll once (or twice if you need it to strictly exceed 1 and roll a 1 first)

Hayaku

sorry I meant a boundary of B, not 1

(i choose 50 arbitrarily)

my original problem has 1

but i want to figure out the dice one first

skullpatrol

Yo @JasperLoy how's it going pal?

user19161

@skullpatrol I got 2 downvotes today, lol.

skullpatrol

@JasperLoy Welcome back to the rollercoaster :-)

J. M.

@TobiasKildetoft It's a common metaphor we use.

@PeterTamaroff I've given a close vote.

Tobias Kildetoft

5:33 PM

After having answered the "2 proper non-trivial subgroups" question, I realized, that one can do essentially the same to classify groups with 3 or 4 proper non-trivial subgroups. But for the ones with 4, it turned out that it simplified everything a lot if one knew that if all proper subgroups of a finite group have prime power order, then at most two distinct primes divide the order of the group

Peter Tamaroff

@J.M. Seems fine.

Question

Tobias Kildetoft

I have not had a close look at the same question for 5 subgroups yet

Peter Tamaroff

How complicated should this question be?

Consider the following board:

skullpatrol

@BabakS. How old are these students?

Babak S.

@J.M.: I see. Thanks for your suggestion. I think of Poles on Earth.

Peter Tamaroff

5:34 PM

Babak S.

21 or higher.

J. M.

@BabakS. Exactly, they ought to cross at antipodal points.

Peter Tamaroff

The question is: in how many ways can we paint it with $11$ colours if we impose that no squares sharing sides are painted with the same colour?

Babak S.

@J.M.: I am not going to involve them in a Calculus version for the first time.

Illustrating with pictures are good first.

J. M.

@BabakS. No calculus, just get a clear plastic ball, if possible.

I've used the ones for hamsters.

Babak S.

5:37 PM

Oh. yes. but my bag doesn't have enough space. Good point. :-)

Tobias Kildetoft

@PeterTamaroff hmm, that seems fairly tricky

Peter Tamaroff

@TobiasKildetoft Yeah, seems like it.

J. M.

@BabakS. That the crossing points are antipodal becomes, if you'll pardon the horrible pun, clear.

Peter Tamaroff

I am trying, though.

Babak S.

I see.

Tobias Kildetoft

5:40 PM

@PeterTamaroff probably some sort of recursive approach will be a good idea

Peter Tamaroff

@TobiasKildetoft Recursive? How?

Tobias Kildetoft

so fixing a colour for one of the most isolated squares gives you a new shape to colour, now with a restriction on the colour of one of the squares

so the number of ways to colour the large one is 11 times the number of ways to colour that new one (minus the ones where you pick the illegal colour)

Peter Tamaroff

@TobiasKildetoft Oh, yes. I am thinking like that too.,

Tobias Kildetoft

hmm, on the other hand, this might not get nice

Peter Tamaroff

@anon Save our souls!!!

Charlie

5:42 PM

@anon ANON!!!!

anon

turn it into a graph, find the chromatic polynomial using deletion-contraction recursion relations

Peter Tamaroff

@anon LOL =)

anon

(I am not joking.)

2

Peter Tamaroff

@anon I know.

@anon OK, the graph is this:

Charlie

@anon This is what I was thinking

J. M.

5:45 PM

@anon ...here we thought that the König does not have to say "I'm not joking". :)

Peter Tamaroff

@anon Could you give me a crash course on that?

anon

The relation is $P_G(t)=P_{G-e}(t)-P_{G\setminus e}(t)$, where $G-e$ is deletion and $G\setminus e$ is contraction. I will have to explain later.

Peter Tamaroff

@anon Oh, but I am supposed to do this with basic combinatorics.

anon

arguing with basic combinatorics would basically parallel the computation I'm sketching but in a more drawn-out way

Peter Tamaroff

@anon OK. I'm thinking about painting one square and then counting from there. I am not sure though.

user19161

5:50 PM

Let me see. I should retire when I reach 1k, lol.

anon

there is also a direct formula involving summing-over-partitions. all of this was covered in my talk on the temperley-lieb algebra and tutte's invariants, as it happens.

user19161

@anon Wow, you sound like a professor!

