Mathematics - 2012-12-24 (page 1 of 3)

12:00 AM

@OldJohn, but z^3 = x^3 - 3 x y^2 + i (3 x^2 y - y^3), so there are your a = x^3 - 3 x y^2, then b = 3 x^2 y - y^3 , and c = x^2 + y^2. It works. You are drunk but not syphilitic

I think it is probably time to go murder some of your relatives in their beds. That always calmed me down.

Old John

@WillJagy I will sleep on it - and kill some relatives whilst sleep-walking - it will all become clear in the morning (unless I have a hangover!)

Will Jagy

@OldJohn, good night.

Old John

@WillJagy Thanks for entertaining me - I'm off to sleep!

Justin

@Charlie Not much, first time in the chat room so..

Ethan

Can somone help me evaluate this integral
$$\int_{0}^1 x\lceil \frac{1}{x} \rceil \{ \frac{1}{x} \} \,dx $$Where {.} is the Fractionalpart function, and $\lceil .\rceil$ is the Floor function.

anon

12:13 AM

Just wrapped christmas presents. Next christmas I want tape that isn't trying to kill me.

What happens when you use the substitution u=1/x?

Ethan

Uhm I get u=1/x

anon

Do you know what substitution means in the context of integrals?

Ethan

Yes

anon

$$\int_0^1x\left\lceil\frac{1}{x}\right\rceil\left\{\frac{1}{x}\right\}dx=\int_1‌^{\infty} \frac{1}{u}\lceil u\rceil\{u\}\frac{du}{u^2}=\sum_{n=1}^\infty (n+1)\int_0^1\frac{udu}{(u+n)^3}$$

Ethan

How did you do that

anon

12:20 AM

mathjax, why must you parse legit latex incorrectly?

oh well

oops, error

Ethan

the last series transformk

where did u get that

anon

$$\int_1^\infty f(x)dx=\sum_{n=1}^\infty\int_n^{n+1} f(x)dx=\sum_{n=1}^\infty \int_0^1f(x+n)dx $$

Ethan

oh

ok i get it

Can we evaluate the last sum? I think its 1/2

ye its 1/2

I got it, thanks

user19161

12:42 AM

Ladies and gentlemen, I have attained 12k on Christmas eve...

Ethan

y do u care

1

Q: Functional Prime Sums

Let $ f(x) $ be an integer function satisfying, atleast $ f(xy) = f(x) + f(y) $ whenever $ \gcd(x,y) = 1 $. How can I prove that $$ \sum_{\substack{p ~ \text{prime} \\ p \leq n}} f(p) \frac{p}{n} \bigg\lceil \frac{n}{p} \bigg\rceil \left\{ \frac{n}{p} \right\} \sim \frac{1}{2} \sum_{\substack{p ...

anon

1:37 AM

@OldJohn, that cancellation answer was hilarious. thank you.

Peter Tamaroff

2:09 AM

@anon Link?=

@WillJagy William, is that you?

anon

@PeterTamaroff it's over here

Peter Tamaroff

@anon Oh. I voted to closed that. What happened?

anon

reopened, as seen in the edit history

user19161

2:43 AM

@Ethan Just because.

user19161

@anon Wrap some for me too!

Peter Tamaroff

3:27 AM

@anon Is there any notation for the indicator of an element $a$ in some subgroup $G'$ of an abelian finite group $G$?

Like ${\rm ind}_{G'}(a)$?

anon

like, indicator function?

Peter Tamaroff

@anon No no, the least $n>0$ such that $a^n\in G'$

anon

yes, |a|

Peter Tamaroff

Really?

anon

yup

in number theory contexts, when working in $(\Bbb Z/n\Bbb Z)^\times$ you'll want $\mathrm{ord}_n(a)$ instead

Peter Tamaroff

3:29 AM

BUt the order is the least $n$ such that $a^n =e$, right?

anon

The reason is that |G| denotes the size of a group/subgroup/subset, and the order of an element is equal to the size of the cyclic subgroup it generates.

Gustavo Bandeira

@GraceNote You again.

anon

Yes, that is the order.

Peter Tamaroff

RIght

anon

BUt, RIght... OKay.

Peter Tamaroff

3:31 AM

@anon ¬¬

I don't do that on purpose, you know?

@anon But how do we emphasize it is wrt to $G'$?

Maybe $|a|_{G'}$?

anon

@PeterTamaroff Its order in the subgroup is equal to its order in the overgroup. There is no need to specify.