Charlie

@JasperLoy STOP IT, NOW

anon

good old jasper.

J. M.

@JasperLoy Are you going to make this phoenix thing a habit?

user19161

5:51 PM

@Charlie You are green like Jayesh. This is a sign, lol.

user19161

@J.M. What is this phoenix thing?

Peter Tamaroff

@anon Hm. I am thinking this, too: I can paint the red parts independently of each other, yes?

Charlie

@JasperLoy I said it to you first

Lord_Farin

@anon Most interesting. I've never done a lot with chromatic polynomials, but I understand that they are a pain to compute in general. My interest has been provoked in the past. Could you provide some links to literature?

Peter Tamaroff

Well, no.

I mean.

I can put in any numbers I want, freely.

So, yes.

user19161

5:52 PM

@PeterTamaroff If you watched Good Will Hunting, they drew such stuff on the blackboard.

Peter Tamaroff

I can input any of the 11 colours in the red dots, so that gives me $11!/3!$ options to start with.

J. M.

@JasperLoy Fine, "zombie". Whatever you kids want to label it. :P

anon

yes, the stuff in the movie was all graph theory, things like 1/det(tI-A) for adjacency matrices A, etc.

Nimza

Hello!

Peter Tamaroff

Now I can look at what happens if dots are all the same, different, or share two colours.

anon

5:53 PM

I don't have a reference Lord Farin, sorry

skullpatrol

http://chat.stackexchange.com/transcript/message/9265993#9265993

@Charlie

user19161

@J.M. Kids? You are not that old, and I am not that young!

J. M.

@JasperLoy S'what you think. :)

Peter Tamaroff

@anon Am I thinking straight?

Tobias Kildetoft

@PeterTamaroff yes, but those possibilities might not each give the same number of colourings

Charlie

5:54 PM

@skullpatrol no, I never watched

Tobias Kildetoft

so it is not really useful to single them out like this

Peter Tamaroff

@TobiasKildetoft Right, that's why I made the last remark.

"Now I can look at what happens if dots are all the same, different, or share two colours."

user19161

So the Ted Shifrin guy is on our site.

Lord_Farin

@anon No problem, thanks anyway :).

Tobias Kildetoft

@PeterTamaroff yeah, thayt would work (so you should count those three possibilities separately)

anon

5:55 PM

Put this as a question on main in about 5 hours and I will explain how to compute the chromatic polynomials (which tells us how many ways there are to color it with a palette of $t$ colors)

Peter Tamaroff

@TobiasKildetoft Right.

J. M.

One mod down, two to go.

Peter Tamaroff

@anon Let me work it out myself first.

anon

5 hours should be good

Charlie

@anon hi anon

you said no hi

anon

5:56 PM

sorry, hello charlie

Charlie

:D

Peter Tamaroff

@Charlie "Why you no say hi?"

skullpatrol

(-:

anon

not you too, skull!

Charlie

@PeterTamaroff you are the most meme guy I've ever seen

J. M.

5:57 PM

@Charlie "meme-iest"?

Charlie

@J.M. yep!

J. M.

(okay, "most memetic")

skullpatrol

memeful

Charlie

memest

Lord_Farin

@J.M. I go for "memest", which can easily be misheard for something else :).

Peter Tamaroff

5:59 PM

@Lord_Farin Hmm, for what?

Lord_Farin

@PeterTamaroff "memest" $\sim$ "meanest" :P

skullpatrol

That is not memingful :-)

Charlie

hahaha

skullpatrol

:D:D:D

Lord_Farin

@skullpatrol Didn't your mother teach you not to speak with your mouth full? :)

Peter Tamaroff

6:01 PM

@Lord_Farin Oh, =P

skullpatrol

@Lord_Farin Yo Mama?