(Exercise.)

Peter Tamaroff

@anon But I'm talking about the indicator.

Not the order.

anon

What's the indicator of an element?

user19161

@anon WTF is an overgroup?

anon

oh, sorry @peter, I didn't fully read your comment on it above

derp

Peter Tamaroff

3:34 AM

I almost shortened finite abelian group to FAG.

anon

@Peter, whenever a subgroup N is normal in G, the least n such that g^n is in N is equal to the order of gN in the quotient group G/N. If that helps.

Peter Tamaroff

@anon Yeah, no idea about normal subgroups yet =P

anon

So you could write $|gN|_{G/N}$ or $\mathrm{ord}_{G/N}(gN)$ or simply $|gN|$.

@PeterTamaroff Normal means closed under conjugation. All you need to know for abelian groups is ... well, every subgroup of an abelian group is normal, so you don't need to bother with it.

Also I haven't seen the term indicator used before, interesting.

Peter Tamaroff

@anon Apostol does.

He is now introducing a way to build "larger" subgroups given $G'\subsetneq G$ and $a\in G,a\notin G'$

he sets $G''=\{xa^k:x\in G',k=0,\dots,h-1\}$

where $h=|a|_{G'}$

Then $G'\subseteq G''\subseteq G$

and $|G''|=|a|_{G'}|G'|$

anon

those apostrophes are very annoying

grrrr

Peter Tamaroff

3:42 AM

@anon HAHA well, you like $G_1$, $G_2$ better?

peoplepower

4:05 AM

@PeterTamaroff Is G' the derived subgroup or something else?

anon

@peoplepower No, he's just using apostrophes to denote different subgroups. He specified G was abelian.

peoplepower

@anon My eyes!

anon

heh heh

peoplepower

So $G''$ is just $\langle G',a\rangle$ right?

No, $h$ is something else...

anon

@peoplepower he's using h for the "indicator" of a, which is the least nonnegative integer such that a^h is in G'.

So, yes, $G''=\langle G',a\rangle$

peoplepower

4:10 AM

Ah, the indicator is just for counting purposes.

I wonder if they proved that $xa^k$ is uniquely defined...

anon

I imagine so.

user19161

4:26 AM

@peoplepower That's what Henry Horton likes to say!

user19161

Are you Henry's other account?

peoplepower

@JasperLoy I don't believe so.

Khromonkey

Hi guys

can you recommend me a good tv show?

After realizing I only watch community because annie is hot I thought I should watch something more substancial

any recommendations?@anon @JasperLoy @JayeshBadwaik @PeterTamaroff

anon

you're a stranger on the internet. I have no idea what you'd like. absurd question!

Gustavo Bandeira

@Khromonkey Rocco's adventures.

Khromonkey

4:39 AM

=(

@anon images2.wikia.nocookie.net/__cb20120817115046/central/images/3/…

@GustavoBandeira you mean this ?? roccosadventures.wordpress.com

I know @PeterTamaroff likes the big bang theory

I think its starting to get worse

do you agree?

I like person of interest, white collar and the middle

I also used to like monk

and numb3rs

Gustavo Bandeira

4:57 AM

@Khromonkey No porn? =(

Khromonkey

no

that is the work of the devil

Gustavo Bandeira

Actually it's the work of porn actors and actresses.

Khromonkey

those actors have parents

Gustavo Bandeira

Perhaps the devil is in the management, hmn....

And the parents have parents? So, everybody have parents ad infinitum. Hmn

Khromonkey

just kidding

I have like 50 gb of it on my computer

Jayesh Badwaik

4:59 AM

@GustavoBandeira While I do not have any compuctions about porn, but recommending it as a TV show? Weird.

Khromonkey

mabye he was going to recommend game of thrones

one of the tv shows I hate the most is two and a half men

It used to be good

now all of the jokes are about sez

sex*

Jayesh Badwaik

Yes, it gets boring after a season or two.

that too, when you take breaks from watching it.

watching a marathon bored me in three to four episodes.

Gustavo Bandeira

I'll make a TV show.

Khromonkey

@JayeshBadwaik do you know what antilia is?

Gustavo Bandeira

I'll film myself while sleeping: The sleeper.

Every episode with me sleeping.

Jayesh Badwaik

5:02 AM

@Khromonkey as in a town or a community?

Khromonkey

@JayeshBadwaik as in a residence in India

Jayesh Badwaik

@Khromonkey okay.

Khromonkey

do you know who mukesh ambani is?

Jayesh Badwaik

Yes. I know his home has that name.

Khromonkey

It is madness dont you agree?

Jayesh Badwaik

5:04 AM

It would be be quite ignorant of someone living in India not to have heard of The Ambani Family.

user19161

@Khromonkey I like to watch the drama series The Wonder Years.

Khromonkey

Here in Mexico we have the richest man in the world and it really pisses me off

user19161

That's because I did not declare my bank account.

Jayesh Badwaik

@Khromonkey I would not build a home like that myself. But, its his money. Yes, it would be more desirable to have the money spent elsewhere, yes, his home is like a slap of arrogance and aristocracy to the slums around it. But in the end, it is his money. He can do whatever he wants with it.

Khromonkey

thats because I decided not to pursue my bright career in adult modelling @JasperLoy

I would be a big name I tell you

user19161

5:07 AM

@Khromonkey Wow! You must look like Taylor Lautner.

Jayesh Badwaik

looking up Taylor Lautner

Khromonkey

Yes only he has an eight pack. I have a 16 pack

Jayesh Badwaik

@Khromonkey is the drug mafia in mexico as bad as I hear on news?

Khromonkey

Yes

definitely

user19161

@Khromonkey Wow, so they come in powers of 2?

Khromonkey

5:11 AM

@JasperLoy Dont tell me you didnt know. In super model fase 4 you get the 16 pack

Jayesh Badwaik

@Khromonkey why is there such a big drug business in mexico though? Any specific reason?

Khromonkey

we are south of the united states

Jayesh Badwaik

ahh...

Khromonkey

We are the only country with borders with United States (except Canada but they dont have pot there)

so all of the drug needs to pass through Mexico

But it is not that horrible in all parts

Jayesh Badwaik

hmm.

Khromonkey

5:16 AM

The capital city is not all that bad

However some states like tamaulipas, Chihuaha or Sonora have ghost cities

Jayesh Badwaik

ghost cities?

Khromonkey

cities that are abandoned

Jayesh Badwaik

okay, by the way, I see a lot of police officers in the drug gang. Are they operating on their own free will? Or is there some sort of extortion or forcing?

@Khromonkey why?

Khromonkey

There are several reasons. Some Cartel leaders force people to plant weed. In others they kidnap people.

Jayesh Badwaik

hmm.

user19161

5:18 AM

Do they force people to do math exercises?

Khromonkey

Something that is really common is for cartels to ask store owners to give a cut of their money,

No, that is called China @JasperLoy

user19161

HAHAHAHAHAHA

Jayesh Badwaik

@Khromonkey ohh, yup that is quite common in some places here too.

Khromonkey

the first part or the second one?@JayeshBadwaik

Being asked for money or being forced to solve math problems?

Jayesh Badwaik

@Khromonkey well, I replied to the first sentence at first. But even the second is somewhat correct. :P Though the pressure is more domestic/economic in the second case rather than legal.

Khromonkey

5:21 AM

Here in Mexico people hate math.

Jayesh Badwaik

Though with a good economy now, the economic factor is weaning off somewhat.

@Khromonkey The phenomena is not limited to mexico I am sure.

Khromonkey

I remember I went to a math competition to another city and I was chatting up a girl and I told her I was in a math competition and about two minutes later she went with her friends.

It was actually really funny even at the moment

Jayesh Badwaik

Ahh, no, the situation is not THAT bad here.

maths is hated, not mathematicians.

though some people do not regard them seriously.

Khromonkey

Im not a mathematician

I just like math

Jayesh Badwaik

In words of G.H. Hardy, anyone who likes and practices math on his own is a mathematician.

Khromonkey

5:26 AM

He was probably talking about Ramanujan

Jayesh Badwaik

no, I think he was talking in general.

but whatever.

Gustavo Bandeira

@JayeshBadwaik I feel uneasy with calling me a mathematician - I think I'll deserve this title when I start to work on something great.

Khromonkey

hey why not

We are now all mathematicians

Jayesh Badwaik

@GustavoBandeira working on something great is a tremendous burden to carry, concentrate on enjoying things, let the other things flow.