Charlie

@Szabolcs Yo, long time no see!

skullpatrol

Charlie

HAHAHAHHAHAAHHAHAHAHHAHAHAHAHHAHAHAHA

Szabolcs

Hey all

skullpatrol

6:05 PM

:D:D:D:D:D:D:D:D:D:D:D:D:D:D:D:D:D:D:D:D:D:D:D

Charlie

@skullpatrol I LOVE THIS MEME

@Szabolcs wassup?

Szabolcs

Learning about maximum likelihood methods, stats.stackexchange.com/questions/16758/…

Hey

you

I know, it's dry.

6:07 PM

take a sad song and make it better...

Charlie

better better better

skullpatrol

Na Na Na Na hey you

Charlie

na na na nanana....nanana heey you

skullpatrol

Hey Jude

user19161

@skullpatrol Jude Law is hot.

Lord_Farin

6:10 PM

;O

Charlie

I'm usually 36,5°

user19161

8

Q: Why is $n^2-\frac{n^2}{2} =\frac{n^2}{2}$?

Could someone please expand on how to get from $\displaystyle n^2-\frac{n^2}{2}$ to $\dfrac{n^2}{2}$? I can't seem to wrap my head around that.

user19161

Guys, this question got so many votes and answers?

Charlie

@JasperLoy NOT AGAIN

skullpatrol

And any time you feel the pain, hey, Jude, refrain
Don't carry the world upon your shoulders
Well don't you know that its a fool who plays it cool
By making his world a little colder

Charlie

6:12 PM

nanananana

J. M.

@JasperLoy If I ask "why is 1-1/2 the same as 1/2?", will I get a crapton of votes, too?

Jayesh Badwaik

Yes!

I bet there is one type of answer missing.

Charlie

OH NOEEEEEEEEEES

@JayeshBadwaik HI JAYESH

Jayesh Badwaik

A long answer that proves this is so using the first order logic.

@Charlie HI!

Charlie

@JayeshBadwaik Long time no see

Jayesh Badwaik

6:14 PM

@Charlie Yeah, have been busy somewhat. :-)

Lord_Farin

6 hours ago, by Lord_Farin

(Intermediary rant: RAAHHH @ this. Will it ever stop?!)

Charlie

@JayeshBadwaik yeah, me too

crazy life

Lord_Farin

It seems we are on the same line :)

J. M.

@JayeshBadwaik I think a close facsimile of Russell's argument has already been posted on main, accompanied with an astounding amount of upvotes.

Charlie

We could have MSE junior

skullpatrol

6:16 PM

@Charlie Rating please

Lord_Farin

@Charlie, please pay close attention to the following:

user19161

@JayeshBadwaik Are you the Hulk?

Charlie

@skullpatrol nicki minaj..i'd give zero just because of the singer

Lord_Farin

:)

@Charlie, got that?

Charlie

@Lord_Farin no

user19161

6:18 PM

@Lord_Farin LOL

J. M.

@Lord_Farin Push comes to shove, give Charlie a Caesar cipher or something. ;)

Charlie

nevermind, F, let it go

let it go, Indiana

Jayesh Badwaik

@JasperLoy YES!

Lord_Farin

@Charlie Whatever. Your choice. Now don't come crying again.

Charlie

@Lord_Farin i'm not crying

Lord_Farin

6:19 PM

@Charlie I didn't say you were.

Charlie

i asked you a simple question, you made it complicated

J. M.

"I'm not crying, I'm just sweating through my eyes..."

Charlie

>8(

you are mocking me

I'll call another moderator

:D

Lord_Farin

@Charlie I'm just a little paranoid. You should have noticed that some time before. Didn't you? :)

Charlie

@Lord_Farin no

you think you are paranoid?

I am afraid of my own shadow

Lord_Farin

6:20 PM

@Charlie To some IMO healthy extent, yes.

Charlie

@skullpatrol 8,0

Lord_Farin

Perhaps "conscientious with one's private information" is a better description. But compared to community standards, I'm paranoid.

skullpatrol

@Charlie :-|

user19161

@Charlie So when are you getting married?

Charlie

@JasperLoy when you decide to stop deleting your accounts

skullpatrol

6:22 PM

(-:

J. M.

@Charlie I guess this gives ammunition for the "Bieber is a zombie" camp...