Khromonkey

Well, I have to go now. Good night guys @JayeshBadwaik @JasperLoy @GustavoBandeira @PeterTamaroff @peoplepower @anon @robjohn Im sorry if I left you out, I dont have time for everyone

Jayesh Badwaik

5:29 AM

@Khromonkey here, if I tell people that I am interested in mathematics, they will think of me as some sort of alien from other planet.

@Khromonkey Good night.

Khromonkey

@JayeshBadwaik also in mexico

Jayesh Badwaik

but then, if you are a nice enough alien who is nice enough otherwise, people would have no problems in dating you here.

Gustavo Bandeira

@Khromonkey You're a businessman! Time's money!

Jayesh Badwaik

anyway, I must go now too. Lang awaits.

@GustavoBandeira bye

good night

Gustavo Bandeira

Good night. =)

peoplepower

5:32 AM

@Khromonkey Good night.

user19161

6:30 AM

@Khromonkey Whoa, you must be feeling full of love to ping everyone!

user19161

@JayeshBadwaik Ah, then that counts me as one too: a banana mathematician!

Ethan

7:01 AM

hello

jelllo

user19161

@Ethan Merry Xmas! I hope you live well.

Ethan

jasper

5

Q: Supposed proof of dirichlets theorem on primes

I think theirs somthing wrong with this proof as it was not hard to create, if someone could find a mistake I would greatly appreiciate it: Define a function $[k\equiv b$ mod a], to be equal to zero if k isn't congruent to b mod a, and 1 if it is. From that definition we have: $$\sum _{k=1}^{\...

prime-numbers proof-strategy proof-writing

copy my post

and repost it

il award you the bounty

also thumbs my post up

the very bottom post

user19161

@Ethan Nah, I don't know this stuff. You should ask anon for his opinion.

Ethan

No, im giveing you the bounty

just repost my answer

everyones being a dick to me, and I dont want to give them the bounty

user19161

@Ethan That would not be appropriate!

Ethan

7:05 AM

?

can I delete a bounty

just do it

it will be funny

give like a, one sentence answer

user19161

@Ethan I don't think so. Are you not happy with Matt E?

Ethan

ya

hes cerial down voting me

user19161

@Ethan How do you know it is him?

Ethan

because hes the only one whose been on that post in days

and he made me sound like a dick in all his posts

user19161

@Ethan How many downvotes did you receive?

Ethan

7:08 AM

5

peoplepower

I say don't take it personally. If someone is an ass but is nevertheless teaching me something, the best thing to do is just learn.

Ethan

I deleted it

Just write like a sentence and I will award you the bounty, im tired of geting notifications from him

user19161

4

A: Show ${156 \choose 87} + {156 \choose 86} = {157 \choose 87}$

I think this is the best way to do it. Using Wolfram Alpha, we see that $$\begin{align} \binom{156}{87} &= 2071356239566075202805184663760798068902414000\\[12pt] \binom{156}{86} &= 2574399897746407752057872367816991885635857400\\[12pt] \binom{157}{87} &= 46457561373124829548630570315...

user19161

Is this a sick joke?

peoplepower

@JasperLoy Awesome.

Ethan

7:10 AM

lmao

yes

user19161

@peoplepower And people actually upvoted him?

Ethan

Dude just calculate 156! by hand, gosh

user19161

Please don't upvote joke answers.

user19161

It is correct of course.

user19161

I flagged a joke answer once for removal but the mods did not remove it.

user19161

7:12 AM

That one was not an answer at all...

peoplepower

@JasperLoy I suppose for computer mathematicians this would be helpful because they can check it immediately.

Ethan

peoplepow

post this as an answer: "The statement that I can freely chose b to adapt to certain conditions involving c, is false, as the value of c may very well be dependent on the value of b I chose originally."

to this

6

Q: Supposed proof of dirichlets theorem on primes

I think theirs somthing wrong with this proof as it was not hard to create, if someone could find a mistake I would greatly appreiciate it: Define a function $[k\equiv b$ mod a], to be equal to zero if k isn't congruent to b mod a, and 1 if it is. From that definition we have: $$\sum _{k=1}^{\...

prime-numbers proof-strategy proof-writing

and il give you 300 points reputation

peoplepower

I am not convinced that the first and second add to the third, and will never be that way.

@Ethan I don't want rep.

Ethan

just do it please, i dont want to award the bounty to the other guy

peoplepower

I don't want the bad "rep" from denying 300 points to a good answer.

Ethan

7:13 AM

its a bad answer

I asked

if it could be adapted

user19161

Rather I don't want to post something I don't understand.

Ethan

all he did was go on and on about how obviously wrong it was

peoplepower

I'm not saying a good answer has been provided yet.

Ethan

The bounty ends in 7 hours

peoplepower

OK.

Ethan

7:14 AM

I wont get 1

plz

peoplepower

I don't want it.

user19161

Have you asked anon?

peoplepower

I don't have a good answer. Therefore, I am out of the picture. QED.

Ethan

yes you do

The statement that I can freely chose b to adapt to certain conditions involving c, is false, as the value of c may very well be dependent on the value of b I chose originally.

write that

and il accept it

experimentX

Hello

what is this symbol \ mean? $$S∪(T∖S)$$

user19161

7:17 AM

@experimentX The first is the set union. The second is the set difference.

user19161

Look them up online.

experimentX

oh!!

thanks!!

user19161

@experimentX A\B just means A-B, or the set of elements in A but not in B.

peoplepower

$S\cup (T-S)=S\cup T$ btw.

Ethan

can someone post this as an answer: The statement that I can freely chose b to adapt to certain conditions involving c, is false, as the value of c may very well be dependent on the value of b I chose originally. ON this page math.stackexchange.com/questions/256444/…

experimentX

7:18 AM

i see ... i never used that symbol

user19161

@Ethan I don't think anyone will oblige you unless they are very crazy.

Ethan

can someone post that for me

dude im legit

just post that answer

user19161

@Ethan So just let the bounty go to whoever it is, never mind!

Ethan

no

never

Can you give yourself the bounty

user19161

You can read more about bounties in the FAQ.

user19161

7:20 AM

I don't know.

anon

Ethan, I think Matt E's is a reasonable answer, that what you put in your answer was already contained in Matt's (albeit nowhere near as concise), and that the making-you-look-like-a-dick part (via extremeness of his courteous tone, which is actually very common among math profs) is only a product of your perception and of course the huge discrepancy between writing styles, neither of which he can be faulted for. I also feel "going on and on about how obviously wrong it is" is very unfair.

Ethan

"Obtensibly"

is that even a word?

anon

ostensibly is a word, yes

Ethan

he wrote

user19161

@Ethan ostensibly from ostensible

Ethan

7:24 AM

Thats not what he wrote

obstensibly

anon

Indeed, he mispelled it with a b.

Ethan

is what

he rejected my edit

lol]

anon

Hmm, can I see a link to the suggested edit?

Old John

@anon It was one of those moments that made teaching less painful :)

user19161

No wonder I got three votes today for that answer, because of that rubbish answer bumping it.

Ethan

7:29 AM

troll

Brian M. Scott

@Ethan The second half of your proposed edit was incorrect $-$ it made Matt E’s grammatical sentence ungrammatical $-$ but I deleted that bit and made the spelling correction.

anon

@Ethan According to your stats, you have had 2 edits accepted, and no edits rejected.

Ethan

Maybe it wasn't rejected

lol

I don't know

anon

Not only were you wrong it was rejected, you immediately accused Matt E. You also accused Matt E of serial downvoting, which I find preposterously out of character.

user19161

I am going to take a nap. I will see all of you in my dreams.

Ethan

7:32 AM

Brian I added the second part because it wouldn't let me edit somthing without typing in more then 6 characters

user19161

@old Good morning there!

Old John

@JasperLoy HI - well done on 12K

Ethan

12k?

user19161

@OldJohn =) Just in time for Xmas!

Old John

@JasperLoy :)

user19161

7:33 AM

But I really hope to have X, Y and Z. Over and out!

Old John

@JasperLoy bye!

Jayesh Badwaik

8:09 AM

I was able to write a parallelized simulation code for self-consistent simulation of nanoelectronic devices modelled by poisson and schrodinger equations. The beautuy of the code was that solver methods were parallelized into two sections :
1. transport equations (schrodinger, diffusion etc)
2. field equation (poisson)

the set of methods you wanted to use for solving a particular region of the device could be parametrized (we used templates in C++ for that) and the device could be divided into different regions each of which could be solved by different set of equations.

experimentX

8:25 AM

hi .. can anyone explain this proof no 2 to me?

proofwiki.org/wiki/Infinite_Set_Equivalent_to_Proper_Subset

hi @OldJohn!!

anon

Which part would you like explained?