Lord_Farin

Ok, I'm seriously astounded.

user19161

@Lord_Farin OMG

Charlie

yes, I cry so what?

J. M.

@Lord_Farin "No, I did not try it." - how beautiful.

Jayesh Badwaik

6:24 PM

@Lord_Farin Dude, delivery failure!

J. M.

@Charlie Nothing wrong with it; people who don't cry at all are the nutty ones.

Charlie

good

user19161

math.stackexchange.com/questions/380495/some-algebra-2-help OMG!!!

Charlie

man...

you are drunk, go home

Lord_Farin

@JasperLoy The most disturbing part is the upvote.

user19161

6:27 PM

@Lord_Farin It is disturbing how I dropped from 20k to 100, lol.

Lord_Farin

@JasperLoy From what I gather, there's only one person to blame for that...

Overall consensus seems to be that we're glad you're back, though. :)

Charlie

Lord Farin, of course

skullpatrol

(-;

J. M.

@JasperLoy It is perfectly alright to be disturbed by self-inflicted losses, my good man. :)

user19161

@J.M. Will you run for mod on MSE?

Charlie

6:29 PM

I will

Lord_Farin

@Jasper meta.math.stackexchange.com/questions/9323/…

J. M.

@JasperLoy I'd rather herd cats than herd mathematicians.

2

...and the luckless students that come along with them.

Lord_Farin

@J.M. ...lured in by said mathematicians and the few glimmers they have encountered elsewhere. So sad, all these lost souls.

J. M.

@Lord_Farin See, you see what I'm seeing! :)

Charlie

so many green avatars lately...

J. M.

6:32 PM

Maybe I'll make my next piece of artwork green...

Brian M. Scott

@Charlie Seasickness epidemic.

Charlie

maybe

Peter Tamaroff

@BrianM.Scott Yo.

Charlie

@BrianM.Scott Hi, by the way, mr. Scott

Brian M. Scott

@PeterTamaroff What’s the problem?

5

@Charlie Hullo, Charlie.

Jayesh Badwaik

6:33 PM

@PeterTamaroff Yo.

Peter Tamaroff

@BrianM.Scott Colouring graphs! =)

J. M.

@BrianM.Scott or algal overgrowth.

Peter Tamaroff

I have done some work, I have!

Charlie

hahahahahahahahahahhahahaa

skullpatrol

:D:D:D:D:D:D:D:D:D:D:D:D:D:D

Brian M. Scott

6:34 PM

@PeterTamaroff Are you sure? Charlie sounds a bit doubtful ... :-)

Peter Tamaroff

@BrianM.Scott To Oblivion with her!

Charlie

@BrianM.Scott I laughed of what you said:"what's the problem?"

skullpatrol

To get there you have to go through hell.

Charlie

:)

Peter Tamaroff

I am counting the ways of colouring the following graph:

Lord_Farin

6:36 PM

Oh noes. The children's swing across the street is being taken over by loudly shouting and screaming teenagers! :(

Peter Tamaroff

Such that no adjacent vertices have the same colour.

Charlie

@Lord_Farin DO SOMETHING!

user19161

@PeterTamaroff What course is this?

Brian M. Scott

@skullpatrol Nah, Lethe borders Elysium

Peter Tamaroff

@JasperLoy Combinatorics and other weeds.

Lord_Farin

6:37 PM

@Charlie I forgot my sniper...

Peter Tamaroff

@BrianM.Scott My work:

Brian M. Scott

@PeterTamaroff How many colors?

Peter Tamaroff

Eleven.

Charlie

@Lord_Farin true men use their hands

Peter Tamaroff

@BrianM.Scott Now, this is my idea.

Lord_Farin

6:37 PM

@Charlie Draw your conclusions. :)

Charlie

@Lord_Farin you have no hands

@Lord_Farin or you are no man

J. M.

@Lord_Farin knowyourmeme.com/photos/533939-gravity-falls

Peter Tamaroff

Label the vertices that way.