Old John

@experimentX Hi

anon

Basically, given S infinite, you split S into a countable part and another part that you ignore. You set h:S->S to do nothing on the ignorable part whereas it "shifts" the elements in the copy of N inside of S.

experimentX

how does that mapping proves that there is bijection between a countably infinite set and it's proper sub-set

hold on ... am uploading

anon

The image of S under h is missing the image of 1 (from the naturals N under v) inside of S.

so h(S) is a proper subset of S, but h is injective so that S and h(S) are equivalent.

experimentX

8:30 AM

peoplepower

(h is automatically surjective on h(S).)

experimentX

so it is shifting index of S

@anon the second map $h(x) = x : x \not{\in } \Im (v)$

anon

so, what do you want to know?

experimentX

on that image we are definite $f:S'U{x_1,x_2, .. } -> S$ also given that $S'U{x_1,x_2, .. } = S$

shouldn't it be like this $S' \cup \{x_1, x_2, ..\} \subset S $ what the point if we are defining a one to one mapping to itself?

anon

We are splitting up the domain. The image is proper, because $x_1$ is not in the image.

Since the map fixes every element of $S'$, and $x_1\mapsto x_2\mapsto x_3\mapsto\cdots$, the image is $S'\cup\{x_2,x_3,\cdots\}$ which does not contain $x_1$.

The notation "f:A->B" does not mean that im(f)=B, only that codom(f)=B and im(f) is a subset of B.

experimentX

8:42 AM

i see!! so $S'$ in that picture is uncountable part?

peoplepower

Not necessarily. :)

Let $S=\mathbb Q$ and enumerate the positive integers.

anon

{x1,x2,...} is any countable subset, and S' is "all the other elements" of S (which can be "small" - in fact S' can be empty if S is countable, and the proof still works)

experimentX

i see thanks!!

@anon do you have any book recommendations for set theory at basic level?

anon

nope

experimentX

ok

Old John

8:49 AM

@experimentX I read Halmos "Naive Set Theory" some years ago, and found it useful

and "The Joy of Sets" had some good reviews, but not read it

experimentX

thanks @OldJohn !!

robjohn

9:10 AM

$\displaystyle\left(1-\frac1x\right)\sqrt[\Large3]{x^2-2x+2}
=(x-1)\sqrt[\Large3]{\frac{x^2-2x+2}{x^3}}$

At $x=1$, $\displaystyle\frac{{x^2-2x+2}}{x^3}=1$ and $\displaystyle\frac{\mathrm{d}}{\mathrm{d}x}\frac{x^2-2x+2}{x^3}=-\frac{(x-2)^2+2}{x^4}\lt0$.

Thus, for $x>0$, $\sqrt[\Large3]{\dfrac{x^2-2x+2}{x^3}}\gtreqqless1$ when $x\lesseqqgtr1$. This shows that $\displaystyle\left(1-\frac1x\right)\sqrt[\Large3]{x^2-2x+2}\le x-1$.

Chris's sister

Hi

robjohn

@Chris'ssister hey there

Chris's sister

@robjohn: hi. How are you? :D

robjohn

@Chris'ssister pretty good. And you?

Chris's sister

@robjohn: I'm preparing the stuff for learning things till late in the night. :-)

Old John

9:17 AM

@WillJagy I see you have now killed off the $n=9$ case by two different methods - along with many other values of $n$. Do we suspect that there are a lot more cases that could be killed off with similar examples?

Matt N.

Srsly?

robjohn

@MattN. It's on its way to deletion

@OldJohn you killed WillJagy!

at least sent him sledding

Dumb Cow

10:17 AM

Hey, wasn't there a way to derive a formula for $\binom{n}{r}$ using the fact that $n!$ is the number of ways to arrange $n$ distinct objects?

anon

10:31 AM

@DumbCow Yes. ${n\choose r}r!$ is the number of ordered lists of length $r$ of entries from a set of size $n$ without repetitions. The number of ways to choose the first entry of the list is $n$, the number of ways to choose the second entry is $(n-1)$, etc., so we derive $${n\choose r}r!=n(n-1)\cdots\big(n-(r-1)\big)=\frac{n!}{(n-r)!}\implies{n\choose r}=\frac{n!}{r!(n-r)!}.$$

Dumb Cow

Hmm... having trouble understanding the first sentence. Thanks @anon!

Oh, got the first sentence.

Thanks :)

Transcript for

$Mathematics$ Mathematics