J. M.

@BrianM.Scott I always found that bit amusing.

Peter Tamaroff

Then, $2,4,7$ are independent of each other, in the sense I can colour them freely, @BrianM.Scott

So, I will consider how things will unveil when I choose to colour those in different manners.

Lord_Farin

6:40 PM

@J.M. Very apt. :)

Brian M. Scott

@PeterTamaroff True; or you could use the larger set $\{1,3,5,6,8\}$.

Peter Tamaroff

@BrianM.Scott Don't ruin my idea =(

J. M.

@Lord_Farin I'm supposed to be too old for cartoons, but I love that show.

skullpatrol

@BrianM.Scott What I meant was beyond hell.

Peter Tamaroff

First, suppose that I colour $2,4$ the same colour and $7$ different. In such a case, I have $10^2 9^3$ options.
Second, suppose that I colour $2,7$ the same colour and $4$ different. In such a case, I have $10^3 9^2$ options. This is the same if I colour $4,7$ the same and $2$ different.
Thirdly, suppose that I colour them all different, then I have $8\cdot 9^2 10^2$ options.

Now, I have to count in how many ways I can get each configuration.

@BrianM.Scott I am all ears to your idea now.

In the two cases, I have to take from $[11]$ a set of the form $a,a,b$. In the third, I want $a,b,c$.

Charlie

6:44 PM

@skullpatrol interesting

skullpatrol

@Charlie yipyipyip

Charlie

@skullpatrol yipyipyip

Peter Tamaroff

@BrianM.Scott

skullpatrol

Let him think...

Charlie

give him a time @peter

Brian M. Scott

6:46 PM

@PeterTamaroff Dealing with someone on main; I may be a few minutes.

Peter Tamaroff

@BrianM.Scott Oh, OK. No problemo.

user19161

No problemo, no petero

Charlie

HAHAHAHAHAH

skullpatrol

:D:D:D:D:D:D:D

user19161

My poem sounds nice.

skullpatrol

6:47 PM

We had a poetry exercise in here yesterday.

Peter Tamaroff

@JasperLoy You don't want to find out what "petero" means in Argentina...

Babak S.

@JasperLoy: Your avatar has lost its color?

Brian M. Scott

@PeterTamaroff Yes, this should work.

user19161

@BabakS. Yes. Be careful, there are several imposters!

Peter Tamaroff

@BrianM.Scott Snoopy dance

user19161

6:48 PM

@PeterTamaroff I think it means banana?

Charlie

@PeterTamaroff Oh my god!

Peter Tamaroff

@JasperLoy No.

@Charlie You googled it?

Charlie

@PeterTamaroff you said it before

user19161

@PeterTamaroff OK, WTF is it?

Brian M. Scott

And the count of the color combinations isn’t bad: you’ve $11\cdot10\cdot9$ ways to assign $3$ distinct colors, and $11\cdot10\cdot3$ ways to assign two.

Hayaku

6:51 PM

can someone help me understand something on mathworld.wolfram.com/UniformSumDistribution.html

Peter Tamaroff

@BrianM.Scott Good. I think I got it. =)

@JasperLoy It means "c**ks***er".

J. M.

@PeterTamaroff Don't tell him; have him sweat for it. ;)

Tsk, damn.

Peter Tamaroff

@J.M. Awww, too late.

@J.M. Did you know?

user19161

@PeterTamaroff OMG!!! Sounds like me, LOL.

Charlie

O.o

Hayaku

6:53 PM

if P_n^(1) = integral from 1 to n of PX1...Xn(u)du - integral from 1 to n-1 of PX1...Pn-1(u)du tells the probability of getting a sum over 1 after n turns, how can i turn this to figure out the expected value of just the last turn?

as opposed to the whole sum

J. M.

@PeterTamaroff Apart from knowing Spanish, I know a number of inelegant words in various dialects...

Peter Tamaroff

@J.M. Heh.

user19161

@J.M. I know some Tamil curse words, LOL

Peter Tamaroff

Kudos?

skullpatrol

(removed)

J. M.